| Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
Exercise 10.3
Question 1 A
Find the coordinates of the point which divides the line segment joining (2,4) and (6,8) in the ratio 1:3 internally and externally.
Sol :
Let P(x,y) be the point which divides the line segment internally.
Using the section formula for the internal division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 1, m2 = 3
(x1, y1) = (2, 4) and (x2, y2) = (6, 8)
Putting the above values in the above formula, we get
$\Rightarrow \mathrm{x}=\frac{1(6)+3(2)}{1+3}, \mathrm{y}=\frac{1(8)+3(4)}{1+3}$
$\Rightarrow x=\frac{6+6}{4}, y=\frac{8+12}{4}$
$\Rightarrow x=\frac{12}{4}, y=\frac{20}{4}$
⇒ x = 3, y = 5
Hence, (3,5) is the point which divides the line segment internally.
Now, Let Q(x,y) be the point which divides the line segment externally.
Using the section formula for the external division, i.e.
$(\mathrm{X}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}-\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}-\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 1, m2 = 3
(x1, y1) = (2, 4) and (x2, y2) = (6, 8)
Putting the above values in the above formula, we get
$\Rightarrow \mathrm{x}=\frac{1(6)-3(2)}{1-3}, \mathrm{y}=\frac{1(8)-3(4)}{1-3}$
$\Rightarrow x=\frac{6-6}{-2}, y=\frac{8-12}{-2}$
$\Rightarrow \mathrm{x}=\frac{0}{-2}, \mathrm{y}=\frac{-4}{-2}$
⇒ x = 0, y = 2
Hence, (0,2) is the point which divides the line segment externally.
Question 1 B
Find the coordinates of the point which divides the join of (-1,7) and (4,-3) internally in the ratio 2:3.
Sol :
Let P(x,y) be the point which divides the line segment internally.
Using the section formula for the internal division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 2, m2 = 3
(x1, y1) = (-1, 7) and (x2, y2) = (4, -3)
Putting the above values in the above formula, we get
$\Rightarrow x=\frac{2(4)+3(-1)}{2+3}, y=\frac{2(-3)+3(7)}{2+3}$
$\Rightarrow x=\frac{8-3}{5}, y=\frac{-6+21}{5}$
$\Rightarrow \mathrm{x}=\frac{5}{5}, \mathrm{y}=\frac{15}{5}$
⇒ x = 1, y = 3
Hence, (1,3) is the point which divides the line segment internally.
Question 1 C
Find the coordinates of the point which divides the line segment joining the points (4,-3) and (8,5) in the ratio 3:1 internally.
Sol :
Let P(x,y) be the point which divides the line segment internally.
Using the section formula for the internal division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 3, m2 = 1
(x1, y1) = (4, -3) and (x2, y2) = (8, 5)
Putting the above values in the above formula, we get
$\Rightarrow \mathrm{x}=\frac{3(8)+1(4)}{3+1}, \mathrm{y}=\frac{3(5)+1(-3)}{3+1}$
$\Rightarrow x=\frac{24+4}{4}, y=\frac{15-3}{4}$
$\Rightarrow x=\frac{28}{4}, y=\frac{12}{4}$
⇒ x = 7, y = 3
Hence, (7,3) is the point which divides the line segment internally.
Question 2 A
Find the coordinates of the points which trisect the line segment joining the points (2,3) and (6,5).
Sol :
Let P and Q be the points of trisection of AB, i.e. AP = PQ = QB
∴ P divides AB internally in the ratio 1: 2.
∴ the coordinates of P, by applying the section formula, are
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 1, m2 = 2
(x1, y1) = (2, 3) and (x2, y2) = (6, 5)
Putting the above values in the above formula, we get
$\Rightarrow \mathrm{x}=\frac{1(6)+2(2)}{1+2}, \mathrm{y}=\frac{1(5)+2(3)}{1+2}$
$\Rightarrow x=\frac{6+4}{3}, y=\frac{5+6}{3}$
$\Rightarrow \mathrm{x}=\frac{10}{3}, \mathrm{y}=\frac{11}{3}$
Now, Q also divides AB internally in the ratio 2: 1. So, the coordinates of Q are
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 2, m2 = 1
(x1, y1) = (2, 3) and (x2, y2) = (6, 5)
Putting the above values in the above formula, we get
$\Rightarrow \mathrm{x}=\frac{2(6)+1(2)}{2+1}, \mathrm{y}=\frac{2(5)+1(3)}{2+1}$
$\Rightarrow x=\frac{12+2}{3}, y=\frac{10+3}{3}$
$\Rightarrow x=\frac{14}{3}, y=\frac{13}{3}$
Therefore, the coordinates of the points of trisection of the line segment joining A and B are $\left(\frac{10}{3}, \frac{11}{3}\right)$ and $\left(\frac{14}{3}, \frac{13}{3}\right)$
Question 2 B
Find the coordinates of the point of trisection of the line segment joining (1,-2) and (-3,4).
Sol :
Let P and Q be the points of trisection of AB, i.e. AP = PQ = QB
∴ P divides AB internally in the ratio 1: 2.
∴ the coordinates of P, by applying the section formula, are
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 1, m2 = 2
(x1, y1) = (1, -2) and (x2, y2) = (-3, 4)
Putting the above values in the above formula, we get
$\Rightarrow \mathrm{x}=\frac{1(-3)+2(1)}{1+2}, \mathrm{y}=\frac{1(4)+2(-2)}{1+2}$
$\Rightarrow x=\frac{-3+2}{3}, y=\frac{4-4}{3}$
$\Rightarrow \mathrm{x}=\frac{-1}{3}, \mathrm{y}=0$
$\Rightarrow \mathrm{x}=\frac{1(-3)+2(1)}{1+2}, \mathrm{y}=\frac{1(4)+2(-2)}{1+2}$
$\Rightarrow x=\frac{-3+2}{3}, y=\frac{4-4}{3}$
$\Rightarrow \mathrm{x}=\frac{-1}{3}, \mathrm{y}=0$
Now, Q also divides AB internally in the ratio 2: 1. So, the coordinates of Q are
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 2, m2 = 1
(x1, y1) = (1, -2) and (x2, y2) = (-3, 4)
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 2, m2 = 1
(x1, y1) = (1, -2) and (x2, y2) = (-3, 4)
Putting the above values in the above formula, we get
$\Rightarrow \mathrm{x}=\frac{2(-3)+1(1)}{2+1}, \mathrm{y}=\frac{2(4)+1(-2)}{2+1}$
$\Rightarrow \mathrm{x}=\frac{-6+1}{3}, \mathrm{y}=\frac{8-2}{3}$
$\Rightarrow x=\frac{-5}{3}, y=\frac{6}{3}=2$
Therefore, the coordinates of the points of trisection of the line segment joining A and B are $\left(\frac{-1}{3}, 0\right)$ and $\left(\frac{-5}{3}, 2\right)$
$\Rightarrow \mathrm{x}=\frac{2(-3)+1(1)}{2+1}, \mathrm{y}=\frac{2(4)+1(-2)}{2+1}$
$\Rightarrow \mathrm{x}=\frac{-6+1}{3}, \mathrm{y}=\frac{8-2}{3}$
$\Rightarrow x=\frac{-5}{3}, y=\frac{6}{3}=2$
Therefore, the coordinates of the points of trisection of the line segment joining A and B are $\left(\frac{-1}{3}, 0\right)$ and $\left(\frac{-5}{3}, 2\right)$
Question 3 A
The coordinates of A and B are (1,2) and (2,3) respectively, If P lies on AB, find the coordinates of P such that $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{4}{3}$
Sol :
Given:
$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{4}{3}$
⇒ m1 = 4 and m2 = 3
and (x1, y1) = (1, 2) ; (x2, y2) = (2, 3)
Using the section formula for the internal division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
$\Rightarrow x=\frac{4(2)+3(1)}{4+3}, y=\frac{4(3)+3(2)}{4+3}$
$\Rightarrow \mathrm{x}=\frac{8+3}{7}, \mathrm{y}=\frac{12+6}{7}$
$\Rightarrow \mathrm{x}=\frac{11}{7}, \mathrm{y}=\frac{18}{7}$
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
$\Rightarrow x=\frac{4(2)+3(1)}{4+3}, y=\frac{4(3)+3(2)}{4+3}$
$\Rightarrow \mathrm{x}=\frac{8+3}{7}, \mathrm{y}=\frac{12+6}{7}$
$\Rightarrow \mathrm{x}=\frac{11}{7}, \mathrm{y}=\frac{18}{7}$
Hence, the coordinates of P are
$\mathrm{P}(\mathrm{x}, \mathrm{y})=\mathrm{P}\left(\frac{11}{7}, \frac{18}{7}\right)$
Question 3 B
If A (4,-8), B (3,6) and C(5,-4) are the vertices of a ΔABC, D is the mid-point of BC and P is a point on AD joined such that $\frac{A P}{P D}=2$, find the coordinates of P.
Sol :
Given: D is the midpoint of BC. So, BD = DC
Then the coordinates of D are
$\mathrm{x}=\left(\frac{3+5}{2}\right), \mathrm{y}=\left(\frac{6+(-4)}{2}\right)$
$\Rightarrow \mathrm{x}=\frac{8}{2}, \mathrm{y}=\frac{2}{2}$
⇒ x = 4 and y = 1
So, coordinates of D are (4, 1)
Now, we have to find the coordinates of P.
Given:
$\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{2}{1}$
⇒ m1 = 2 and m2 = 1
and (x1, y1) = (4, -8) ; (x2, y2) = (4, 1)
Using the section formula for the internal division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
$\Rightarrow x=\frac{2(4)+1(4)}{2+1}, y=\frac{2(1)+1(-8)}{2+1}$
$\Rightarrow x=\frac{8+4}{3}, y=\frac{2-8}{3}$
$\Rightarrow \mathrm{x}=\frac{12}{3}, \mathrm{y}=\frac{-6}{3}$
⇒ x = 4, y = -2
Hence, the coordinates of P are P(x,y) = P(4, -2)
Given:
$\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{3}{7}$
$\Rightarrow \mathrm{AP}=\frac{3}{7} \mathrm{AB}$
$\Rightarrow \mathrm{AP}=\frac{3}{7}(\mathrm{AP}+\mathrm{PB})$
⇒ 7AP = 3AP + 3PB
⇒ 7AP – 3AP = 3PB
⇒ 4AP = 3PB
$\Rightarrow \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{3}{4}$
Hence, the point P divides AB in the ratio of 3:4
⇒ m1 = 3 and m2 = 4
and (x1, y1) = (-2, -2) ; (x2, y2) = (2, -4)
Using the section formula for the internal division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
$\Rightarrow \mathrm{x}=\frac{3(2)+4(-2)}{3+4}, \mathrm{y}=\frac{3(-4)+4(-2)}{3+4}$
$\Rightarrow x=\frac{6-8}{7}, y=\frac{-12-8}{7}$
$\Rightarrow \mathrm{x}=\frac{-2}{7}, \mathrm{y}=\frac{-20}{7}$
Hence, the coordinates of P are $\mathrm{P}(\mathrm{x}, \mathrm{y})=\mathrm{P}\left(\frac{-2}{7}, \frac{-20}{7}\right)$
Given: AP = AB + BP and AP = 10
Firstly, we find the distance between A and B
d(A,B) = √(x2 – x1)2 + (y2 – y1)2
= √(4 – 1)2 + (8 – 4)2
= √(3)2 + (4)2
= √9 + 16
= √25
= 5
So, AB = 5
It is given that AP = AB + BP
⇒ 10 = 5 + BP
⇒ 10 – 5 = BP
⇒ BP = 5
⇒ A, B and P are collinear
and since AB = BP
⇒ B is the midpoint of AP
Let the coordinates of P = (x,y)
$\Rightarrow\left(\frac{x+1}{2}, \frac{y+4}{2}\right)=(4,8)$
$\Rightarrow \frac{x+1}{2}=4$ and $\frac{y+4}{2}=8$
⇒ x + 1 = 8 and y + 4 = 16
⇒ x = 7 and y = 12
Hence, the coordinates of P are (7, 12)
Let P and Q be the required new ends
Coordinates of P
Let AP = k
∴ AB = AP = k
and PB = AP + AB = k + k = 2k
$\therefore \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{k}}{2 \mathrm{k}}=\frac{1}{2}$
∴ P divides AB externally in the ratio 1:2
Using the section formula for the external division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}-\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}-\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 1, m2 = 2
(x1, y1) = (2, 3) and (x2, y2) = (-3, 5)
Putting the above values in the above formula, we get
$\Rightarrow \mathrm{x}=\frac{1(-3)-2(2)}{1-2}, \mathrm{y}=\frac{1(5)-2(3)}{1-2}$
⇒ x = 7, y = 1
∴Coordinates of P are (7, 1)
Coordinates of Q.
Q divides AB externally in the ratio 2:1
Again, Using the section formula for the external division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}-\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}-\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 2, m2 = 1
(x1, y1) = (2, 3) and (x2, y2) = (-3, 5)
Putting the above values in the above formula, we get
$\Rightarrow x=\frac{2(-3)-1(2)}{2-1}, y=\frac{2(5)-1(3)}{2-1}$
∴Coordinates of Q are (-8, 7)
Let P and Q be the required new ends
Coordinates of P
Let AP = k
∴ AB = 2AP = 2k
and PB = AP + AB = k + 2k = 3k
$\therefore \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{k}}{3 \mathrm{k}}=\frac{1}{3}$
∴ P divides AB externally in the ratio 1:3
Using the section formula for the external division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}-\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}-\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 1, m2 = 3
(x1, y1) = (6, 3) and (x2, y2) = (-1, -4)
Putting the above values in the above formula, we get
∴Coordinates of P are $\left(\frac{19}{2}, \frac{13}{2}\right)$
Coordinates of Q.
Q divides AB externally in the ratio 3:1
Again, Using the section formula for the external division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}-\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}-\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 3, m2 = 1
(x1, y1) = (6, 3) and (x2, y2) = (-1, -4)
Putting the above values in the above formula, we get
∴Coordinates of Q are $\left(\frac{-9}{2}, \frac{-15}{2}\right)$
Now, Let P(x,y) be the point which lies on extended line AB
Using the section formula for the external division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}-\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}-\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 1, m2 = 3
(x1, y1) = (-1, 4) and (x2, y2) = (5, 1)
Putting the above values in the above formula, we get
Let the coordinates of the point be (x,y)
Let A = (5, -4) and B = (3, -2)
Here, the point divides the line segment in the ratio 4:3
So, m1 = 4 and m2 = 3
Using section formula,
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
Hence, the coordinates of P are $\mathrm{P}(\mathrm{x}, \mathrm{y})=\mathrm{P}\left(\frac{27}{7}, \frac{-20}{7}\right)$
Now, the distance from the origin (0,0) is
Consider a ΔABC with A(x1, y1), B(x2, y2) and C(x3, y3). If P(1, 1), Q(2, 3) and R(4, 1) are the midpoints of AB, BC, and CA. Then,
Adding (i), (iii) and (v), we get
x1 + x2 + x2 + x3 + x1 + x3 = 2 + 4 + 8
⇒ 2(x1 + x2 + x3) = 14
⇒ x1 + x2 + x3 = 7 …(vii)
From (i) and (vii), we get
x3 = 7 – 2 = 5
From (iii) and (vii), we get
x1 = 7 – 4 = 3
From (v) and (vii), we get
x2 = 7 – 8 = -1
Now adding (ii), (iv) and (vi), we get
y1 + y2 + y2 + y3 + y1 + y3 = 2 + 6 + 2
⇒ 2(y1 + y2 + y3) = 10
⇒ y1 + y2 + y3 = 5 …(viii)
From (ii) and (viii), we get
y3 = 5 – 2 = 3
From (iv) and (vii), we get
y1 = 5 – 6 = -1
From (vi) and (vii), we get
y2 = 5 – 2 = 3
Hence, the vertices of ΔABC are A(3, -1), B(-1, 3) and C(5, 3)
Consider a ΔABC with A(x1, y1), B(x2, y2) and C(x3, y3). If P(10, 5), Q(8, 4) and R(6, 6) are the midpoints of AB, BC, and CA. Then,
So, coordinates of D are (4, 1)
Now, we have to find the coordinates of P.
Given:
$\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{2}{1}$
⇒ m1 = 2 and m2 = 1
and (x1, y1) = (4, -8) ; (x2, y2) = (4, 1)
Using the section formula for the internal division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
$\Rightarrow x=\frac{2(4)+1(4)}{2+1}, y=\frac{2(1)+1(-8)}{2+1}$
$\Rightarrow x=\frac{8+4}{3}, y=\frac{2-8}{3}$
$\Rightarrow \mathrm{x}=\frac{12}{3}, \mathrm{y}=\frac{-6}{3}$
⇒ x = 4, y = -2
Hence, the coordinates of P are P(x,y) = P(4, -2)
Question 3 C
If p divides the join of A (-2,-2) and B (2,-4) such that $\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{3}{7}$, find the coordinates of P.
Sol :Given:
$\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{3}{7}$
$\Rightarrow \mathrm{AP}=\frac{3}{7} \mathrm{AB}$
$\Rightarrow \mathrm{AP}=\frac{3}{7}(\mathrm{AP}+\mathrm{PB})$
⇒ 7AP = 3AP + 3PB
⇒ 7AP – 3AP = 3PB
⇒ 4AP = 3PB
$\Rightarrow \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{3}{4}$
Hence, the point P divides AB in the ratio of 3:4
⇒ m1 = 3 and m2 = 4
and (x1, y1) = (-2, -2) ; (x2, y2) = (2, -4)
Using the section formula for the internal division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
$\Rightarrow \mathrm{x}=\frac{3(2)+4(-2)}{3+4}, \mathrm{y}=\frac{3(-4)+4(-2)}{3+4}$
$\Rightarrow x=\frac{6-8}{7}, y=\frac{-12-8}{7}$
$\Rightarrow \mathrm{x}=\frac{-2}{7}, \mathrm{y}=\frac{-20}{7}$
Hence, the coordinates of P are $\mathrm{P}(\mathrm{x}, \mathrm{y})=\mathrm{P}\left(\frac{-2}{7}, \frac{-20}{7}\right)$
Question 3 D
A (1,4) and B (4,8) are two points. P is a point on AB such that AP = AB + BP. If AP = 10 find the coordinates of P.
Sol :Given: AP = AB + BP and AP = 10
Firstly, we find the distance between A and B
d(A,B) = √(x2 – x1)2 + (y2 – y1)2
= √(4 – 1)2 + (8 – 4)2
= √(3)2 + (4)2
= √9 + 16
= √25
= 5
So, AB = 5
It is given that AP = AB + BP
⇒ 10 = 5 + BP
⇒ 10 – 5 = BP
⇒ BP = 5
⇒ A, B and P are collinear
and since AB = BP
⇒ B is the midpoint of AP
Let the coordinates of P = (x,y)
$\Rightarrow\left(\frac{x+1}{2}, \frac{y+4}{2}\right)=(4,8)$
$\Rightarrow \frac{x+1}{2}=4$ and $\frac{y+4}{2}=8$
⇒ x + 1 = 8 and y + 4 = 16
⇒ x = 7 and y = 12
Hence, the coordinates of P are (7, 12)
Question 4
The line segment joining A (2,3) and B(-3,5) is extended through each end by a length equal to its original length. Find the coordinates of the new ends.
Sol :Let P and Q be the required new ends
Coordinates of P
Let AP = k
∴ AB = AP = k
and PB = AP + AB = k + k = 2k
$\therefore \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{k}}{2 \mathrm{k}}=\frac{1}{2}$
∴ P divides AB externally in the ratio 1:2
Using the section formula for the external division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}-\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}-\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 1, m2 = 2
(x1, y1) = (2, 3) and (x2, y2) = (-3, 5)
Putting the above values in the above formula, we get
$\Rightarrow \mathrm{x}=\frac{1(-3)-2(2)}{1-2}, \mathrm{y}=\frac{1(5)-2(3)}{1-2}$
$\Rightarrow x=\frac{-3-4}{-1}, y=\frac{5-6}{-1}$
$\Rightarrow x=\frac{-7}{-1}, y=\frac{-1}{-1}$
⇒ x = 7, y = 1
∴Coordinates of P are (7, 1)
Coordinates of Q.
Q divides AB externally in the ratio 2:1
Again, Using the section formula for the external division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}-\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}-\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 2, m2 = 1
(x1, y1) = (2, 3) and (x2, y2) = (-3, 5)
Putting the above values in the above formula, we get
$\Rightarrow x=\frac{2(-3)-1(2)}{2-1}, y=\frac{2(5)-1(3)}{2-1}$
$\Rightarrow \mathrm{x}=\frac{-6-2}{1}, \mathrm{y}=\frac{10-3}{1}$
$\Rightarrow \mathrm{x}=\frac{-8}{1}, \mathrm{y}=\frac{7}{1}$
Question 5
The line segment joining A(6,3) to B(-1,-4) is doubled in length by having half its length added to each end. Find the coordinates of the new ends.
Sol :Let P and Q be the required new ends
Coordinates of P
Let AP = k
∴ AB = 2AP = 2k
and PB = AP + AB = k + 2k = 3k
$\therefore \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{k}}{3 \mathrm{k}}=\frac{1}{3}$
∴ P divides AB externally in the ratio 1:3
Using the section formula for the external division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}-\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}-\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 1, m2 = 3
(x1, y1) = (6, 3) and (x2, y2) = (-1, -4)
Putting the above values in the above formula, we get
$\Rightarrow x=\frac{1(-1)-3(6)}{1-3}, y=\frac{1(-4)-3(3)}{1-3}$
$\Rightarrow x=\frac{-1-18}{-2}, y=\frac{-4-9}{-2}$
$\Rightarrow x=\frac{-19}{-2}, y=\frac{-13}{-2}$
$\Rightarrow x=\frac{19}{2}, y=\frac{13}{2}$
∴Coordinates of P are $\left(\frac{19}{2}, \frac{13}{2}\right)$
Coordinates of Q.
Q divides AB externally in the ratio 3:1
Again, Using the section formula for the external division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}-\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}-\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 3, m2 = 1
(x1, y1) = (6, 3) and (x2, y2) = (-1, -4)
Putting the above values in the above formula, we get
$\Rightarrow \mathrm{x}=\frac{3(-1)-1(6)}{3-1}, \mathrm{y}=\frac{3(-4)-1(3)}{3-1}$
$\Rightarrow \mathrm{x}=\frac{-3-6}{2}, \mathrm{y}=\frac{-12-3}{2}$
$\Rightarrow \mathrm{x}=\frac{-9}{2}, \mathrm{y}=\frac{-15}{2}$
∴Coordinates of Q are $\left(\frac{-9}{2}, \frac{-15}{2}\right)$
Question 6
The coordinates of two points A and B are (-1,4) and (5,1) respectively. Find the coordinates of the point P which lies on extended line AB such that it is three times as far from B as from A.
Sol :Now, Let P(x,y) be the point which lies on extended line AB
Using the section formula for the external division, i.e.
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}-\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}-\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}-\mathrm{m}_{2}}\right)$ …(i)
Here, m1 = 1, m2 = 3
(x1, y1) = (-1, 4) and (x2, y2) = (5, 1)
Putting the above values in the above formula, we get
$\Rightarrow x=\frac{1(5)-3(-1)}{1-3}, y=\frac{1(1)-3(4)}{1-3}$
$\Rightarrow x=\frac{5+3}{-2}, y=\frac{1-12}{-2}$
$\Rightarrow x=\frac{8}{-2}, y=\frac{-11}{-2}$
$\Rightarrow x=-4, y=\frac{11}{2}$
Question 7
Find the distances of that point from the origin which divides the line segment joining the points (5,-4) and (3,-2) in the ration 4:3.
Sol :Let the coordinates of the point be (x,y)
Let A = (5, -4) and B = (3, -2)
Here, the point divides the line segment in the ratio 4:3
So, m1 = 4 and m2 = 3
Using section formula,
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
$\Rightarrow x=\frac{4(3)+3(5)}{4+3}, y=\frac{4(-2)+3(-4)}{4+3}$
$\Rightarrow x=\frac{12+15}{7}, y=\frac{-8-12}{7}$
$\Rightarrow x=\frac{27}{7}, y=\frac{-20}{7}$
Hence, the coordinates of P are $\mathrm{P}(\mathrm{x}, \mathrm{y})=\mathrm{P}\left(\frac{27}{7}, \frac{-20}{7}\right)$
Now, the distance from the origin (0,0) is
$D=\sqrt{\left(\frac{27}{7}-0\right)^{2}+\left(\frac{-20}{7}-0\right)^{2}}$
$=\sqrt{\frac{729}{49}+\frac{400}{49}}$
$=\sqrt{\frac{1129}{49}}$
$=\frac{\sqrt{1129}}{7}$
Question 8 A
The coordinates of the middle points of the sides of a triangle are (1,1), (2,3) and (4,1), find the coordinates of its vertices.
Sol :Consider a ΔABC with A(x1, y1), B(x2, y2) and C(x3, y3). If P(1, 1), Q(2, 3) and R(4, 1) are the midpoints of AB, BC, and CA. Then,
$1=\frac{x_{1}+x_{2}}{2} \Rightarrow x_{1}+x_{2}=2$ …(i)
$1=\frac{y_{1}+y_{2}}{2} \Rightarrow y_{1}+y_{2}=2$ …(ii)
$2=\frac{x_{2}+x_{3}}{2} \Rightarrow x_{2}+x_{3}=4$ …(iii)
$3=\frac{y_{2}+y_{3}}{2} \Rightarrow y_{2}+y_{3}=6$ …(iv)
$4=\frac{x_{1}+x_{3}}{2} \Rightarrow x_{1}+x_{3}=8$ …(v)
$1=\frac{y_{1}+y_{2}}{2} \Rightarrow y_{1}+y_{3}=2$ …(vi)Adding (i), (iii) and (v), we get
x1 + x2 + x2 + x3 + x1 + x3 = 2 + 4 + 8
⇒ 2(x1 + x2 + x3) = 14
⇒ x1 + x2 + x3 = 7 …(vii)
From (i) and (vii), we get
x3 = 7 – 2 = 5
From (iii) and (vii), we get
x1 = 7 – 4 = 3
From (v) and (vii), we get
x2 = 7 – 8 = -1
Now adding (ii), (iv) and (vi), we get
y1 + y2 + y2 + y3 + y1 + y3 = 2 + 6 + 2
⇒ 2(y1 + y2 + y3) = 10
⇒ y1 + y2 + y3 = 5 …(viii)
From (ii) and (viii), we get
y3 = 5 – 2 = 3
From (iv) and (vii), we get
y1 = 5 – 6 = -1
From (vi) and (vii), we get
y2 = 5 – 2 = 3
Hence, the vertices of ΔABC are A(3, -1), B(-1, 3) and C(5, 3)
Question 8 B
If the points (10,5),(8,4) and (6,6) are the mid-points of the sides of a triangle, find its vertices.
Sol :Consider a ΔABC with A(x1, y1), B(x2, y2) and C(x3, y3). If P(10, 5), Q(8, 4) and R(6, 6) are the midpoints of AB, BC, and CA. Then,
$10=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2} \Rightarrow \mathrm{x}_{1}+\mathrm{x}_{2}=20$…(i)
$5=\frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2} \Rightarrow \mathrm{y}_{1}+\mathrm{y}_{2}=10$…(ii)
$8=\frac{\mathrm{x}_{2}+\mathrm{x}_{3}}{2} \Rightarrow \mathrm{x}_{2}+\mathrm{x}_{3}=16$ …(iii)
$4=\frac{\mathrm{y}_{2}+\mathrm{y}_{3}}{2} \Rightarrow \mathrm{y}_{2}+\mathrm{y}_{3}=8$ …(iv)
$6=\frac{\mathrm{x}_{1}+\mathrm{x}_{3}}{2} \Rightarrow \mathrm{x}_{1}+\mathrm{x}_{3}=12$ …(v)
$6=\frac{\mathrm{y}_{1}+\mathrm{y}_{3}}{2} \Rightarrow \mathrm{y}_{1}+\mathrm{y}_{3}=12$…(vi)Adding (i), (iii) and (v), we get
x1 + x2 + x2 + x3 + x1 + x3 = 20 + 16 + 12
⇒ 2(x1 + x2 + x3) = 48
⇒ x1 + x2 + x3 = 24 …(vii)
From (i) and (vii), we get
x3 = 24 – 20 = 4
From (iii) and (vii), we get
x1 = 24 – 16 = 8
From (v) and (vii), we get
x2 = 24 – 12 = 12
Now adding (ii), (iv) and (vi), we get
y1 + y2 + y2 + y3 + y1 + y3 = 10 + 8 + 12
⇒ 2(y1 + y2 + y3) = 30
⇒ y1 + y2 + y3 = 15 …(viii)
From (ii) and (viii), we get
y3 = 15 – 10 = 5
From (iv) and (vii), we get
y1 = 15 – 8 = 7
From (vi) and (vii), we get
y2 = 15 – 12 = 3
Hence, the vertices of ΔABC are A(8, 7), B(12, 3) and C(4, 5)
Question 8 C
The mid-points of the sides of a triangle are (3,4),(4,6) and (5,7). Find the coordinates of the vertices of the triangle.
Sol :
Consider a ΔABC with A(x1, y1), B(x2, y2) and C(x3, y3). If P(3, 4), Q(4, 6) and R(5, 7) are the midpoints of AB, BC, and CA. Then,
$3=\frac{x_{1}+x_{2}}{2} \Rightarrow x_{1}+x_{2}=6$ …(i)
$4=\frac{y_{1}+y_{2}}{2} \Rightarrow y_{1}+y_{2}=8$ …(ii)
$4=\frac{x_{2}+x_{3}}{2} \Rightarrow x_{2}+x_{3}=8$ …(iii)
$5=\frac{y_{2}+y_{3}}{2} \Rightarrow y_{2}+y_{3}=10$ …(iv)
$6=\frac{x_{1}+x_{3}}{2} \Rightarrow x_{1}+x_{3}=12$ …(v)
$7=\frac{y_{1}+y_{3}}{2} \Rightarrow y_{1}+y_{3}=14$ …(vi)
Adding (i), (iii) and (v), we get
x1 + x2 + x2 + x3 + x1 + x3 = 6 + 8 + 10
⇒ 2(x1 + x2 + x3) = 24
⇒ x1 + x2 + x3 =12 …(vii)
From (i) and (vii), we get
x3 = 12 – 6 = 6
From (iii) and (vii), we get
x1 = 12 – 8 = 4
From (v) and (vii), we get
x2 = 12 – 10 = 2
Now adding (ii), (iv) and (vi), we get
y1 + y2 + y2 + y3 + y1 + y3 = 8 + 12 + 14
⇒ 2(y1 + y2 + y3) = 34
⇒ y1 + y2 + y3 = 17 …(viii)
From (ii) and (viii), we get
y3 = 17 – 8 = 9
From (iv) and (vii), we get
y1 = 17 – 12 = 5
From (vi) and (vii), we get
y2 = 17 – 14 = 3
Hence, the vertices of ΔABC are A(4, 5), B(2, 3) and C(6, 9)
Question 9
A(1,-2) and B(2,5) are two points. The lines OA, OB are produced to C and D respectively such that OC = 2OA and OD = 2OB. Find CD.
Sol :
Given:
A(1, -2) and B(2, 5) are two points.
OC = 2OA …(i)
and OD = 2OB …(ii)
Adding (i) and (ii), we get
OC + OD = 2OA + 2OB
⇒ CD = 2[OA + OB]
⇒ CD = 2[AB] …(iii)
Now, we find the distance between A and B
d(A,B) = √(x2 – x1)2 + (y2 – y1)2
= √(2 – 1)2 + {5 – (-2)}2
= √(1)2 + (5 + 2)2
= √1 + 49
= √50
= 5√2
Putting the value in eq. (iii), we get
CD = 2 × 5√2
= 10√2
Question 10
Find the length of the medians of the triangle whose vertices are (-1,3),(1,-1) and (5,1).
Sol :
Let the given points of a triangle be A(-1, 3), B(1, -1) and C(5,1)
Let D, E and F are the midpoints of the sides BC, CA and AB respectively.
The coordinates of D are:
$\mathrm{D}=\left[\frac{5+1}{2}, \frac{1+(-1)}{2}\right]$
$\mathrm{D}=\left[\frac{6}{2}, \frac{0}{2}\right]$
D = (3, 0)
The coordinates of E are:
$\mathrm{E}=\left[\frac{5+(-1)}{2}, \frac{1+3}{2}\right]$
$\mathrm{E}=\left[\frac{4}{2}, \frac{4}{2}\right]$
E = (2, 2)
The coordinates of F are:
$\mathrm{F}=\left[\frac{1+(-1)}{2}, \frac{-1+3}{2}\right]$
$\mathrm{F}=\left[\frac{0}{2}, \frac{2}{2}\right]$
F = (0, 1)Now, we have to find the lengths of the medians.
d(A,D) = √(x2 – x1)2 + (y2 – y1)2
= √{3 – (-1)2} + {0 – 3}2
= √(3 + 1)2 + (-3)2
= √16 + 9
= √25
= 5 units
d(B,E) = √(x2 – x1)2 + (y2 – y1)2
= √(2 – 1)2 + {2 – (-1)}2
= √(1)2 + (2 + 1)2
= √1 + 9
= √10 units
d(C,F) = √(x2 – x1)2 + (y2 – y1)2
= √(5 – 0)2 + {1 – 1}2
= √(5)2 + (0)2
= √25
= 5 units
Hence, the length of the medians AD, BE and CF are 5, √10, 5 units respectively.
Question 11
If A(1,5), B (-2,1) and C(4,1) be the vertices of ΔABC and the internal bisector of ∠A meets BC and D, find AD.
Sol :
Given: A(1, 5), B(-2, 1) and C(4,1) are the vertices of ΔABC
Using angle bisector theorem, which states that:
The ratio of the length of the line segment BD to the length of segment DC is equal to the ratio of the length of side AB to the length of side AC:{\displaystyle {\frac {|BD|}{|DC|}}={\frac {|AB|}{|AC|}},}
$\frac{|\mathrm{BD}|}{|\mathrm{DC}|}=\frac{|\mathrm{AB}|}{|\mathrm{AC}|}$
$\Rightarrow \frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\sqrt{(-2-1)^{2}+(1-5)^{2}}}{\sqrt{(4-1)^{2}+(1-5)^{2}}}$
$\Rightarrow \frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\sqrt{9+16}}{\sqrt{9+16}}$
$\Rightarrow \frac{\mathrm{BD}}{\mathrm{DC}}=\frac{1}{1}$
⇒ BD = DC⇒ D is the midpoint of BC
So, the coordinates of D are:
$D=\left[\frac{-2+4}{2}, \frac{1+1}{2}\right]$
$D=\left[\frac{2}{2}, \frac{2}{2}\right]$
D = (1, 1)
Now, AD = √(x2 – x1)2 + (y2 – y1)2
= √(1 – 1)2 + {5 – 1}2
= √(0)2 + (4)2
= √16
= 4 units
Hence, AD = 4 units
Question 12
If the middle point of the line segment joining (3,4) and (k,7) is (x,y) and 2x+2y+1=0, find the value of k.
Sol :
Let P be the midpoint of the line segment joining (3, 4) and (k, 7)
So, the coordinates of P are:
$x=\frac{3+\mathrm{k}}{2}, \mathrm{y}=\frac{4+7}{2}$
$\mathrm{x}=\frac{3+\mathrm{k}}{2}, \mathrm{y}=\frac{11}{2}$
Again,
2x + 2y + 1 = 0
$\Rightarrow 2\left(\frac{3+\mathrm{k}}{2}\right)+2\left(\frac{11}{2}\right)+1=0$
⇒ 3 + k + 11 + 1 = 0
⇒ k + 15 = 0
⇒ k = -15
Question 13 A
one end of a diameter of a circle is at (2,3) and the center is (-2,5), find the coordinates of the other end of the diameter.
Sol :
Let the coordinates of the other end be (x,y).
Since (-2, 5) is the midpoint of the line joining (2,3) and (x,y)
$\therefore(-2,5)=\left(\frac{2+\times}{2}, \frac{3+\mathrm{y}}{2}\right)$
$\Rightarrow \frac{2+x}{2}=-2$ and $\frac{3+y}{2}=5$
⇒ x + 2 = -4 and y + 3 = 10
⇒ x = -4 – 2 and y = 10 – 3
⇒ x = -6 and y = 7
Hence, the coordinates of the other end are (-6, 7)
Question 13 B
Find the coordinates of a point A, where AB is the diameter of a circle whose center is (2,-3), and B is (1,4)
Sol :
Let the coordinates of the A be (x,y).
Since 2, -3) is the midpoint of the line joining (1, 4) and (x,y)
$\therefore(2,-3)=\left(\frac{1+\times}{2}, \frac{4+y}{2}\right)$
$\Rightarrow \frac{1+x}{2}=2$ and $\frac{4+y}{2}=-3$
⇒ x + 1 = 4 and y + 4 = -6
⇒ x = 4 – 1 and y = -6 – 4
⇒ x = 3 and y = -10
Hence, the coordinates of A are (3, -10)
Question 14
If the point C (-1,2) divides internally the line segment joining A (2,5) and B in the ratio 3:4. Find the coordinates of B.
Sol :
Let the coordinates of B are (x, y)
It is given that the line segment divide in the ratio 3:4
So, m1 = 3 and m2 = 4
and (x’, y’) =(-1, 2); (x1, y1) = (2,5); (x2, y2) = (x, y)
Using section formula for the internal division, we get
$\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$ …(i)
$\Rightarrow(-1,2)=\frac{3(\mathrm{x})+4(2)}{3+4}, \mathrm{y}=\frac{3(\mathrm{y})+4(5)}{3+4}$
$\Rightarrow-1=\frac{3 x+8}{7}, 2=\frac{3 y+20}{7}$
⇒ 3x + 8 = -7 and 3y + 20 = 14
⇒ 3x = -7 – 8 and 3y = 14 – 20
⇒ 3x = -15 and 3y = -6
⇒ x = -5 and y = -2
Hence, the coordinates of B are (-5, -2)
Question 15 A
Find the ratio in which (-8,3) divides the line segment joining the points (2,-2) and (-4,1).
Sol :Let C(-8, 3) divides the line segment AB in the ratio m:n
$\therefore \mathrm{x}=\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{m}+\mathrm{n}}$ and $\mathrm{y}=\frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{m}+\mathrm{n}}$
Here, (x, y) = (-8, 3); (x1, y1) = (2, -2) and (x2, y2) = (-4,1)
So, $-8=\frac{\mathrm{m}(-4)+\mathrm{n}(2)}{\mathrm{m}+\mathrm{n}}$ and $3=\frac{\mathrm{m}(1)+\mathrm{n}(-2)}{\mathrm{m}+\mathrm{n}}$
⇒ -8m -8n = -4m + 2n and 3m + 3n = m - 2n
⇒ -8m + 4m - 8n - 2n = 0 and 3m – m + 3n + 2n = 0
⇒ -4m – 10n = 0 and 2m + 5n = 0
⇒ -2m – 5n = 0 and 2m + 5n = 0
⇒ 2m + 5n = 0 and 2m + 5n = 0
⇒ 2m = -5n
$\Rightarrow \frac{\mathrm{m}}{\mathrm{n}}=\frac{-5}{2}$
Hence, the ratio is 5:2 and this negative sign shows that the division is external.
Question 15 B
In what ratio does the point (-4,6) divide the line segment joining the point A(-6,10) and B (3,-8)?
Sol :Let C(-4, 6) divides the line segment AB in the ratio m:n
$\therefore \mathrm{x}=\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{m}+\mathrm{n}}$ and $\mathrm{y}=\frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{m}+\mathrm{n}}$
Here, (x, y) = (-4, 6); (x1, y1) = (-6, 10) and (x2, y2) = (3, -8)
So, $-4=\frac{\mathrm{m}(3)+\mathrm{n}(-6)}{\mathrm{m}+\mathrm{n}}$ and $6=\frac{\mathrm{m}(-8)+\mathrm{n}(10)}{\mathrm{m}+\mathrm{n}}$
⇒ -4m -4n = 3m - 6n and 6m + 6n = -8m + 10n
⇒ -4m – 3m – 4n + 6n = 0 and 6m + 8m + 6n – 10n = 0
⇒ -7m + 2n = 0 and 14m - 4n = 0
⇒ -7m = -2n and 14m = 4n
⇒ 7m = 2n and 7m = 2n
$\Rightarrow \frac{\mathrm{m}}{\mathrm{n}}=\frac{2}{7}$
Hence, the ratio is 2:7 and the division is internal.
Question 15 C
Find the ratio in which the line segment joining (-3,10) and (6,-8) is divided by (-1,6)
Sol :Let C(-1, 6) divides the line segment AB in the ratio m:n
$\therefore \mathrm{x}=\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{m}+\mathrm{n}}$ and $\mathrm{y}=\frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{m}+\mathrm{n}}$
Here, (x, y) = (-1, 6); (x1, y1) = (-3, 10) and (x2, y2) = (6, -8)
So, $-1=\frac{\mathrm{m}(6)+\mathrm{n}(-3)}{\mathrm{m}+\mathrm{n}}$ and $6=\frac{\mathrm{m}(-8)+\mathrm{n}(10)}{\mathrm{m}+\mathrm{n}}$
⇒ -m – n = 6m - 3n and 6m + 6n = -8m + 10n
⇒ -m – 6m – n + 3n = 0 and 6m + 8m + 6n – 10n = 0
⇒ -7m + 2n = 0 and 14m - 4n = 0
⇒ 2n = 7m and 4n = 14m
⇒ 7m = 2n
$\Rightarrow \frac{\mathrm{m}}{\mathrm{n}}=\frac{2}{7}$
Hence, the ratio is 2:7 and the division is internal.
Question 15 D
Find the ratio in which the line segment joining (-3,-4) and (3,5) is divided by (x,2). Also, find x.
Sol :Let C(x, 2) divides the line segment AB in the ratio m:n
$\therefore \mathrm{x}=\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{m}+\mathrm{n}}$ and $\mathrm{y}=\frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{m}+\mathrm{n}}$
Here, (x, y) = (x, 2); (x1, y1) = (-3, -4) and (x2, y2) = (3, 5)
So, $2=\frac{\mathrm{m}(5)+\mathrm{n}(-4)}{\mathrm{m}+\mathrm{n}}$
⇒ 2m + 2n = 5m – 4n
⇒ 2m – 5m = -4n – 2n
⇒ -3m = -6n
⇒ m = 2n
$\Rightarrow \frac{\mathrm{m}}{\mathrm{n}}=\frac{2}{1}$
Now, the ratio is 2:1
Now,$\mathrm{x}=\frac{\mathrm{m}(3)+\mathrm{n}(-3)}{\mathrm{m}+\mathrm{n}}$
$\Rightarrow x=\frac{2(3)+1(-3)}{2+1}$
⇒ 3x = 6 – 3
⇒ 3x = 3
⇒ x = 1
Hence, the ratio is 2:1 and the division is internal and the value of x = 1
Question 16 A
In what ratio does the x-axis divide the line segment joining the points (2,-3) and (5,6).
Sol :Let the line segment A(2, -3) and B(5, 6) is divided at point P(x,0) by x-axis in ratio m:n
$\therefore \mathrm{x}=\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{m}+\mathrm{n}}$ and $\mathrm{y}=\frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{m}+\mathrm{n}}$
Here, (x, y) = (x, 0); (x1, y1) = (2, -3) and (x2, y2) = (5, 6)
So, $0=\frac{\mathrm{m}(6)+\mathrm{n}(-3)}{\mathrm{m}+\mathrm{n}}$
⇒ 0 = 6m – 3n
⇒ -6m = -3n
$\Rightarrow \frac{\mathrm{m}}{\mathrm{n}}=\frac{1}{2}$
Hence, the ratio is 1:2 and the division is internal.
Question 16 B
Find the ratio in which the line segment joining A(1,-5) and B(-4,5) is divided by the x-axis. Also, find the coordinates of the point of division.
Sol :Let the line segment A(1, -5) and B(-4, 5) is divided at point P(x,0) by x-axis in ratio m:n
$\therefore \mathrm{x}=\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{m}+\mathrm{n}}$ and $\mathrm{y}=\frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{m}+\mathrm{n}}$
Here, (x, y) = (x, 0); (x1, y1) = (1, -5) and (x2, y2) = (-4, 5)
So, $0=\frac{\mathrm{m}(5)+\mathrm{n}(-5)}{\mathrm{m}+\mathrm{n}}$
⇒ 0 = 5m – 5n
⇒ 5m = 5n
$\Rightarrow \frac{\mathrm{m}}{\mathrm{n}}=\frac{1}{1}$
Hence, the ratio is 1:1 and the division is internal.
Now,
$\mathrm{x}=\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{m}+\mathrm{n}}$
$\Rightarrow \mathrm{x}=\frac{1(-4)+1(1)}{1+1}$
$\Rightarrow \mathrm{x}=\frac{-3}{2}$
Hence, the coordinates of the point of division is $\left(\frac{-3}{2}, 0\right)$
Question 16 C
Find the ratio in which the y-axis divides the line segment joining points (5,-6) and (-1,-4). Also, find the point of intersection.Sol :
Let the line segment A(5, -6) and B(-1, -4) is divided at point P(0, y) by y-axis in ratio m:n
$\therefore \mathrm{x}=\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{m}+\mathrm{n}}$ and $\mathrm{y}=\frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{m}+\mathrm{n}}$
Here, (x, y) = (0, y); (x1, y1) = (5, -6) and (x2, y2) = (-1, -4)
So, $0=\frac{\mathrm{m}(-1)+\mathrm{n}(5)}{\mathrm{m}+\mathrm{n}}$
⇒ 0 = -m + 5n
⇒ m = 5n
$\Rightarrow \frac{\mathrm{m}}{\mathrm{n}}=\frac{5}{1}$
Hence, the ratio is 5:1 and the division is internal.
Now,
$\mathrm{y}=\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{m}+\mathrm{n}}$
$\Rightarrow \mathrm{y}=\frac{5(-4)+1(-6)}{5+1}$
$\Rightarrow \mathrm{y}=\frac{-20-6}{6}=\frac{-26}{6}=\frac{-13}{3}$
Hence, the coordinates of the point of division is $\left(0, \frac{-13}{3}\right)$
Question 17
Find the centroid of the triangle whose vertices are (2,4), (6,4), (2,0).
Sol :Here, x1 = 2, x2 = 6, x3 = 2
and y1 = 4, y2 = 4, y3 = 0
Let the coordinates of the centroid be(x,y)
So,
Centroid of triangle (x,y)=
$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}\right)$
$=\left(\frac{2+6+2}{3}, \frac{4+4+0}{3}\right)$
$=\left(\frac{10}{3}, \frac{8}{3}\right)$
Hence, the centroid of a triangle is $\left(\frac{10}{3}, \frac{8}{3}\right)$
Let (2, 2), (0, 6) and (8, 10) be the vertices A, B and C of the triangle respectively. Let AD, BE, CF be the medians
The coordinates of D are:
D = (4, 8)
The coordinates of E are:
E = (5, 6)
The coordinates of F are:
F = (1, 4)
Let P be the trisection point of the median AD which is nearer to the opposite side BC
∴ P divides DA in the ratio 1:2 internally
$\therefore \mathrm{P}=\left(\frac{1(2)+2(4)}{1+2}, \frac{1(2)+2(8)}{1+2}\right)$
Let Q be the trisection point of the median BE which is nearer to the opposite side CA
∴ Q divides EB in the ratio 1:2 internally
$\therefore \mathrm{Q}=\left(\frac{1(0)+2(5)}{1+2}, \frac{1(6)+2(6)}{1+2}\right)$
Let R be the trisection point of the median CF which is nearer to the opposite side AB
∴ R divides FC in the ratio 1:2 internally
$\therefore \mathrm{R}=\left(\frac{1(8)+2(1)}{1+2}, \frac{1(10)+2(4)}{1+2}\right)$
Therefore, Coordinates of required trisection points are $\left(\frac{10}{3}, 6\right),\left(\frac{10}{3}, 6\right)$ and $\left(\frac{10}{3}, 6\right)$
Let the third vertex of a triangle be(x,y)
Here, x1 = 1, x2 = 5, x3 = x
and y1 = 4, y2 = 2, y3 = y
and the coordinates of the centroid is (0, -3)
We know that
Centroid of triangle $(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}\right)$
$(0,-3)=\left(\frac{1+5+x}{3}, \frac{4+2+y}{3}\right)$
$(0,-3)=\left(\frac{6+\times}{3}, \frac{6+y}{3}\right)$
$\Rightarrow 0=\frac{6+\mathrm{x}}{3}$ and $-3=\frac{6+\mathrm{y}}{3}$
⇒ 6 + x = 0 and 6 + y = -9
⇒ x = -6 and y = -15
Hence, the third vertex of a triangle is (-6, -15)
Let the third vertex of a triangle be (x, y)
Here, x1 = 2√3, x2 = 2√3, x3 = x
and y1 = -1, y2 = 5, y3 = y
and the coordinates of the centroid is (√3, 2)
We know that
Centroid of triangle $(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}\right)$
⇒ 4√3 + x = 3√3 and 4 + y = 6
⇒ x = -√3 and y = 2
Hence, the third vertex of a triangle is (-√3, 2)
The vertices of a triangle are A (9,2), B(1,10) and C(-7,-6)
Here, x1 = 9, x2 = 1, x3 = -7
and y1 = 2, y2 = 10, y3 = -6
Let the coordinates of the centroid be(x,y)
So,
= (1,2)
Hence, the centroid of a triangle is (1, 2)
Now,
Let D, E and F are the midpoints of the sides BC, CA and AB respectively.
The coordinates of D are:
D = (-3, 2)
The coordinates of E are:
E = (1, -2)
The coordinates of F are:
F = (5, 6)
Now, we find the centroid of a triangle formed by joining these middle points D, E, and F as shown in figure
Let P be the trisection point of the median AD which is nearer to the opposite side BC
∴ P divides DA in the ratio 1:2 internally
$\therefore \mathrm{P}=\left(\frac{1(9)+2(-3)}{1+2}, \frac{1(2)+2(2)}{1+2}\right)$
$=\left(\frac{9-6}{3}, \frac{2+4}{3}\right)$
$=\left(\frac{3}{3}, \frac{6}{3}\right)$
= (1, 2)
Let Q be the trisection point of the median BE which is nearer to the opposite side CA
∴ Q divides EB in the ratio 1:2 internally
= (1, 2)
Let R be the trisection point of the median CF which is nearer to the opposite side AB
∴ R divides FC in the ratio 1:2 internally
Yes, the triangle has the same centroid, i.e. (1,2)
Consider a ΔABC with A(x1, y1), B(x2, y2) and C(x3, y3). If P(1, 2), Q(0, -1) and R(2, -1) are the midpoints of AB, BC and CA. Then,
$1=\frac{x_{1}+x_{2}}{2} \Rightarrow x_{1}+x_{2}=2$ …(i)
$2=\frac{y_{1}+y_{2}}{2} \Rightarrow y_{1}+y_{2}=4$ …(ii)
$0=\frac{x_{2}+x_{3}}{2} \Rightarrow x_{2}+x_{3}=0$ …(iii)
$-1=\frac{y_{2}+y_{3}}{2} \Rightarrow y_{2}+y_{3}=-2$ …(iv)
$2=\frac{x_{1}+x_{3}}{2} \Rightarrow x_{1}+x_{3}=4$ …(v)
$-1=\frac{y_{1}+y_{3}}{2} \Rightarrow y_{1}+y_{3}=-2$ …(vi)
Adding (i), (iii) and (v), we get
x1 + x2 + x2 + x3 + x1 + x3 = 2 + 0 + 4
⇒ 2(x1 + x2 + x3) = 6
⇒ x1 + x2 + x3 = 3 …(vii)
From (i) and (vii), we get
x3 = 3 – 2 = 1
From (iii) and (vii), we get
x1 = 3 – 0 = 3
From (v) and (vii), we get
x2 = 3 – 4 = -1
Now adding (ii), (iv) and (vi), we get
y1 + y2 + y2 + y3 + y1 + y3 = 4 + (-2) + (-2)
⇒ 2(y1 + y2 + y3) = 0
⇒ y1 + y2 + y3 = 0 …(viii)
From (ii) and (viii), we get
y3 = 0 – 4 = -4
From (iv) and (vii), we get
y1 = 0 – (-2) = 2
From (vi) and (vii), we get
y2 = 0 – (-2) = 2
Hence, the vertices of ΔABC are A(3, 2), B(-1, 2) and C(1, -4)
Now, we have to find the centroid of a triangle
The vertices of a triangle are A(3, 2), B(-1, 2) and C(1, -4)
Here, x1 = 3, x2 = -1, x3 = 1
and y1 = 2, y2 = 2, y3 = -4
Let the coordinates of the centroid be(x,y)
So,
Centroid of triangle $(x, y)=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$
$=\left(\frac{3+(-1)+1}{3}, \frac{2+2+(-4)}{3}\right)$
$=\left(\frac{3}{3}, \frac{0}{3}\right)$
= (1,0)
Hence, the centroid of a triangle is (1, 0)
Note that to show that a quadrilateral is a rhombus, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD have the same midpoint.
(b) a pair of adjacent edges are equal
(c) the diagonal AC and BD are not equal.
Let A(-3, 2), B(-5,-5), C(2,-3) and D(4,4) are the vertices of a rhombus.
Coordinates of the midpoint of AC are
$\left(\frac{-3+2}{2}, \frac{2-3}{2}\right)=\left(\frac{-1}{2}, \frac{-1}{2}\right)$
Coordinates of the midpoint of BD are
$\left(\frac{-5+4}{2}, \frac{-5+4}{2}\right)=\left(\frac{-1}{2}, \frac{-1}{2}\right)$
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, using Distance Formula
d(A,B)= AB = √(-5 + 3)2 + (-5 – 2)2
⇒ AB = √(-2)2 + (-7)2
⇒ AB = √4 +49
⇒ AB = √53 units
d(B,C)= BC = √(-5 – 2)2 + (-5 + 3)2
⇒ BC = √(-7)2 + (-2)2
⇒ BC = √49 +4
⇒ BC = √53 units
d(C,D) = CD = √(4 – 2)2 + (4 + 3)2
⇒ CD = √(2)2 + (7)2
⇒ CD = √4 +49
⇒ CD = √53 units
d(A,D) = AD =√(4 + 3)2 +(4 – 2)2
⇒ AD = √(7)2 + (2)2
⇒ AD = √49 +4
⇒ AD = √53 units
Therefore, AB = BC = CD = AD = √53 units
Now, check for the diagonals
AC = √(2 + 3)2 + (-3 – 2)2
= √(5)2 + (-5)2
= √25 + 25
= √50
and
BD = √(4 + 5)2 + (4 + 5)2
⇒ BD = √(9)2 + (9)2
⇒ BD = √81 + 81
⇒ BD = √162
⇒ Diagonal AC ≠ Diagonal BD
Hence, ABCD is a rhombus.
Note that to show that a quadrilateral is a square, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD bisect each other
(b) a pair of adjacent edges are equal
(c) the diagonal AC and BD are equal.
Let the vertices of a quadrilateral are A(3, 2), B(0,5), C(-3, 2) and D(0, -1).
Coordinates of the midpoint of AC are
$\left(\frac{3+(-3)}{2}, \frac{2+2}{2}\right)=\left(\frac{0}{2}, \frac{4}{2}\right)=(0,2)$
Coordinates of the midpoint of BD are
$\left(\frac{0+0}{2}, \frac{5+(-1)}{2}\right)=\left(\frac{0}{2}, \frac{4}{2}\right)=(0,2)$
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, Using Distance Formula, we get
AB = √(x2 – x1)2 + (y2 – y1)2
= √[(0 – 3)2 + (5 - 2)2]
= √(-3)2 + (3)2
= √(9 + 9)
= √18 units
BC = √[(-3 – 0)2 + (2 - 5)2]
= √(-3)2 + (-3)2
= √(9 + 9)
= √18 units
Therefore, AB = BC = √18 units
Now, check for the diagonals
AC = √(-3 – 3)2 + (2 – 2)2
= √(-6)2 + (0)2
= √36
= 6 units
and
BD = √(0 - 0)2 + (-1 – 5)2
⇒ BD = √(0)2 + (-6)2
⇒ BD = √36
⇒ BD = 6 units
∴ AC = BD
Hence, ABCD is a square.
Note that to show that a quadrilateral is a parallelogram, it is sufficient to show that the diagonals of the quadrilateral bisect each other.
Let A(-2, -1), B(1, 0), C(4, 3) and D(1, 2) are the vertices of a parallelogram.
Let M be the midpoint of AC, then the coordinates of M are given by
$\left(\frac{-2+4}{2}, \frac{-1+3}{2}\right)=\left(\frac{2}{2}, \frac{2}{2}\right)=(1,1)$
Let N be the midpoint of BD, then the coordinates of N are given by
$\left(\frac{1+1}{2}, \frac{2+0}{2}\right)=\left(\frac{2}{2}, \frac{2}{2}\right)=(1,1)$
Thus, AC and BD have the same midpoint.
In other words, AC and BD bisect each other.
Hence, ABCD is a parallelogram.
Note that to show that a quadrilateral is a rhombus, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD have the same midpoint.
(b) a pair of adjacent edges are equal
Let A(1, 0), B(5, 3), C(2, 7) and D(-2, 4) are the vertices of a rhombus.
Coordinates of the midpoint of AC are
$\left(\frac{1+2}{2}, \frac{0+7}{2}\right)=\left(\frac{3}{2}, \frac{7}{2}\right)$
Coordinates of the midpoint of BD are
$\left(\frac{5-2}{2}, \frac{3+4}{2}\right)=\left(\frac{3}{2}, \frac{7}{2}\right)$
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, using Distance Formula
d(A,B)= AB = √(5 – 1)2 + (3 – 0)2
⇒ AB = √(4)2 + (3)2
⇒ AB = √16 + 9
⇒ AB = √25 = 5 units
d(B,C)= BC = √(2 – 5)2 + (7 – 3)2
⇒ BC = √(-3)2 + (4)2
⇒ BC = √9 + 16
⇒ BC = √25 = 5 units
Therefore, adjacent sides are equal.
Hence, ABCD is a rhombus.
Note that to show that a quadrilateral is a rectangle, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD bisect each other and,
(b) the diagonal AC and BD are equal
Let A(4, 8), B(0, 2), C(3, 0) and D(7, 6) are the vertices of a rectangle.
Coordinates of the midpoint of AC are
$\left(\frac{4+3}{2}, \frac{8+0}{2}\right)=\left(\frac{7}{2}, 4\right)$
Coordinates of the midpoint of BD are
$\left(\frac{4+3}{2}, \frac{8+0}{2}\right)=\left(\frac{7}{2}, 4\right)$
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, check for the diagonals by using the distance formula
AC = √(3 – 4)2 + (0 – 8)2
= √(-1)2 + (-8)2
= √1 + 64
= √65 units
and
BD = √(7 - 0)2 + (6 – 2)2
⇒ BD = √(7)2 + (4)2
⇒ BD = √49 + 16
⇒ BD = √65 units
∴ AC = BD
Hence, ABCD is a rectangle.
Note that to show that a quadrilateral is a square, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD bisect each other
(b) a pair of adjacent edges are equal
(c) the diagonal AC and BD are equal.
Let the vertices of a quadrilateral are A(4, 3), B(6, 4), C(5, 6) and D(3, 5).
Coordinates of the midpoint of AC are
$\left(\frac{4+5}{2}, \frac{3+6}{2}\right)=\left(\frac{9}{2}, \frac{9}{2}\right)$
Coordinates of the midpoint of BD are
$\left(\frac{6+3}{2}, \frac{4+5}{2}\right)=\left(\frac{9}{2}, \frac{9}{2}\right)$
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, Using Distance Formula, we get
AB = √(x2 – x1)2 + (y2 – y1)2
= √[(6 – 4)2 + (4 - 3)2]
= √(2)2 + (1)2
= √(4 + 1)
= √5 units
BC = √[(5 – 6)2 + (6 - 4)2]
= √(-1)2 + (2)2
= √(1 + 4)
= √5 units
Therefore, AB = BC = √5 units
Now, check for the diagonals
AC = √(5 – 4)2 + (6 – 3)2
= √(1)2 + (3)2
= √1 + 9
= √10 units
and
BD = √(3 - 6)2 + (5 – 4)2
⇒ BD = √(-3)2 + (1)2
⇒ BD = √9 + 1
⇒ BD = √10 units
∴ AC = BD
Hence, ABCD is a square.
Sol :
Let the coordinates of the fourth vertex D be (x, y).
We know that diagonals of a parallelogram bisect each other.
∴ Midpoint of AC = Midpoint of BD …(i)
Coordinates of the midpoint of AC are
$\left(\frac{6-2}{2}, \frac{8-2}{2}\right)=\left(\frac{4}{2}, \frac{6}{2}\right)=(2,3)$
Coordinates of the midpoint of BD are
$\left(\frac{3+x}{2}, \frac{7+y}{2}\right)$
So, according to eq. (i), we have
⇒ 3 + x = 4 and 7 + y = 6
⇒ x = 1 and y = -1
Thus, the coordinates of the vertex D are (1, -1)
Let the coordinates of the fourth vertex D be (x, y).
We know that diagonals of a rhombus bisect each other.
∴ Midpoint of AC = Midpoint of BD …(i)
Coordinates of the midpoint of AC are
$\left(\frac{5-2}{2}, \frac{3+4}{2}\right)=\left(\frac{3}{2}, \frac{7}{2}\right)$
Coordinates of the midpoint of BD are
$\left(\frac{2+x}{2}, \frac{7+y}{2}\right)$
So, according to eq. (i), we have
$\Rightarrow\left(\frac{3}{2}, \frac{7}{2}\right)=\left(\frac{2+x}{2}, \frac{7+y}{2}\right)$
$\Rightarrow \frac{3}{2}=\frac{2+x}{2}$ and $\frac{7}{2}=\frac{7+y}{2}$
⇒ 2 + x = 3 and 7 + y = 7
⇒ x = 1 and y = 0
Thus, the coordinates of the vertex D are (1, 0)
Let the vertices of quadrilateral be P(-4,2), Q(2,6), R(8,5) and S(9,-7)
Let A, B, C and D are the midpoints of PQ, QR, RS and SP respectively.
Now, since A is the midpoint of P(-4, 2) and Q(2, 6)
∴ Coordinates of A are
$\left(\frac{-4+2}{2}, \frac{2+6}{2}\right)=\left(\frac{-2}{2}, \frac{8}{2}\right)=(-1,4)$
Coordinates of B are
$\left(\frac{2+8}{2}, \frac{6+5}{2}\right)=\left(\frac{10}{2}, \frac{11}{2}\right)=\left(5, \frac{11}{2}\right)$
Coordinates of C are
$\left(\frac{8+9}{2}, \frac{5-7}{2}\right)=\left(\frac{17}{2}, \frac{-2}{2}\right)=\left(\frac{17}{2},-1\right)$
and
Coordinates of D are
$\left(\frac{9-4}{2}, \frac{-7+2}{2}\right)=\left(\frac{5}{2}, \frac{-5}{2}\right)$
Now,
we find the distance between A and B
$\mathrm{d}(\mathrm{A}, \mathrm{B})=\sqrt{(-1-5)^{2}+\left(4-\frac{11}{2}\right)^{2}}$
$=\sqrt{36+\frac{9}{4}}=\sqrt{\frac{144+9}{4}}=\sqrt{\frac{153}{4}}$
Hence, the centroid of a triangle is $\left(\frac{10}{3}, \frac{8}{3}\right)$
Question 18
The vertices of a triangle are at (2,2), (0,6) and (8,10). Find the coordinates of the trisection point of each median which is nearer the opposite side.
Sol :Let (2, 2), (0, 6) and (8, 10) be the vertices A, B and C of the triangle respectively. Let AD, BE, CF be the medians
The coordinates of D are:
$\mathrm{D}=\left[\frac{0+8}{2}, \frac{6+10}{2}\right]$
$\mathrm{D}=\left[\frac{8}{2}, \frac{16}{2}\right]$
D = (4, 8)
The coordinates of E are:
$\mathrm{E}=\left[\frac{8+2}{2}, \frac{10+2}{2}\right]$
$\mathrm{E}=\left[\frac{10}{2}, \frac{12}{2}\right]$
E = (5, 6)
The coordinates of F are:
$\mathrm{F}=\left[\frac{2+0}{2}, \frac{2+6}{2}\right]$
$\mathrm{F}=\left[\frac{2}{2}, \frac{8}{2}\right]$
F = (1, 4)
Let P be the trisection point of the median AD which is nearer to the opposite side BC
∴ P divides DA in the ratio 1:2 internally
$\therefore \mathrm{P}=\left(\frac{1(2)+2(4)}{1+2}, \frac{1(2)+2(8)}{1+2}\right)$
$=\left(\frac{2+8}{3}, \frac{2+16}{3}\right)$
$=\left(\frac{10}{3}, 6\right)$
Let Q be the trisection point of the median BE which is nearer to the opposite side CA
∴ Q divides EB in the ratio 1:2 internally
$\therefore \mathrm{Q}=\left(\frac{1(0)+2(5)}{1+2}, \frac{1(6)+2(6)}{1+2}\right)$
$=\left(\frac{0+10}{3}, \frac{6+12}{3}\right)$
$=\left(\frac{10}{3}, 6\right)$
Let R be the trisection point of the median CF which is nearer to the opposite side AB
∴ R divides FC in the ratio 1:2 internally
$\therefore \mathrm{R}=\left(\frac{1(8)+2(1)}{1+2}, \frac{1(10)+2(4)}{1+2}\right)$
$=\left(\frac{8+2}{3}, \frac{10+8}{3}\right)$
$=\left(\frac{10}{3}, 6\right)$
Therefore, Coordinates of required trisection points are $\left(\frac{10}{3}, 6\right),\left(\frac{10}{3}, 6\right)$ and $\left(\frac{10}{3}, 6\right)$
Question 19
Two vertices of a triangle are (1,4) and (5,2). If its centroid is (0,-3), find the third vertex.
Sol :Let the third vertex of a triangle be(x,y)
Here, x1 = 1, x2 = 5, x3 = x
and y1 = 4, y2 = 2, y3 = y
and the coordinates of the centroid is (0, -3)
We know that
Centroid of triangle $(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}\right)$
$(0,-3)=\left(\frac{1+5+x}{3}, \frac{4+2+y}{3}\right)$
$(0,-3)=\left(\frac{6+\times}{3}, \frac{6+y}{3}\right)$
$\Rightarrow 0=\frac{6+\mathrm{x}}{3}$ and $-3=\frac{6+\mathrm{y}}{3}$
⇒ 6 + x = 0 and 6 + y = -9
⇒ x = -6 and y = -15
Hence, the third vertex of a triangle is (-6, -15)
Question 20
The coordinates of the centroid of a triangle are (√3,2), and two of its vertices are (2√3,-1) and (2√3,5). Find the third vertex of the triangle.
Sol :Let the third vertex of a triangle be (x, y)
Here, x1 = 2√3, x2 = 2√3, x3 = x
and y1 = -1, y2 = 5, y3 = y
and the coordinates of the centroid is (√3, 2)
We know that
Centroid of triangle $(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}\right)$
$(\sqrt{3}, 2)=\left(\frac{2 \sqrt{3}+2 \sqrt{3}+x}{3}, \frac{-1+5+y}{3}\right)$
$(\sqrt{3}, 2)=\left(\frac{4 \sqrt{3}+x}{3}, \frac{4+y}{3}\right)$
$\Rightarrow \sqrt{3}=\frac{4 \sqrt{3}+x}{3}$ and $2=\frac{4+y}{3}$
⇒ 4√3 + x = 3√3 and 4 + y = 6
⇒ x = -√3 and y = 2
Hence, the third vertex of a triangle is (-√3, 2)
Question 21
Find the centroid of the triangle ABC whose vertices are A (9,2), B(1,10) and C(-7,-6). Find the coordinates of the middle points of its sides and hence find the centroid of the triangle formed by joining these middle points. Do the two triangles have the same centroid?
Sol :The vertices of a triangle are A (9,2), B(1,10) and C(-7,-6)
Here, x1 = 9, x2 = 1, x3 = -7
and y1 = 2, y2 = 10, y3 = -6
Let the coordinates of the centroid be(x,y)
So,
Centroid of triangle $(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}\right)$
$=\left(\frac{9+1+(-7)}{3}, \frac{2+10+(-6)}{3}\right)$
$=\left(\frac{10-7}{3}, \frac{12-6}{3}\right)$
$=\left(\frac{3}{3}, \frac{6}{3}\right)$
= (1,2)
Hence, the centroid of a triangle is (1, 2)
Now,
Let D, E and F are the midpoints of the sides BC, CA and AB respectively.
The coordinates of D are:
$D=\left[\frac{-7+1}{2}, \frac{-6+10}{2}\right]$
$D=\left[\frac{-6}{2}, \frac{4}{2}\right]$
D = (-3, 2)
The coordinates of E are:
$\mathrm{E}=\left[\frac{-7+9}{2}, \frac{-6+2}{2}\right]$
$\mathrm{E}=\left[\frac{2}{2}, \frac{-4}{2}\right]$
E = (1, -2)
The coordinates of F are:
$\mathrm{F}=\left[\frac{1+9}{2}, \frac{10+2}{2}\right]$
$\mathrm{F}=\left[\frac{10}{2}, \frac{12}{2}\right]$
F = (5, 6)
Now, we find the centroid of a triangle formed by joining these middle points D, E, and F as shown in figure
Let P be the trisection point of the median AD which is nearer to the opposite side BC
∴ P divides DA in the ratio 1:2 internally
$\therefore \mathrm{P}=\left(\frac{1(9)+2(-3)}{1+2}, \frac{1(2)+2(2)}{1+2}\right)$
$=\left(\frac{9-6}{3}, \frac{2+4}{3}\right)$
$=\left(\frac{3}{3}, \frac{6}{3}\right)$
= (1, 2)
Let Q be the trisection point of the median BE which is nearer to the opposite side CA
∴ Q divides EB in the ratio 1:2 internally
$\therefore Q=\left(\frac{1(1)+2(1)}{1+2}, \frac{1(10)+2(-2)}{1+2}\right)$
$=\left(\frac{1+2}{3}, \frac{10-4}{3}\right)$
$=\left(\frac{3}{3}, \frac{6}{3}\right)$
= (1, 2)
Let R be the trisection point of the median CF which is nearer to the opposite side AB
∴ R divides FC in the ratio 1:2 internally
$\therefore \mathrm{R}=\left(\frac{1(-7)+2(5)}{1+2}, \frac{1(-6)+2(6)}{1+2}\right)$
$=\left(\frac{-7+10}{3}, \frac{-6+12}{3}\right)$
$=\left(\frac{3}{3}, \frac{6}{3}\right)$
= (1, 2)Yes, the triangle has the same centroid, i.e. (1,2)
Question 22
If (1,2), (0,-1) and (2,-1) are the middle points of the sides of the triangle, find the coordinates of its centroid.
Sol :Consider a ΔABC with A(x1, y1), B(x2, y2) and C(x3, y3). If P(1, 2), Q(0, -1) and R(2, -1) are the midpoints of AB, BC and CA. Then,
$1=\frac{x_{1}+x_{2}}{2} \Rightarrow x_{1}+x_{2}=2$ …(i)
$2=\frac{y_{1}+y_{2}}{2} \Rightarrow y_{1}+y_{2}=4$ …(ii)
$0=\frac{x_{2}+x_{3}}{2} \Rightarrow x_{2}+x_{3}=0$ …(iii)
$-1=\frac{y_{2}+y_{3}}{2} \Rightarrow y_{2}+y_{3}=-2$ …(iv)
$2=\frac{x_{1}+x_{3}}{2} \Rightarrow x_{1}+x_{3}=4$ …(v)
$-1=\frac{y_{1}+y_{3}}{2} \Rightarrow y_{1}+y_{3}=-2$ …(vi)
Adding (i), (iii) and (v), we get
x1 + x2 + x2 + x3 + x1 + x3 = 2 + 0 + 4
⇒ 2(x1 + x2 + x3) = 6
⇒ x1 + x2 + x3 = 3 …(vii)
From (i) and (vii), we get
x3 = 3 – 2 = 1
From (iii) and (vii), we get
x1 = 3 – 0 = 3
From (v) and (vii), we get
x2 = 3 – 4 = -1
Now adding (ii), (iv) and (vi), we get
y1 + y2 + y2 + y3 + y1 + y3 = 4 + (-2) + (-2)
⇒ 2(y1 + y2 + y3) = 0
⇒ y1 + y2 + y3 = 0 …(viii)
From (ii) and (viii), we get
y3 = 0 – 4 = -4
From (iv) and (vii), we get
y1 = 0 – (-2) = 2
From (vi) and (vii), we get
y2 = 0 – (-2) = 2
Hence, the vertices of ΔABC are A(3, 2), B(-1, 2) and C(1, -4)
Now, we have to find the centroid of a triangle
The vertices of a triangle are A(3, 2), B(-1, 2) and C(1, -4)
Here, x1 = 3, x2 = -1, x3 = 1
and y1 = 2, y2 = 2, y3 = -4
Let the coordinates of the centroid be(x,y)
So,
Centroid of triangle $(x, y)=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$
$=\left(\frac{3+(-1)+1}{3}, \frac{2+2+(-4)}{3}\right)$
$=\left(\frac{3}{3}, \frac{0}{3}\right)$
= (1,0)
Hence, the centroid of a triangle is (1, 0)
Question 23
Show that A(-3,2), B(-5,-5), C(2,-3) and D(4,4) are the vertices of a rhombus.
Sol :Note that to show that a quadrilateral is a rhombus, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD have the same midpoint.
(b) a pair of adjacent edges are equal
(c) the diagonal AC and BD are not equal.
Let A(-3, 2), B(-5,-5), C(2,-3) and D(4,4) are the vertices of a rhombus.
Coordinates of the midpoint of AC are
$\left(\frac{-3+2}{2}, \frac{2-3}{2}\right)=\left(\frac{-1}{2}, \frac{-1}{2}\right)$
Coordinates of the midpoint of BD are
$\left(\frac{-5+4}{2}, \frac{-5+4}{2}\right)=\left(\frac{-1}{2}, \frac{-1}{2}\right)$
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, using Distance Formula
d(A,B)= AB = √(-5 + 3)2 + (-5 – 2)2
⇒ AB = √(-2)2 + (-7)2
⇒ AB = √4 +49
⇒ AB = √53 units
d(B,C)= BC = √(-5 – 2)2 + (-5 + 3)2
⇒ BC = √(-7)2 + (-2)2
⇒ BC = √49 +4
⇒ BC = √53 units
d(C,D) = CD = √(4 – 2)2 + (4 + 3)2
⇒ CD = √(2)2 + (7)2
⇒ CD = √4 +49
⇒ CD = √53 units
d(A,D) = AD =√(4 + 3)2 +(4 – 2)2
⇒ AD = √(7)2 + (2)2
⇒ AD = √49 +4
⇒ AD = √53 units
Therefore, AB = BC = CD = AD = √53 units
Now, check for the diagonals
AC = √(2 + 3)2 + (-3 – 2)2
= √(5)2 + (-5)2
= √25 + 25
= √50
and
BD = √(4 + 5)2 + (4 + 5)2
⇒ BD = √(9)2 + (9)2
⇒ BD = √81 + 81
⇒ BD = √162
⇒ Diagonal AC ≠ Diagonal BD
Hence, ABCD is a rhombus.
Question 24
Show that the point (3,2),(0,5),(-3,2) and (0,-1) are the vertices of a square.
Sol :Note that to show that a quadrilateral is a square, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD bisect each other
(b) a pair of adjacent edges are equal
(c) the diagonal AC and BD are equal.
Let the vertices of a quadrilateral are A(3, 2), B(0,5), C(-3, 2) and D(0, -1).
Coordinates of the midpoint of AC are
$\left(\frac{3+(-3)}{2}, \frac{2+2}{2}\right)=\left(\frac{0}{2}, \frac{4}{2}\right)=(0,2)$
Coordinates of the midpoint of BD are
$\left(\frac{0+0}{2}, \frac{5+(-1)}{2}\right)=\left(\frac{0}{2}, \frac{4}{2}\right)=(0,2)$
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, Using Distance Formula, we get
AB = √(x2 – x1)2 + (y2 – y1)2
= √[(0 – 3)2 + (5 - 2)2]
= √(-3)2 + (3)2
= √(9 + 9)
= √18 units
BC = √[(-3 – 0)2 + (2 - 5)2]
= √(-3)2 + (-3)2
= √(9 + 9)
= √18 units
Therefore, AB = BC = √18 units
Now, check for the diagonals
AC = √(-3 – 3)2 + (2 – 2)2
= √(-6)2 + (0)2
= √36
= 6 units
and
BD = √(0 - 0)2 + (-1 – 5)2
⇒ BD = √(0)2 + (-6)2
⇒ BD = √36
⇒ BD = 6 units
∴ AC = BD
Hence, ABCD is a square.
Question 25
Prove that the points (-2,-1), (1,0),(4,3) and (1,2) are the vertices of a parallelogram.
Sol :Note that to show that a quadrilateral is a parallelogram, it is sufficient to show that the diagonals of the quadrilateral bisect each other.
Let A(-2, -1), B(1, 0), C(4, 3) and D(1, 2) are the vertices of a parallelogram.
Let M be the midpoint of AC, then the coordinates of M are given by
$\left(\frac{-2+4}{2}, \frac{-1+3}{2}\right)=\left(\frac{2}{2}, \frac{2}{2}\right)=(1,1)$
Let N be the midpoint of BD, then the coordinates of N are given by
$\left(\frac{1+1}{2}, \frac{2+0}{2}\right)=\left(\frac{2}{2}, \frac{2}{2}\right)=(1,1)$
Thus, AC and BD have the same midpoint.
In other words, AC and BD bisect each other.
Hence, ABCD is a parallelogram.
Question 26
Show that the points A(1,0), B(5,3), C(2,7) and D(-2,4) are the vertices of a rhombus.
Sol :Note that to show that a quadrilateral is a rhombus, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD have the same midpoint.
(b) a pair of adjacent edges are equal
Let A(1, 0), B(5, 3), C(2, 7) and D(-2, 4) are the vertices of a rhombus.
Coordinates of the midpoint of AC are
$\left(\frac{1+2}{2}, \frac{0+7}{2}\right)=\left(\frac{3}{2}, \frac{7}{2}\right)$
Coordinates of the midpoint of BD are
$\left(\frac{5-2}{2}, \frac{3+4}{2}\right)=\left(\frac{3}{2}, \frac{7}{2}\right)$
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, using Distance Formula
d(A,B)= AB = √(5 – 1)2 + (3 – 0)2
⇒ AB = √(4)2 + (3)2
⇒ AB = √16 + 9
⇒ AB = √25 = 5 units
d(B,C)= BC = √(2 – 5)2 + (7 – 3)2
⇒ BC = √(-3)2 + (4)2
⇒ BC = √9 + 16
⇒ BC = √25 = 5 units
Therefore, adjacent sides are equal.
Hence, ABCD is a rhombus.
Question 27
Prove that the point (4,8), (0,2), (3,0) and (7,6) are the vertices of a rectangle.
Sol :Note that to show that a quadrilateral is a rectangle, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD bisect each other and,
(b) the diagonal AC and BD are equal
Let A(4, 8), B(0, 2), C(3, 0) and D(7, 6) are the vertices of a rectangle.
Coordinates of the midpoint of AC are
$\left(\frac{4+3}{2}, \frac{8+0}{2}\right)=\left(\frac{7}{2}, 4\right)$
Coordinates of the midpoint of BD are
$\left(\frac{4+3}{2}, \frac{8+0}{2}\right)=\left(\frac{7}{2}, 4\right)$
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, check for the diagonals by using the distance formula
AC = √(3 – 4)2 + (0 – 8)2
= √(-1)2 + (-8)2
= √1 + 64
= √65 units
and
BD = √(7 - 0)2 + (6 – 2)2
⇒ BD = √(7)2 + (4)2
⇒ BD = √49 + 16
⇒ BD = √65 units
∴ AC = BD
Hence, ABCD is a rectangle.
Question 28
Prove that the points (4,3), (6,4), (5,6) and (3,5) are the vertices of a square.
Sol :Note that to show that a quadrilateral is a square, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD bisect each other
(b) a pair of adjacent edges are equal
(c) the diagonal AC and BD are equal.
Let the vertices of a quadrilateral are A(4, 3), B(6, 4), C(5, 6) and D(3, 5).
Coordinates of the midpoint of AC are
$\left(\frac{4+5}{2}, \frac{3+6}{2}\right)=\left(\frac{9}{2}, \frac{9}{2}\right)$
Coordinates of the midpoint of BD are
$\left(\frac{6+3}{2}, \frac{4+5}{2}\right)=\left(\frac{9}{2}, \frac{9}{2}\right)$
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, Using Distance Formula, we get
AB = √(x2 – x1)2 + (y2 – y1)2
= √[(6 – 4)2 + (4 - 3)2]
= √(2)2 + (1)2
= √(4 + 1)
= √5 units
BC = √[(5 – 6)2 + (6 - 4)2]
= √(-1)2 + (2)2
= √(1 + 4)
= √5 units
Therefore, AB = BC = √5 units
Now, check for the diagonals
AC = √(5 – 4)2 + (6 – 3)2
= √(1)2 + (3)2
= √1 + 9
= √10 units
and
BD = √(3 - 6)2 + (5 – 4)2
⇒ BD = √(-3)2 + (1)2
⇒ BD = √9 + 1
⇒ BD = √10 units
∴ AC = BD
Hence, ABCD is a square.
Question 29
If (6,8), (3,7) and (-2,-2) be the coordinates of the three consecutive vertices of a parallelogram, find coordinates of the fourth vertex.Sol :
Let the coordinates of the fourth vertex D be (x, y).
We know that diagonals of a parallelogram bisect each other.
∴ Midpoint of AC = Midpoint of BD …(i)
Coordinates of the midpoint of AC are
$\left(\frac{6-2}{2}, \frac{8-2}{2}\right)=\left(\frac{4}{2}, \frac{6}{2}\right)=(2,3)$
Coordinates of the midpoint of BD are
$\left(\frac{3+x}{2}, \frac{7+y}{2}\right)$
So, according to eq. (i), we have
$\Rightarrow(2,3)=\left(\frac{3+x}{2}, \frac{7+y}{2}\right)$
$\Rightarrow 2=\frac{3+x}{2}$ and $3=\frac{7+y}{2}$
⇒ 3 + x = 4 and 7 + y = 6
⇒ x = 1 and y = -1
Thus, the coordinates of the vertex D are (1, -1)
Question 30
Three consecutive vertices of a rhombus are (5,3), (2,7) and (-2,4). Find the fourth vertex.
Sol :Let the coordinates of the fourth vertex D be (x, y).
We know that diagonals of a rhombus bisect each other.
∴ Midpoint of AC = Midpoint of BD …(i)
Coordinates of the midpoint of AC are
$\left(\frac{5-2}{2}, \frac{3+4}{2}\right)=\left(\frac{3}{2}, \frac{7}{2}\right)$
Coordinates of the midpoint of BD are
$\left(\frac{2+x}{2}, \frac{7+y}{2}\right)$
So, according to eq. (i), we have
$\Rightarrow\left(\frac{3}{2}, \frac{7}{2}\right)=\left(\frac{2+x}{2}, \frac{7+y}{2}\right)$
$\Rightarrow \frac{3}{2}=\frac{2+x}{2}$ and $\frac{7}{2}=\frac{7+y}{2}$
⇒ 2 + x = 3 and 7 + y = 7
⇒ x = 1 and y = 0
Thus, the coordinates of the vertex D are (1, 0)
Question 31
A quadrilateral has the vertices at the point (-4,2), (2,6), (8,5) and (9,-7). Show that the mid-point of the sides of this quadrilateral are the vertices of a parallelogram.
Sol :Let the vertices of quadrilateral be P(-4,2), Q(2,6), R(8,5) and S(9,-7)
Let A, B, C and D are the midpoints of PQ, QR, RS and SP respectively.
Now, since A is the midpoint of P(-4, 2) and Q(2, 6)
∴ Coordinates of A are
$\left(\frac{-4+2}{2}, \frac{2+6}{2}\right)=\left(\frac{-2}{2}, \frac{8}{2}\right)=(-1,4)$
Coordinates of B are
$\left(\frac{2+8}{2}, \frac{6+5}{2}\right)=\left(\frac{10}{2}, \frac{11}{2}\right)=\left(5, \frac{11}{2}\right)$
Coordinates of C are
$\left(\frac{8+9}{2}, \frac{5-7}{2}\right)=\left(\frac{17}{2}, \frac{-2}{2}\right)=\left(\frac{17}{2},-1\right)$
and
Coordinates of D are
$\left(\frac{9-4}{2}, \frac{-7+2}{2}\right)=\left(\frac{5}{2}, \frac{-5}{2}\right)$
Now,
we find the distance between A and B
$\mathrm{d}(\mathrm{A}, \mathrm{B})=\sqrt{(-1-5)^{2}+\left(4-\frac{11}{2}\right)^{2}}$
$=\sqrt{36+\frac{9}{4}}=\sqrt{\frac{144+9}{4}}=\sqrt{\frac{153}{4}}$
$\mathrm{d}(\mathrm{C}, \mathrm{D})=\sqrt{\left(\frac{17}{2}-\frac{5}{2}\right)^{2}+\left(-1+\frac{5}{2}\right)^{2}}$
$=\sqrt{36+\frac{9}{4}}=\sqrt{\frac{144+9}{4}}=\sqrt{\frac{153}{4}}$
$\mathrm{d}(\mathrm{A}, \mathrm{D})=\sqrt{\left(-1-\frac{5}{2}\right)^{2}+\left(4+\frac{5}{2}\right)^{2}}$
$=\sqrt{\frac{49}{4}+\frac{169}{4}}=\sqrt{\frac{218}{4}}$
$\mathrm{d}(\mathrm{B}, \mathrm{C})=\sqrt{\left(5-\frac{17}{2}\right)^{2}+\left(\frac{11}{2}+1\right)^{2}}$
$=\sqrt{\frac{49}{4}+\frac{169}{4}}=\sqrt{\frac{218}{4}}$
Now, since length of opposite sides of the quadrilateral formed by the midpoints of the given quadrilateral are equal .i.e.
AB = CD and AD = BC
∴ it is a parallelogram
Hence Proved
Let the points be A(6,1), B(8,2), C(9,4) and D(p,3)
We know that diagonals of parallelogram bisect each other.
∴ Midpoint of AC = Midpoint of BD …(i)
Coordinates of the midpoint of AC are
$\left(\frac{6+9}{2}, \frac{1+4}{2}\right)=\left(\frac{15}{2}, \frac{5}{2}\right)$
Coordinates of the midpoint of BD are
$\left(\frac{8+p}{2}, \frac{2+3}{2}\right)=\left(\frac{8+p}{2}, \frac{5}{2}\right)$
So, according to eq. (i), we have
$\Rightarrow\left(\frac{15}{2}, \frac{5}{2}\right)=\left(\frac{8+p}{2}, \frac{5}{2}\right)$
$\Rightarrow \frac{15}{2}=\frac{8+\mathrm{p}}{2}$
⇒ 8 + p = 15
⇒ p = 15 – 8 = 7
Hence, the value of p is 7
We take O as the origin and OX and OY as the x and y axis respectively.
Let BC = 2a, then B = (-a, 0) and C = (a, 0)
Let A = (b, c), if E and F are the midpoints of sides AC and AB respectively.
Coordinates of midpoint of AC are
$\left(\frac{\mathrm{b}+\mathrm{a}}{2}, \frac{\mathrm{c}+0}{2}\right)=\left(\frac{\mathrm{a}+\mathrm{b}}{2}, \frac{\mathrm{c}}{2}\right)$
Coordinates of the midpoint of AB are
$\left(\frac{b-a}{2}, \frac{c-0}{2}\right)=\left(\frac{b-a}{2}, \frac{c}{2}\right)$
Now, distance between F and E is
d(F,E) = √(x2 – x1)2 + (y2 – y1)2
$=\sqrt{\left(\frac{\mathrm{a}+\mathrm{b}}{2}-\frac{\mathrm{b}-\mathrm{a}}{2}\right)^{2}+\left(\frac{\mathrm{c}}{2}-\frac{\mathrm{c}}{2}\right)^{2}}$
= a …(i)
and Length of BC = 2a …(ii)
From (i) and (ii), we can say that
$\mathrm{FE}=\frac{1}{2} \mathrm{BC}$
Hence Proved
Let P, Q, R be the midpoints of sides BC, CA and AB respectively
Construct a ΔPQR by joining these three midpoints of the sides.
This is called the medial triangle
Since, PQ, QR and PR are midsegments of BC, AB and AC respectively
So,
$\mathrm{PQ}=\frac{1}{2} \mathrm{BC} ; \mathrm{QR}=\frac{1}{2} \mathrm{AB}$ and $\mathrm{PR}=\frac{1}{2} \mathrm{AC}$
Since the corresponding sides are proportional
∴ ΔPQR ≅ ΔABC
Now, we have to prove that the centroid of the triangle ABC and PQR coincide.
For that we must show that the medians of ΔABC pass through the midpoints of three sides of the medial triangle ΔPQR.
Since PQ is a midsegment of ΔABC,
⇒ PQ || BC, so PQ || BR.
And since QR is a midsegment of AB,
⇒ AB || QR, so QR || PB.
By definition, a quadrilateral PQRB is a parallelogram.
The medians BQ and CP are in fact the diagonals of the parallelogram PQRB.
And we know that the diagonals of a parallelogram bisect each other, so PD = DR.
In other words, D is the midpoint of PR.
In the similar manner, we can show that F and E are midpoints of RQ and PQ respectively.
Hence, the centroid of the triangle ABC and PQR coincide.
$=\sqrt{36+\frac{9}{4}}=\sqrt{\frac{144+9}{4}}=\sqrt{\frac{153}{4}}$
$\mathrm{d}(\mathrm{A}, \mathrm{D})=\sqrt{\left(-1-\frac{5}{2}\right)^{2}+\left(4+\frac{5}{2}\right)^{2}}$
$=\sqrt{\frac{49}{4}+\frac{169}{4}}=\sqrt{\frac{218}{4}}$
$\mathrm{d}(\mathrm{B}, \mathrm{C})=\sqrt{\left(5-\frac{17}{2}\right)^{2}+\left(\frac{11}{2}+1\right)^{2}}$
$=\sqrt{\frac{49}{4}+\frac{169}{4}}=\sqrt{\frac{218}{4}}$
Now, since length of opposite sides of the quadrilateral formed by the midpoints of the given quadrilateral are equal .i.e.
AB = CD and AD = BC
∴ it is a parallelogram
Hence Proved
Question 32
If the points A(6,1), B(8,2), C(9,4) and D(p,3) are the vertices of a parallelogram taken in order, find the value of p.
Sol :Let the points be A(6,1), B(8,2), C(9,4) and D(p,3)
We know that diagonals of parallelogram bisect each other.
∴ Midpoint of AC = Midpoint of BD …(i)
Coordinates of the midpoint of AC are
$\left(\frac{6+9}{2}, \frac{1+4}{2}\right)=\left(\frac{15}{2}, \frac{5}{2}\right)$
Coordinates of the midpoint of BD are
$\left(\frac{8+p}{2}, \frac{2+3}{2}\right)=\left(\frac{8+p}{2}, \frac{5}{2}\right)$
So, according to eq. (i), we have
$\Rightarrow\left(\frac{15}{2}, \frac{5}{2}\right)=\left(\frac{8+p}{2}, \frac{5}{2}\right)$
$\Rightarrow \frac{15}{2}=\frac{8+\mathrm{p}}{2}$
⇒ 8 + p = 15
⇒ p = 15 – 8 = 7
Hence, the value of p is 7
Question 33
Prove that the line segment joining the middle points of two sides of a triangle is half the third side.
Sol :We take O as the origin and OX and OY as the x and y axis respectively.
Let BC = 2a, then B = (-a, 0) and C = (a, 0)
Let A = (b, c), if E and F are the midpoints of sides AC and AB respectively.
Coordinates of midpoint of AC are
$\left(\frac{\mathrm{b}+\mathrm{a}}{2}, \frac{\mathrm{c}+0}{2}\right)=\left(\frac{\mathrm{a}+\mathrm{b}}{2}, \frac{\mathrm{c}}{2}\right)$
Coordinates of the midpoint of AB are
$\left(\frac{b-a}{2}, \frac{c-0}{2}\right)=\left(\frac{b-a}{2}, \frac{c}{2}\right)$
Now, distance between F and E is
d(F,E) = √(x2 – x1)2 + (y2 – y1)2
$=\sqrt{\left(\frac{\mathrm{a}+\mathrm{b}}{2}-\frac{\mathrm{b}-\mathrm{a}}{2}\right)^{2}+\left(\frac{\mathrm{c}}{2}-\frac{\mathrm{c}}{2}\right)^{2}}$
$=\sqrt{\left(\frac{a+b-b+a}{2}\right)^{2}}$
$=\sqrt{\left(\frac{2 a}{2}\right)^{2}}$
= a …(i)
and Length of BC = 2a …(ii)
From (i) and (ii), we can say that
$\mathrm{FE}=\frac{1}{2} \mathrm{BC}$
Hence Proved
Question 34
If P,Q,R divide the side BC,CA and AB of ΔABC in the same ratio, prove that the centroid of the triangle ABC and PQR coincide.
Sol :Let P, Q, R be the midpoints of sides BC, CA and AB respectively
Construct a ΔPQR by joining these three midpoints of the sides.
This is called the medial triangle
Since, PQ, QR and PR are midsegments of BC, AB and AC respectively
So,
$\mathrm{PQ}=\frac{1}{2} \mathrm{BC} ; \mathrm{QR}=\frac{1}{2} \mathrm{AB}$ and $\mathrm{PR}=\frac{1}{2} \mathrm{AC}$
Since the corresponding sides are proportional
∴ ΔPQR ≅ ΔABC
Now, we have to prove that the centroid of the triangle ABC and PQR coincide.
For that we must show that the medians of ΔABC pass through the midpoints of three sides of the medial triangle ΔPQR.
Since PQ is a midsegment of ΔABC,
⇒ PQ || BC, so PQ || BR.
And since QR is a midsegment of AB,
⇒ AB || QR, so QR || PB.
By definition, a quadrilateral PQRB is a parallelogram.
The medians BQ and CP are in fact the diagonals of the parallelogram PQRB.
And we know that the diagonals of a parallelogram bisect each other, so PD = DR.
In other words, D is the midpoint of PR.
In the similar manner, we can show that F and E are midpoints of RQ and PQ respectively.
Hence, the centroid of the triangle ABC and PQR coincide.
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
Thankyou ❤️ so much
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