| Exercise
10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
Exercise 10.4
Question 1 A
Find the area of the triangle whose vertices are
(3, -4), (7, 5), (-1, 10)
Sol :(3, -4), (7, 5), (-1, 10)
Given: (3, -4), (7, 5), (-1, 10)
Let us Assume A(x1, y1) = (3, -4)
Let us Assume B(x2, y2) = (7, 5)
Let us Assume C(x3, y3) = (-1, 10)
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Now,
Area of given triangle $\left.=\frac{1}{2}[\{3(5-(10))\}+(7)\{(10+4)\}-1(-4-5)\}\right]$
$=\frac{1}{2}\{-15+98+9\}$
$=\frac{92}{2}$
= 46 square unitsQuestion 1 B
Find the area of the triangle whose vertices are
(-1.5, 3), (6, -2), (-3, 4)Sol :
Given (-1.5, 3), (6, -2), (-3, 4)
Let us Assume A(x1, y1) = (-1.5, 3)
Let us Assume B(x2, y2) = (6, -2)
Let us Assume C(x3, y3) = (-3, 4)
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Now,
Area of given triangle $\left.=\frac{1}{2}[\{-1.5(-2-(4))\}+(6)\{(4-3)\}-3(3+2)\}\right]$
$=\frac{1}{2}\{+9+6-15\}$
$=\frac{0}{2}$
= 0 square unitsQuestion 1 C
Find the area of the triangle whose vertices are
(-5, -1), (3, -5), (5, 2)Sol :
Given (-5, -1), (3, -5), (5, 2)
Let us Assume A(x1, y1) = (-5, -1)
Let us Assume B(x2, y2) = (3, -5)
Let us Assume C(x3, y3) = (5, 2)
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Now,
Area of given triangle $\left.=\frac{1}{2}[\{-5(-5-2)\}+3\{(2+1)\}+5(-1+5)\}\right]$
$=\frac{1}{2}\{35+9+20\}$
$=\frac{64}{2}$
= 32 square unitsQuestion 1 D
Find the area of the triangle whose vertices are
(5, 2), (4, 7), (7, -4)
Sol :(5, 2), (4, 7), (7, -4)
Given (5, 2), (4, 7), (7, -4)
Let us Assume A(x1, y1) = (5, 2)
Let us Assume B(x2, y2) = (4, 7)
Let us Assume C(x3, y3) = (7, -4)
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Now,
Area of given triangle $=\frac{1}{2}[\{5(7-(-4))\}+4\{(-4-2)+7(2-7)\}]$
$=\frac{1}{2}\{55-24-35\}$
$=\frac{4}{2}$
= 2 square unitsQuestion 1 E
Find the area of the triangle whose vertices are
(2, 3), (-1, 0), (2, -4)
Sol :(2, 3), (-1, 0), (2, -4)
Given (2, 3), (-1, 0), (2, -4)
Let us Assume A(x1, y1) = (2, 3)
Let us Assume B(x2, y2) = (-1, 0)
Let us Assume C(x3, y3) = (2, -4)
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Now,
Area of given triangle $=\frac{1}{2}[\{2(0-(-4))\}+(-1)\{(-4-3)+2(3-0)\}]$
$=\frac{1}{2}\{8+7+6\}$
$=\frac{21}{2}$Square unitsQuestion 1 F
Find the area of the triangle whose vertices are
(1, -1), (-4, 6), (-3, -5)
Sol :(1, -1), (-4, 6), (-3, -5)
Given (1, -1), (-4, 6), (-3, -5)
Let us Assume A(x1, y1) = (1, -1)
Let us Assume B(x2, y2) = (-4, 6)
Let us Assume C(x3, y3) = (-3, -5)
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Now,
Area of given triangle $=\frac{1}{2}[\{1(6-(-5))\}+(-4)\{(-5+1)-3(-1-6)\}]$
$=\frac{1}{2}\{30+16+21\}$
$=\frac{67}{2}$ Square units
Question 1 G
Find the area of the triangle whose vertices are
$\left(a t_{1}^{2}, 2 a t_{1}\right),\left(a t_{2}^{2}, 2 a t_{2}\right),\left(a t_{3}^{2}, 2 a t_{3}\right)$
Sol :$\left(a t_{1}^{2}, 2 a t_{1}\right),\left(a t_{2}^{2}, 2 a t_{2}\right),\left(a t_{3}^{2}, 2 a t_{3}\right)$
Given $\left(a t_{1}^{2}, 2 a t_{1}\right),\left(a t_{2}^{2}, 2 a t_{2}\right),\left(a t_{3}^{2}, 2 a t_{3}\right)$
Let us Assume A(x1, y1) $=\left(a t_{1}^{2}, 2 a t_{1}\right)$
Let us Assume B(x2, y2) $=\left(a t_{2}^{2}, 2 a t_{2}\right)$
Let us Assume C(x3, y3) $=\left(a t_{3}^{2}, 2 a t_{3}\right)$
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Now,
Area of given triangle $=\frac{1}{2}\left[\left\{\mathrm{at}_{1}^{2}\left(2 \mathrm{at}_{2}-2 \mathrm{at}_{3}\right)\right\}+\mathrm{at}_{2}^{2}\left\{\left(2 \mathrm{at}_{3}-2 \mathrm{at}_{1}\right)-\mathrm{at}_{3}^{2}\left(2 \mathrm{at}_{1}-2 \mathrm{at}_{2}\right)\right\}\right]$
$=\frac{1}{2}\{30+16+21\}$
$=\frac{67}{2}$ Square units
Question 1 H
Find the area of the triangle whose vertices are
(-5, 7), (-4, -5), (4, 5)
Sol :(-5, 7), (-4, -5), (4, 5)
Given (-5, 7), (-4, -5), (4, 5)
Let us Assume A(x1, y1) = (-5, 7)
Let us Assume B(x2, y2) = (-4, -5)
Let us Assume C(x3, y3) = (4, 5)
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Now,
Area of given triangle $=\frac{1}{2}[\{-5(-5-(5))\}+(-4)\{(5-7)+4(7+5)\}]$
$=\frac{1}{2}\{50+8+48\}$
$=\frac{106}{2}$
= 53 square unitsQuestion 2 A
Find the area of the quadrilateral whose vertices are
(1, 1), (7, -3), (12, 2) and (7, 21)
Sol :(1, 1), (7, -3), (12, 2) and (7, 21)
Given (1, 1), (7, -3), (12, 2) and (7, 21)
Let us Assume A(x1, y1) = (1, 1)
Let us Assume B(x2, y2) = (7, -3)
Let us Assume C(x3, y3) = (12, 2)
Let us Assume D(x4, y4) = (7, 21)
Let us join Ac to from two triangles ∆ABC and ∆ACD
Now
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
Then,
Area of given triangle ABC $=\frac{1}{2}[1(-3-2)+7(2-1)+12(1+3)]$
$=\frac{1}{2}[-5+7+48]$
$=\frac{50}{2}$
= 25 square unitsArea of given triangle ABC $=\frac{1}{2}[1(2-21)+12(21-1)+7(1-2)]$
$=\frac{1}{2}[-19+240-7]$
$=\frac{214}{2}$
= 107 square units
Area of quadrilateral ABCD = Area of ABC + Area of ACD
= 25 + 107
= 132 sq units.
Question 2 B
Find the area of the quadrilateral whose vertices are
(-4, 5), (0, 7), (5, -5), and (-4, -2)
Sol :(-4, 5), (0, 7), (5, -5), and (-4, -2)
Given (-4, 5), (0, 7), (5, -5), and (-4, -2)
To Find: Find the area of quadrilateral.
Let us Assume A(x1, y1) = (-4, 5)
Let us Assume B(x2, y2) = (0, 7)
Let us Assume C(x3, y3) = (5, -5)
Let us Assume D(x4, y4) = (4, -2)
Let us join Ac to from two triangles ∆ABC and ∆ACD
Now
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of given triangle ABC $=\frac{1}{2}|-4(7+5)+0(-5-5)+5(5-7)|$
$=\frac{1}{2}[-48+0-10]$
$=\frac{58}{2}$
Area of given triangle ABC $=\frac{1}{2}|-4(-5+2)+5(-2-5)-4(5+5)|$
$=\frac{1}{2}|-4(-3)+5(-7)-4(10)|$
$=\frac{1}{2}|12-35-40|$
$=\frac{1}{2}|-63|$
$=\frac{63}{2}$ square unitsArea of quadrilateral ABCD = Area of ABC + Area of ACD
$=\left|\frac{58}{2}+\frac{63}{2}\right|$
Hence, Area of Quadrilateral ABCD $=\frac{121}{2}$ sq units.
Question 2 C
Find the area of the quadrilateral whose vertices are
Given (-5, 7), (-4, -5), (-1, -6) and (4, 5)
Sol :Given (-5, 7), (-4, -5), (-1, -6) and (4, 5)
To Find: Find the area of quadrilateral.
Let us Assume A(x1, y1) = (-5, 7)
Let us Assume B(x2, y2) = (-4, -5)
Let us Assume C(x3, y3) = (-1, -6)
Let us Assume D(x4, y4) = (4, 5)
Let us join Ac to from two triangles ∆ABC and ∆ACD
Now
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of given triangle ABC $=\frac{1}{2}|-5(-5+6)-4(-6-7)-1(7+5)|$
$=\frac{1}{2}|-5(1)-4(-13)-1(12)|$
$=\frac{1}{2}|-5+52-12|$
$=\frac{35}{2}$
Area of given triangle ABC $=\frac{1}{2}|-5(-6-5)-1(5-7)+4(7+6)|$
Area of quadrilateral ABCD = Area of ABC + Area of ACD
$=\left|\frac{35}{2}+56\right|$
Hence, Area $=\frac{147}{2}$ sq units.
To Find: Find the area of quadrilateral.
Let us Assume A(x1, y1) = (0, 0)
Let us Assume B(x2, y2) = (6, 0)
Let us Assume C(x3, y3) = (4, 3)
Let us Assume D(x4, y4) = (0, 3)
Let us join Ac to from two triangles ∆ABC and ∆ACD
Now
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of given triangle ABC $=\frac{1}{2}|0(0-3)+6(3-0)+4(0-0)|$
Area of given triangle ABC $=\frac{1}{2}|0(3-3)+4(3-0)+0(0-3)|$
Area of quadrilateral ABCD = Area of ABC + Area of ACD
= 9 + 6
Hence, Area = 15 sq units.
To Find: Find the area of the quadrilateral.
Let us Assume A(x1, y1) = (1, 0)
Let us Assume B(x2, y2) = (5, 3)
Let us Assume C(x3, y3) = (2, 7)
Let us Assume D(x4, y4) = (-2, 4)
Let us join Ac to from two triangles ∆ABC and ∆ACD
Now
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of given triangle ABC $=\frac{1}{2}|1(3-7)+5(7-0)+2(0-3)|$
Area of given triangle ABC $=\frac{1}{2} \mid 1(7-4)+2(4-0)-2(0-7)$
Area of quadrilateral ABCD = Area of ABC + Area of ACD
$=\frac{25}{2}+\frac{25}{2}$
Hence, Area = 25 sq units.
Given: The vertices of the quadrilateral be A(-4, -2), B(-3, -5), C(3, -2) and D(2, 3).
Let join AC to form two triangles,
Now, We know that
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of triangle ABC $\left.=\frac{1}{2} \mid(-4)\{(-5+2)\}+(-3)\{((-2)+2)\}+3\{((-2)+2)\}\right] \mid$
$=\frac{1}{2}[12+0+9]$
$=\frac{21}{2}$ Square Units
Now, Area of triangle ACD$=\frac{1}{2}|[(-4)\{(-2+3)\}+(3)\{((3)+2)\}+2\{((-2)+2)\}]|$
$=\frac{1}{2}[20+15+0]$
$=\frac{35}{2}$ Square Units
Area of quadrilateral ABCD = Area of triangle ABC+ Area of triangle ACD
Hence, Area of quadrilateral ABCD = 28 square Units
Given a triangle whose vertices A(1, 2), B(2, 5), C(3, 1)
Let AD is the median on side BC
D will be the mid-point of segment BC. Therefore,
Coordinate of D
$=\frac{1}{2}|-5(-11)-1(-2)+4(13)|$
$=\frac{1}{2}|55+2+52|$
$=\frac{112}{2}$
= 56 square unitsArea of quadrilateral ABCD = Area of ABC + Area of ACD
$=\left|\frac{35}{2}+56\right|$
Hence, Area $=\frac{147}{2}$ sq units.
Question 2 D
Find the area of the quadrilateral whose vertices are
Given (0, 0), (6, 0), (4, 3), and (0, 3)
Sol :Given (0, 0), (6, 0), (4, 3), and (0, 3)
To Find: Find the area of quadrilateral.
Let us Assume A(x1, y1) = (0, 0)
Let us Assume B(x2, y2) = (6, 0)
Let us Assume C(x3, y3) = (4, 3)
Let us Assume D(x4, y4) = (0, 3)
Let us join Ac to from two triangles ∆ABC and ∆ACD
Now
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of given triangle ABC $=\frac{1}{2}|0(0-3)+6(3-0)+4(0-0)|$
$=\frac{1}{2}|0+18+0|$
$=\frac{1}{2}|18|$
= 9 Square unitsArea of given triangle ABC $=\frac{1}{2}|0(3-3)+4(3-0)+0(0-3)|$
$=\frac{1}{2}|0+12+0|$
$=\frac{1}{2}|12|$
= 6 square unitsArea of quadrilateral ABCD = Area of ABC + Area of ACD
= 9 + 6
Hence, Area = 15 sq units.
Question 2 E
Find the area of the quadrilateral whose vertices are
Given (1, 0), (5, 3), (2, 7) and (-2, 4)
Sol :Given (1, 0), (5, 3), (2, 7) and (-2, 4)
To Find: Find the area of the quadrilateral.
Let us Assume A(x1, y1) = (1, 0)
Let us Assume B(x2, y2) = (5, 3)
Let us Assume C(x3, y3) = (2, 7)
Let us Assume D(x4, y4) = (-2, 4)
Let us join Ac to from two triangles ∆ABC and ∆ACD
Now
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of given triangle ABC $=\frac{1}{2}|1(3-7)+5(7-0)+2(0-3)|$
$=\frac{1}{2}|-4+35-6|$
$=\frac{1}{2}|25|$
$=\frac{25}{2}$ Square unitsArea of given triangle ABC $=\frac{1}{2} \mid 1(7-4)+2(4-0)-2(0-7)$
$=\frac{1}{2}|3+8+14|$
$=\frac{1}{2}|25|$
$=\frac{25}{2}$ square unitsArea of quadrilateral ABCD = Area of ABC + Area of ACD
$=\frac{25}{2}+\frac{25}{2}$
Hence, Area = 25 sq units.
Question 3
Find the area of the quadrilateral whose vertices taken in order are (- 4, -2), (-3, -5), (3, -2) and (2,
3).
Sol :Given: The vertices of the quadrilateral be A(-4, -2), B(-3, -5), C(3, -2) and D(2, 3).
Let join AC to form two triangles,
Now, We know that
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of triangle ABC $\left.=\frac{1}{2} \mid(-4)\{(-5+2)\}+(-3)\{((-2)+2)\}+3\{((-2)+2)\}\right] \mid$
$=\frac{1}{2}[12+0+9]$
$=\frac{21}{2}$ Square Units
Now, Area of triangle ACD$=\frac{1}{2}|[(-4)\{(-2+3)\}+(3)\{((3)+2)\}+2\{((-2)+2)\}]|$
$=\frac{1}{2}[20+15+0]$
$=\frac{35}{2}$ Square Units
Area of quadrilateral ABCD = Area of triangle ABC+ Area of triangle ACD
$=\frac{21}{2}+\frac{35}{2}$
$=\frac{56}{2}$
Hence, Area of quadrilateral ABCD = 28 square Units
Question 4
A median of a triangle divides it into two triangles of equal area. Verify this result for 
ABC whose vertices are A(1, 2), B(2, 5), C(3, 1).
Sol :ABC whose vertices are A(1, 2), B(2, 5), C(3, 1).
Given a triangle whose vertices A(1, 2), B(2, 5), C(3, 1)
Let AD is the median on side BC
D will be the mid-point of segment BC. Therefore,
Coordinate of D
$=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2},
\frac{\mathrm{y}_{1} \mathrm{y}_{2}}{2}\right]$
Area of triangle
$=\left[\frac{2+3}{2}, \frac{5+1}{2}\right]$
$=\left[\frac{5}{2}, \frac{6}{2}\right]$
$=\left[\frac{5}{2}, 3\right]$
$=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
Then,
Area of triangle ABD $=\frac{1}{2}\left|\left[1(5-3)+2(3-2)+\frac{5}{2}(2-5)\right]\right|$
Area of triangle ACD $=\frac{1}{2}\left|\left[1(1-3)+3(3-2)+\frac{5}{2}(2-1)\right]\right|$
Hence, ∆ABD = ∆ACD
Given: ABC is a triangle with points (-1, 5), (3, 1), (5, 7)
To Find ∆ABC=4∆DEF
We know that
Area of triangle $=\frac{1}{2} \mid \mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)$
Then,
Area of triangle ABC $=\frac{1}{2}|(-1)\{(1-7)\}+3(7-5)+5(5-1)|$
$=\frac{1}{2}[6+6+20]$
$=\frac{32}{2}$
= 16
Now we have to find point D, E, and F.
Hence D is the midpoint of side BC then,
Coordinates of D $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{3+5}{2}, \frac{1+7}{2}\right]$
= (4, 4 )
Hence E is the midpoint of side AC then,
Coordinates of E $=\left[\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right]$
$=\left[\frac{-1+5}{2}, \frac{5+7}{2}\right]$
= (2, 6)
Hence F is the midpoint of side AB then,
Coordinates of F $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{-1+3}{2}, \frac{5+1}{2}\right]$
= (1, 3)
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+1\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Now Area of triangle DEF $=\frac{1}{2}[4(6-3)+2(3-4)+1(4-6)]$
Therefore Area of ∆ABC= 4 Area of ∆DEF.
Hence Proved.
Given: ABC is a triangle with points (1, 2), (-3, 6), (5, 4)
To prove: The area of triangle ABC is four times the area of triangle DEF
We know that
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
Then,
Area of triangle ABC $=\frac{1}{2}|(1)\{(6-4)\}-3(4-2)+5(2-6)|$
Now we have to find point D, E, F
Hence D is the midpoint of side BC then,
Coordinates of D $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{-3+5}{2}, \frac{6+4}{2}\right]$
= (1, 5 )
Hence E is the midpoint of side AC then,
Coordinates of E $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{1+5}{2}, \frac{2+4}{2}\right]$
= (3, 3)
Hence F is the midpoint of side AB then,
Coordinates of F $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{1-3}{2}, \frac{2+6}{2}\right]$
= (-1, 4)
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
Now Area of triangle DEF $=\frac{1}{2}|1(3-4)+3(4-5)-1(5-3)|$
Therefore Area of ∆ABC = 4 Area of ∆DEF.
Hence, Proved.
Let ABC is a triangle with points (0, -1), (2, 1), (0, 3)
To Find: Ratio of area of triangle ABC to triangle DEF
We know that
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
Then,
Area of triangle ABC $=\frac{1}{2}|0(1-3)+2(3+1)+0(-1-1)|$
$=\frac{1}{2}[8]$
= 4
Now we have to find point D, E, and F.
Hence D is the midpoint of side BC then,
Coordinates of D $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{2+0}{2}, \frac{1+3}{2}\right]$
= (1, 2 )
Hence E is the midpoint of side AC then,
Coordinates of E $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{0+0}{2}, \frac{-1+3}{2}\right]$
= (0, 1)
Hence F is the midpoint of side AB then,
Coordinates of F $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{0+2}{2}, \frac{-1+1}{2}\right]$
= (1, 0)
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
Now Area of triangle DEF $=\frac{1}{2}|1(1-0)+0(0-2)+1(2-1)|$
Therefore Area of ∆ABC= 4 Area of ∆DEF.
Then, The ratio of ∆DEF and ∆ABC = 1:4
Given: Coordinates of middle points are D(-1, -2), E(6, 1) and F(3, 5).
To find: Area of triangle ABC
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
Now Area of triangle DEF
Then,
Area of triangle ABD $=\frac{1}{2}\left|\left[1(5-3)+2(3-2)+\frac{5}{2}(2-5)\right]\right|$
$=\frac{1}{2}\left[2+2-\frac{15}{2}\right]$
$=\frac{1}{2}\left|-\frac{7}{2}\right|$
$=\frac{7}{4}$ sq unitsArea of triangle ACD $=\frac{1}{2}\left|\left[1(1-3)+3(3-2)+\frac{5}{2}(2-1)\right]\right|$
$=\frac{1}{2} \|\left[-2+3+\frac{5}{2}\right] \mid$
$=\frac{1}{2}\left|\frac{7}{2}\right|$
$=\frac{7}{4}$sq unitsHence, ∆ABD = ∆ACD
Question 5
If A, B, C are the points (-1, 5), (3, 1), (5, 7) respectively and D, E, F are the middle points of BC, CA
and AB respectively, prove that ΔABC = 4ΔDEF.
Sol :Given: ABC is a triangle with points (-1, 5), (3, 1), (5, 7)
To Find ∆ABC=4∆DEF
We know that
Area of triangle $=\frac{1}{2} \mid \mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)$
Then,
Area of triangle ABC $=\frac{1}{2}|(-1)\{(1-7)\}+3(7-5)+5(5-1)|$
$=\frac{1}{2}[6+6+20]$
$=\frac{32}{2}$
= 16
Now we have to find point D, E, and F.
Hence D is the midpoint of side BC then,
Coordinates of D $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{3+5}{2}, \frac{1+7}{2}\right]$
= (4, 4 )
Hence E is the midpoint of side AC then,
Coordinates of E $=\left[\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right]$
$=\left[\frac{-1+5}{2}, \frac{5+7}{2}\right]$
= (2, 6)
Hence F is the midpoint of side AB then,
Coordinates of F $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{-1+3}{2}, \frac{5+1}{2}\right]$
= (1, 3)
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+1\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Now Area of triangle DEF $=\frac{1}{2}[4(6-3)+2(3-4)+1(4-6)]$
$=\frac{1}{2}[12-2-2]$
$=\frac{1}{2}[8]$
= 4Therefore Area of ∆ABC= 4 Area of ∆DEF.
Hence Proved.
Question 6 A
Three vertices of a triangle are A(1, 2), B(-3, 6) and C(5, 4). If D, E, and C, respectively, show that the
area of triangle ABC is four times the area of triangle DEF.
Sol :Given: ABC is a triangle with points (1, 2), (-3, 6), (5, 4)
To prove: The area of triangle ABC is four times the area of triangle DEF
We know that
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
Then,
Area of triangle ABC $=\frac{1}{2}|(1)\{(6-4)\}-3(4-2)+5(2-6)|$
$=\frac{1}{2}|2-6-20|$
$=\frac{-24}{2}$
= 12Now we have to find point D, E, F
Hence D is the midpoint of side BC then,
Coordinates of D $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{-3+5}{2}, \frac{6+4}{2}\right]$
= (1, 5 )
Hence E is the midpoint of side AC then,
Coordinates of E $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{1+5}{2}, \frac{2+4}{2}\right]$
= (3, 3)
Hence F is the midpoint of side AB then,
Coordinates of F $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{1-3}{2}, \frac{2+6}{2}\right]$
= (-1, 4)
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
Now Area of triangle DEF $=\frac{1}{2}|1(3-4)+3(4-5)-1(5-3)|$
$=\frac{1}{2}[-1-3-2]$
$=\frac{1}{2}|-6|$
= 3Therefore Area of ∆ABC = 4 Area of ∆DEF.
Hence, Proved.
Question 6 B
Find the area of the triangle formed by joining the mid-points of the sides of the triangles whose vertices
are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Sol :Let ABC is a triangle with points (0, -1), (2, 1), (0, 3)
To Find: Ratio of area of triangle ABC to triangle DEF
We know that
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
Then,
Area of triangle ABC $=\frac{1}{2}|0(1-3)+2(3+1)+0(-1-1)|$
$=\frac{1}{2}[8]$
= 4
Now we have to find point D, E, and F.
Hence D is the midpoint of side BC then,
Coordinates of D $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{2+0}{2}, \frac{1+3}{2}\right]$
= (1, 2 )
Hence E is the midpoint of side AC then,
Coordinates of E $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{0+0}{2}, \frac{-1+3}{2}\right]$
= (0, 1)
Hence F is the midpoint of side AB then,
Coordinates of F $=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right]$
$=\left[\frac{0+2}{2}, \frac{-1+1}{2}\right]$
= (1, 0)
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
Now Area of triangle DEF $=\frac{1}{2}|1(1-0)+0(0-2)+1(2-1)|$
$=\frac{1}{2}[1+1]$
$=\frac{1}{2}[2]$
= 1Therefore Area of ∆ABC= 4 Area of ∆DEF.
Then, The ratio of ∆DEF and ∆ABC = 1:4
Question 7
Find the area of a triangle ABC if the coordinates of the middle points of the sides of the triangle are
(-1, - 2), (6, 1) and (3, 5).
Sol :Given: Coordinates of middle points are D(-1, -2), E(6, 1) and F(3, 5).
To find: Area of triangle ABC
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
Now Area of triangle DEF
$=\frac{1}{2}|-1(1-5)+6(5+2)+3(-2-1)|$
Hence the area of ABC is $=\frac{37}{2}$ square units
Given, ABC is a triangle with vertices A(3, 0), B(0, 6) and C (6, 9)
To find: $\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{ADC}}=9$
We know that
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
Now Area of triangle DEF $=\frac{1}{2}|3(6-9)+0(9-0)+6(0-6)|$
Now, According to the question,
DE internally divides AB in the ratio 1:2 hence
Coordinates of D $=\left[\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m} 1+\mathrm{m} 2}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m} 1+\mathrm{m} 2}\right]$
E internally divides AC in the ratio 1:2 hence
Coordinates of D $=\left[\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m} 1+\mathrm{m} 2}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m} 1+\mathrm{m} 2}\right]$
$=\left[\frac{1 \times 6+2 \times 3}{2+1}, \frac{1 \times 9+2 \times 0}{1+2}\right]$
$=\left[\frac{12}{3}, \frac{9}{3}\right]$
= (4, 3)
Now Area of triangle ADE $=\frac{1}{2}|3(2-3)+2(3-0)+4(0-2)|$
Therefore, Area of ∆ABC $=\frac{45}{2}$ sq. units
Hence, Area of ABC = 9. Area of ADE
Given a triangle with vertices (t, t-2), (t+3, t) and (t+2, t+2)
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
= 2 sq units
Hence, t is not dependant variable in the triangle.
Given: A triangle with vertices A(x, y), B(1, 2) and C (2, 1)
To find: x + y = 15 or -9
The area is 6 square units.
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
$\frac{1}{2}|\mathrm{x}(2-1)+1(1-\mathrm{y})+2(\mathrm{y}-2)|=6$
[x + 1 – y + 2y - 4] = 12
[x + y - 3] = 12
x + y =15
Hence, Proved
Given: A(a, b + c), B(b, c + a) and C(c, a + b)
To prove : Given points are collinear
We know the points are collinear if area(∆ABC)=0
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|=0$
Then,
Area $=\frac{1}{2}[\mathrm{a}\{(\mathrm{c}+\mathrm{a})-(\mathrm{a}+\mathrm{b})\}+\mathrm{b}\{(\mathrm{a}+\mathrm{b})-(\mathrm{b}+\mathrm{c})\}+\mathrm{c}\{(\mathrm{b}+\mathrm{c})-(\mathrm{c}+\mathrm{a})\}]$
= 0
Hence, Points are collinear.
Sol :
Given : A(x1, y1), B(x2, y2) and C(x3, y3)
We know the points are collinear if area(∆ABC)=0
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|=0$
Then, Area $=\frac{1}{2}\left\{\mathrm{x}_{1} \mathrm{y}_{2}-\mathrm{x}_{1} \mathrm{y}_{3}+\mathrm{x}_{2} \mathrm{y}_{3}-\mathrm{x}_{2} \mathrm{y}_{1}+\mathrm{x}_{3} \mathrm{y}_{1}-\mathrm{x}_{3} \mathrm{y}_{2}\right\}=0$
Now, Divide by $\mathrm{x}_{1} \mathrm{x}_{2} \mathrm{x}_{3}$
$\Rightarrow \frac{1}{2}\left\{\frac{\left(x_{1} y_{2}-x_{1} y_{3}\right)}{x_{1} x_{2} x_{3}}+\frac{\left(x_{2} y_{3}-x_{2} y_{1}\right)}{x_{1} x_{2} x_{3}}+\frac{\left(x_{3} y_{1}-x_{3} y_{2}\right)}{x_{1} x_{2} x_{3}}\right\}=0$
Taking common $\mathrm{x}_{1}, \mathrm{x}_{2}$ and $\mathrm{x}_{3}$ respectively
Hence, Proved.
We know the points are collinear if area(∆ABC)=0
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|=0$
Then, Area $=\frac{1}{2}\left\{\mathrm{a}\left(\mathrm{b}_{1}-\left(\mathrm{b}-\mathrm{b}_{1}\right)+\mathrm{a}_{1}\left\{\left(\mathrm{b}-\mathrm{b}_{1}\right)-\mathrm{b}\right\}+\left(\mathrm{a}-\mathrm{a}_{1}\right)\left(\mathrm{b}-\mathrm{b}_{1}\right\}=0\right.\right.$
$\frac{1}{2}\left\{a b_{1}-\left(a b-a b_{1}\right)+a_{1} b-a_{1} b_{1}-a_{1} b\right\}+\left(a b-a b_{1}-a_{1} b+a_{1} b_{1}\right\}=0$
{ab + a1 b1} = ab – ab1 – a1 b + a1 b1
- ab1 - a1b = 0
-ab1 = a1b
Therofore, We can write as
$\frac{a}{a_{1}}=\frac{b}{b_{1}}$
Hence, Proved.
Given: (a, 0), (0, b) and (1, 1)
We know the points are collinear if area(∆ABC)=0
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|=0$
Then, Area $=\frac{1}{2}\{a(b-1)+0\{1-0\}+1(0-b)\}=0$
$\frac{1}{2}[\mathrm{ab}-\mathrm{a}+0+0-\mathrm{b}]=0$
ab – a – b = 0
ab = a + b
Since, $\frac{1}{2}+\frac{1}{b}=1$
$\frac{a+b}{a b}=1$
Then, a + b = ab
$\frac{1}{a}+\frac{1}{b}=1$
Hence, Proved.
Given (2x, 2x), (3, 2x+1) and (1, 0)
We know the points are collinear if the area(∆ABC)=0
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|=0$
Then, Area $=\frac{1}{2}|2 x(2 x+1-0)+3(0-2 x)+1(2 x-2 x-1)|=0$
$\frac{1}{2}\left[4 \mathrm{x}^{2}+2 \mathrm{x}-6 \mathrm{x}-1\right]=0$
4x2 - 4x – 1 = 0
4x2 - 2x - 2x – 1 = 0
2x(2x - 1) - 1(2x - 1) = 0
(2x-1)(2x-1)=0
Hence, $x=\frac{1}{2}$
Given A(2, 3), B(4, k) and C(6, -3) are collinear
To find: Find the value of K
So, The given points are collinear, if are (∆ABC)=0
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}=0$
Then, $\frac{1}{2}[\{2(\mathrm{k}+3)\}+\{4(-3-3)\}+\{6(3-\mathrm{k})\}]=0$
= 2k+6 - 24 +18 - 6k = 0
= -4k = 0
Hence, K = 0
Given A(7, -2), B(5, 1) and C(3, k) are collinear
To find: Find the value of K
So, The given points are collinear, if are (∆ABC)=0
= Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}=0$
Then, $\frac{1}{2}[\{7(1-\mathrm{k})\}+\{5(\mathrm{k}+2)\}+\{3(-2-1)\}]=0$
= 7 – 7k +5k+10-9 = 0
= -2k + 8 = 0
Hence, K = 4
Given A(8, 1), B(k, -4) and C(2, -5) are collinear
To Find: Find the value of k
So, The given points are collinear, if are (∆ABC)=0
= Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}=0$
Then, $\frac{1}{2}[\{8(-4+5)\}+\{\mathrm{k}(-5-1)\}+\{2(-1+4)\}]=0$
= 8(1)+k(-6)+2(3) = 0
= 8 - 6k + 6 = 0
= -6k = -14
Hence, K =
Given A(2, 1), B(p, -1) and C(-1, 3) are collinear
To find: Find the value of p
So, The given points are collinear, if are (∆ABC)=0
= Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}=0$
Then, $\frac{1}{2}[\{2(-1-3)\}+\{p(3-1)\}+\{-1(1+1)\}]=0$
= 2(-4)+p(2)-2=0
= -8 +2p -2 = 0
= 2p = 10
Hence, p =
Given, A(0, -1) B (15, 2) divides the line on points C(-1, 2) and D(4, -5)
To Prove. Straight line divides in the ratio 2:3 internally
The equation of line $=\left(\mathrm{y}-\mathrm{y}_{1}\right)=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\left(\mathrm{x}-\mathrm{x}_{1}\right)$
Now, Equation of line BC $=(\mathrm{y}+1)=\frac{2+1}{15+0}(\mathrm{x}-0)$
Therefore, x – 5y = 5 ---(1)
Now, Equation of line BC $=(y-2)=\frac{-5-2}{4+1}(x+1)$
$=\frac{1}{2}[4+42-9]$
$=\frac{1}{2}[37]$
$=\frac{37}{2}$ square unitsHence the area of ABC is $=\frac{37}{2}$ square units
Question 8
The vertices of ΔABC are A(3, 0), B(0, 6) and (6, 9). A straight line DE divides AB and AC in the ratio 1:2
at D and E respectively, prove that $\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{ADE}}=9$
Sol :Given, ABC is a triangle with vertices A(3, 0), B(0, 6) and C (6, 9)
To find: $\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{ADC}}=9$
We know that
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
Now Area of triangle DEF $=\frac{1}{2}|3(6-9)+0(9-0)+6(0-6)|$
$=\frac{1}{2}[-9-36]$
$=\frac{1}{2}[-45]$
$=\frac{45}{2}$ square unitsNow, According to the question,
DE internally divides AB in the ratio 1:2 hence
Coordinates of D $=\left[\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m} 1+\mathrm{m} 2}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m} 1+\mathrm{m} 2}\right]$
$=\left[\frac{1 \times 0+2 \times 3}{2}, \frac{1 \times 6+2 \times 0}{1+2}\right]$
$=\left[\frac{6}{3}, \frac{6}{3}\right]$
= (2, 2)E internally divides AC in the ratio 1:2 hence
Coordinates of D $=\left[\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m} 1+\mathrm{m} 2}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m} 1+\mathrm{m} 2}\right]$
$=\left[\frac{1 \times 6+2 \times 3}{2+1}, \frac{1 \times 9+2 \times 0}{1+2}\right]$
$=\left[\frac{12}{3}, \frac{9}{3}\right]$
= (4, 3)
Now Area of triangle ADE $=\frac{1}{2}|3(2-3)+2(3-0)+4(0-2)|$
$=\frac{1}{2}[-3+6-8]$
$=\frac{1}{2}[-5]$
$=\frac{5}{2}$ square unitsTherefore, Area of ∆ABC $=\frac{45}{2}$ sq. units
Hence, Area of ABC = 9. Area of ADE
Question 9
If (t, t - 2), (t + 3, t) and (t + 2, t + 2) are the vertices of a triangle, show that its area is
independent of t.
Sol :Given a triangle with vertices (t, t-2), (t+3, t) and (t+2, t+2)
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
$=\frac{1}{2}|t(t-(t+2))+(t+3)((t+2)-(t-2))+(t+2)((t-2)-t)|$
$=\frac{1}{2}[t(2)+(t+2)\{0\}+(t+2)\{-2\}]$
$=\frac{1}{2}[2 t-2 t-4]$
$=\frac{4}{2}$
= 2 sq units
Hence, t is not dependant variable in the triangle.
Question 10
If A(x, y), B(1, 2) and C(2, 1) are the vertices of a triangle of area 6 square unit, show that x+y=15
or-9
Sol :Given: A triangle with vertices A(x, y), B(1, 2) and C (2, 1)
To find: x + y = 15 or -9
The area is 6 square units.
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
$\frac{1}{2}|\mathrm{x}(2-1)+1(1-\mathrm{y})+2(\mathrm{y}-2)|=6$
[x + 1 – y + 2y - 4] = 12
[x + y - 3] = 12
x + y =15
Hence, Proved
Question 11
Prove that the points (a, b+c), (b, c+a) and (c, a+b) are collinear.
Sol :Given: A(a, b + c), B(b, c + a) and C(c, a + b)
To prove : Given points are collinear
We know the points are collinear if area(∆ABC)=0
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|=0$
Then,
Area $=\frac{1}{2}[\mathrm{a}\{(\mathrm{c}+\mathrm{a})-(\mathrm{a}+\mathrm{b})\}+\mathrm{b}\{(\mathrm{a}+\mathrm{b})-(\mathrm{b}+\mathrm{c})\}+\mathrm{c}\{(\mathrm{b}+\mathrm{c})-(\mathrm{c}+\mathrm{a})\}]$
$=\frac{1}{2}[\mathrm{a}\{\mathrm{c}+\mathrm{a}-\mathrm{a}-\mathrm{b}\}+\mathrm{b}\{\mathrm{a}+\mathrm{b}-\mathrm{b}-\mathrm{c}\}+\mathrm{c}\{\mathrm{b}+\mathrm{c}-\mathrm{c}-\mathrm{a}\}$
$=\frac{1}{2}[\mathrm{a}\{\mathrm{c}-\mathrm{b}\}+\mathrm{b}\{\mathrm{a}-\mathrm{c}\}+\mathrm{c}\{\mathrm{b}-\mathrm{a}\}]$
= 0
Hence, Points are collinear.
Question 12
If the points (x1y1), (x2, y2) and (x3,
y3) be collinear, show that
$\frac{\mathrm{y}_{2}-\mathrm{y}_{3}}{\mathrm{x}_{2}
\mathrm{x}_{3}}+\frac{\mathrm{y}_{3}-\mathrm{y}_{1}}{\mathrm{x}_{3}
\mathrm{y}_{1}}+\frac{\mathrm{y}_{1}-\mathrm{y}_{2}}{\mathrm{x}_{1} \mathrm{x}_{2}}=0$Sol :
Given : A(x1, y1), B(x2, y2) and C(x3, y3)
We know the points are collinear if area(∆ABC)=0
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|=0$
Then, Area $=\frac{1}{2}\left\{\mathrm{x}_{1} \mathrm{y}_{2}-\mathrm{x}_{1} \mathrm{y}_{3}+\mathrm{x}_{2} \mathrm{y}_{3}-\mathrm{x}_{2} \mathrm{y}_{1}+\mathrm{x}_{3} \mathrm{y}_{1}-\mathrm{x}_{3} \mathrm{y}_{2}\right\}=0$
Now, Divide by $\mathrm{x}_{1} \mathrm{x}_{2} \mathrm{x}_{3}$
$\Rightarrow \frac{1}{2}\left\{\frac{\left(x_{1} y_{2}-x_{1} y_{3}\right)}{x_{1} x_{2} x_{3}}+\frac{\left(x_{2} y_{3}-x_{2} y_{1}\right)}{x_{1} x_{2} x_{3}}+\frac{\left(x_{3} y_{1}-x_{3} y_{2}\right)}{x_{1} x_{2} x_{3}}\right\}=0$
Taking common $\mathrm{x}_{1}, \mathrm{x}_{2}$ and $\mathrm{x}_{3}$ respectively
$\Rightarrow \frac{1}{2}\left\{\frac{x_{1}\left(y_{2}-y_{3}\right)}{x_{1} x_{2}
x_{3}}+\frac{x_{2}\left(y_{3}-y_{1}\right)}{x_{1} x_{2} x_{3}}+\frac{x_{3}\left(y_{1}-y_{2}\right)}{x_{1} x_{2}
x_{3}}\right\}=0$
$\Rightarrow\left\{\frac{\left(y_{2}-y_{3}\right)}{x_{2}
x_{3}}+\frac{\left(y_{3}-y_{1}\right)}{x_{1} x_{3}}+\frac{\left(y_{1}-y_{2}\right)}{x_{1} x_{2}}\right\}$
Hence, Proved.
Question 13
If the points (a, b), (a1, b1) and (a-a1, b-b1) are collinear,
show that $\frac{a}{a_{1}}=\frac{b}{b_{1}}$
Sol :
Given (a, b), (a1, b1) and (a-a1, b-b1)We know the points are collinear if area(∆ABC)=0
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|=0$
Then, Area $=\frac{1}{2}\left\{\mathrm{a}\left(\mathrm{b}_{1}-\left(\mathrm{b}-\mathrm{b}_{1}\right)+\mathrm{a}_{1}\left\{\left(\mathrm{b}-\mathrm{b}_{1}\right)-\mathrm{b}\right\}+\left(\mathrm{a}-\mathrm{a}_{1}\right)\left(\mathrm{b}-\mathrm{b}_{1}\right\}=0\right.\right.$
$\frac{1}{2}\left\{a b_{1}-\left(a b-a b_{1}\right)+a_{1} b-a_{1} b_{1}-a_{1} b\right\}+\left(a b-a b_{1}-a_{1} b+a_{1} b_{1}\right\}=0$
{ab + a1 b1} = ab – ab1 – a1 b + a1 b1
- ab1 - a1b = 0
-ab1 = a1b
Therofore, We can write as
$\frac{a}{a_{1}}=\frac{b}{b_{1}}$
Hence, Proved.
Question 14
Show that the point (a, 0), (0, b) and (1, 1) are collinear
if $\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}=1$
Sol :Given: (a, 0), (0, b) and (1, 1)
We know the points are collinear if area(∆ABC)=0
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|=0$
Then, Area $=\frac{1}{2}\{a(b-1)+0\{1-0\}+1(0-b)\}=0$
$\frac{1}{2}[\mathrm{ab}-\mathrm{a}+0+0-\mathrm{b}]=0$
ab – a – b = 0
ab = a + b
Since, $\frac{1}{2}+\frac{1}{b}=1$
$\frac{a+b}{a b}=1$
Then, a + b = ab
$\frac{1}{a}+\frac{1}{b}=1$
Hence, Proved.
Question 15 A
Find the values of x if the points (2x, 2x), (3, 2x+1) and (1, 0) are collinear.
Sol :Given (2x, 2x), (3, 2x+1) and (1, 0)
We know the points are collinear if the area(∆ABC)=0
Area of triangle $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|=0$
Then, Area $=\frac{1}{2}|2 x(2 x+1-0)+3(0-2 x)+1(2 x-2 x-1)|=0$
$\frac{1}{2}\left[4 \mathrm{x}^{2}+2 \mathrm{x}-6 \mathrm{x}-1\right]=0$
4x2 - 4x – 1 = 0
4x2 - 2x - 2x – 1 = 0
2x(2x - 1) - 1(2x - 1) = 0
(2x-1)(2x-1)=0
Hence, $x=\frac{1}{2}$
Question 15 B
Find the value of K if the points A(2, 3), B(4, k) and C(6, -3) are collinear.
Sol :Given A(2, 3), B(4, k) and C(6, -3) are collinear
To find: Find the value of K
So, The given points are collinear, if are (∆ABC)=0
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}=0$
Then, $\frac{1}{2}[\{2(\mathrm{k}+3)\}+\{4(-3-3)\}+\{6(3-\mathrm{k})\}]=0$
= 2k+6 - 24 +18 - 6k = 0
= -4k = 0
Hence, K = 0
Question 15 C
Find the value of K for which the points (7, -2), (5, 1), (3, k) are collinear.
Sol :Given A(7, -2), B(5, 1) and C(3, k) are collinear
To find: Find the value of K
So, The given points are collinear, if are (∆ABC)=0
= Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}=0$
Then, $\frac{1}{2}[\{7(1-\mathrm{k})\}+\{5(\mathrm{k}+2)\}+\{3(-2-1)\}]=0$
= 7 – 7k +5k+10-9 = 0
= -2k + 8 = 0
Hence, K = 4
Question 15 D
Find the value of K for which the points (8, 1), (k, -4), (2, -5) are collinear?
Sol :Given A(8, 1), B(k, -4) and C(2, -5) are collinear
To Find: Find the value of k
So, The given points are collinear, if are (∆ABC)=0
= Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}=0$
Then, $\frac{1}{2}[\{8(-4+5)\}+\{\mathrm{k}(-5-1)\}+\{2(-1+4)\}]=0$
= 8(1)+k(-6)+2(3) = 0
= 8 - 6k + 6 = 0
= -6k = -14
Hence, K =
Question 15 E
Find the value of P are the points (2, 1), (p, -1) and (-1, 3)collinear?
Sol :Given A(2, 1), B(p, -1) and C(-1, 3) are collinear
To find: Find the value of p
So, The given points are collinear, if are (∆ABC)=0
= Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}=0$
Then, $\frac{1}{2}[\{2(-1-3)\}+\{p(3-1)\}+\{-1(1+1)\}]=0$
= 2(-4)+p(2)-2=0
= -8 +2p -2 = 0
= 2p = 10
Hence, p =
Question 16
Show that the straight line joining the points A(0, -1) and B(15, 2) divides the line joining the points
C(-1, 2)and D(4, -5) internally in the ratio 2:3.
Sol :Given, A(0, -1) B (15, 2) divides the line on points C(-1, 2) and D(4, -5)
To Prove. Straight line divides in the ratio 2:3 internally
The equation of line $=\left(\mathrm{y}-\mathrm{y}_{1}\right)=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\left(\mathrm{x}-\mathrm{x}_{1}\right)$
Now, Equation of line BC $=(\mathrm{y}+1)=\frac{2+1}{15+0}(\mathrm{x}-0)$
$\Rightarrow(y+1)=\frac{3 x}{15}$
$\Rightarrow(y+1)=\frac{x}{5}$
⇒ 5y + 5 = xTherefore, x – 5y = 5 ---(1)
Now, Equation of line BC $=(y-2)=\frac{-5-2}{4+1}(x+1)$
$\Rightarrow(\mathrm{y}-2)=\frac{-7}{5}(\mathrm{x}+1)$
⇒ 5(y - 2) = -7(x + 1)
⇒ 5y - 10 = -7x - 7
Therefore, 7x +5y = 3 ---(2)
On solving equation (1) and (2)
X = 1 y = $=-\frac{4}{5}$
Now, Point of the intersection of AB and CD is O $\left(1,-\frac{4}{5}\right)$
Let us Assume that AB divides CD at O in the ratio m:n, then
x coordinate of O $=\frac{m \cdot x_{2}+n \cdot x_{1}}{m+n}$
$1=\frac{4 \mathrm{m}-\mathrm{n}}{\mathrm{m}+\mathrm{n}}$
= 4m – n = m+n
= 4m – m = n+n
= 3m = 2n
$=\frac{m}{n}=\frac{2}{3}$ Hence Proved
Given, A triangle whose vertices are A ((a+1)(a+2), (a+2))
B ((a+2)(a+3), (a+3)) and C = ((a+b)(a+4), (a+4)).
To find: Find the area of a triangle.
Since, Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}=0$
Then, =$\frac{1}{2}[(a+1)(a+2)\{(a+3)-(a+4)\}+(a+2)(a+3)\{(a+4)-(a+2)\}+(a+3)(a+4)\{(a+2)-(a+3)\}]$
⇒ 5y - 10 = -7x - 7
Therefore, 7x +5y = 3 ---(2)
On solving equation (1) and (2)
X = 1 y = $=-\frac{4}{5}$
Now, Point of the intersection of AB and CD is O $\left(1,-\frac{4}{5}\right)$
Let us Assume that AB divides CD at O in the ratio m:n, then
x coordinate of O $=\frac{m \cdot x_{2}+n \cdot x_{1}}{m+n}$
$1=\frac{4 \mathrm{m}-\mathrm{n}}{\mathrm{m}+\mathrm{n}}$
= 4m – n = m+n
= 4m – m = n+n
= 3m = 2n
$=\frac{m}{n}=\frac{2}{3}$ Hence Proved
Question 17
Find the area of the triangle whose vertices are
((a+1) (a+2), (a+2)), ((a+2) (a+3), (a+3)) and ((a, 3) (a+4), (a+4))
Sol :((a+1) (a+2), (a+2)), ((a+2) (a+3), (a+3)) and ((a, 3) (a+4), (a+4))
Given, A triangle whose vertices are A ((a+1)(a+2), (a+2))
B ((a+2)(a+3), (a+3)) and C = ((a+b)(a+4), (a+4)).
To find: Find the area of a triangle.
Since, Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}=0$
Then, =$\frac{1}{2}[(a+1)(a+2)\{(a+3)-(a+4)\}+(a+2)(a+3)\{(a+4)-(a+2)\}+(a+3)(a+4)\{(a+2)-(a+3)\}]$
$=\frac{1}{2}\left[\left(a^{2}+3 a+2\right)(-1)+\left(a^{2}+5
a+6\right)(2)+\left(a^{2}+7 a+12\right)(-1)\right]$
$=\frac{1}{2}\left[-\mathrm{a}^{2}-3 \mathrm{a}-2+2 \mathrm{a}^{2}+10 \mathrm{a}+12-\mathrm{a}^{2}+7 \mathrm{a}+12\right]$
Common terms will be canceled out
$=\frac{1}{2}[2]$
Hence, = 1 sq unit
Question 18
The point A divides the join of P(-5, 1) and Q(3, 5) in the ratio k:1. Find the two values of k for which
the area of ΔABC, where B is (1, 5) and C is (7, - 2) is equal to 2 units in magnitude.
Sol :Given: A divides the join of P(-5, 1) and Q(3, 5) in the ratio k:1
To Find: Two values of K
A divides join of PQ in the ratio k:1 hence
Coordinates of A $=\left[\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m} 1+\mathrm{m} 2}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m} 1+\mathrm{m} 2}\right]$
$=\left[\frac{3 \mathrm{k}-5}{\mathrm{k}+1}, \frac{5 \mathrm{k}+1}{\mathrm{k}+1}\right]$
Now, We have A $\left[\frac{3 k-5}{k+1}, \frac{5 k+1}{k+1}\right]$, B (1, 5), and C(7, -2)
Now, The area of ABC is equal to the magnitude 2 (Given)
Area of ABC $=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|$
$\Rightarrow \frac{1}{2}\left|\frac{3 \mathrm{k}-5}{\mathrm{k}+1}(5+2)+1\left(-2-\left(\frac{5 \mathrm{k}+1}{\mathrm{k}+1}\right)\right)+7\left(\left(\frac{5 \mathrm{k}+1}{\mathrm{k}+1}\right)-5\right)\right|=2$
$\Rightarrow \frac{1}{2}\left|\frac{3 \mathrm{k}-5}{\mathrm{k}+1}(7)+1\left(\frac{-2 \mathrm{k}-1-5 \mathrm{k}-1}{\mathrm{k}+1}\right)+7\left(\left(\frac{5 \mathrm{k}+1-5 \mathrm{k}-5}{\mathrm{k}+1}\right)\right)\right|=2$
$\Rightarrow\left|\frac{21 \mathrm{k}-35}{\mathrm{k}+1}+\left(\frac{-7 \mathrm{k}-2}{\mathrm{k}+1}\right)+7\left(\left(\frac{-4}{\mathrm{k}+1}\right)\right)\right|=4$
$\Rightarrow\left|\frac{21 \mathrm{k}-35}{\mathrm{k}+1}+\left(\frac{-7 \mathrm{k}-2}{\mathrm{k}+1}\right)+\left(\left(\frac{-28}{\mathrm{k}+1}\right)\right)\right|=4$
$\Rightarrow\left|\frac{21 \mathrm{k}-35+(-7 \mathrm{k}-2)(-28)}{\mathrm{k}+1}\right|=4$
14k - 66 = 4k + 4
14k - 66 = -4k - 4
10k = 70
18k = 62
Hence, k = 7 and $\mathrm{k}=\frac{31}{9}$
Question 19
The coordinates of A, B, C, D are (6, 3), (-3, 5), (4, -2) and (x, 3x) respectively. If $\frac{\Delta
\mathrm{DBC}}{\Delta \mathrm{ABC}}=\frac{1}{2}$, find x.
Sol :Given A, B, C, and D are (6, 3), (-3, 5), (4, -2) and (x, 3x) respectively.
and ∆DBC = 2∆ABC
To find: Find x.
Since Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Now, The area of ∆DBC $=\frac{1}{2}[x(5+2)-3(-2-3 x)+4(3 x-5)$
$=\frac{1}{2}[5 x+2 x+6+3 x+12 x-20]$
$=\frac{1}{2}[22 \mathrm{x}-14]$
= 11x – 7 sq units
Now, The area of ∆DBC
$=\frac{1}{2}[6(5+2)-3(-2-3)+4(3-5)$
According to question, ∆DBC = 2∆ABC
$\frac{49}{2}=2(11 x-7)$
⇒ 49 = 4(11x – 7)
⇒ 49 = 44x – 28
⇒ 44x = 77
$\Rightarrow x=\frac{77}{44}$
Hence, $x=\frac{7}{4}$
Given: vertices of the quadrilateral be A(1, 2), B(-5, 6), C(7, -4) and D(h, -2).
Let join AC to form two triangles,
Now, We know that
Area of triangle = $\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of triangle ABC = $\frac{1}{2}[1(6+4)-5(-4-2)+7(2-6)]$
= $\frac{1}{2}[10+30-28]$
= 6 sq units
Now, Area of triangle ADC = $\frac{1}{2}[1(-2+4)+\mathrm{h}(-4-2)+7(2+2)]$
Area of quadrilateral ABCD = Area of triangle ABC+ Area of triangle ADC
= 3h – 15 + 6
= 3h = 9
= h = 3
Hence, h is 3
Given: A triangle whose vertices A (3, 4) B(-4, 3), C(8, 6)
To find: Find the area of Triangle and length of AD
Area of triangle = $\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of triangle ABC = $\frac{1}{2}|3(3-6)-4(6-4)+8(4-3)|$
= $\frac{1}{2}|-9-8+8|$
= $=\frac{8}{2}$
= 4 square units
We need to find the length of AD on BC
Hence, We need to find the slope first,
The slope $=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Now the slope of BC $=\frac{8+2}{-1+3}=\frac{10}{2}=5$
If AD perpendicular BC then the slope of AD is $=-\frac{1}{5}$
Therefore, The equation of is (y - y1) = m(x - x1)
$=\frac{1}{2}[42+15-8]$
$=\frac{1}{2}[49]$
$=\frac{49}{2}$
According to question, ∆DBC = 2∆ABC
$\frac{49}{2}=2(11 x-7)$
⇒ 49 = 4(11x – 7)
⇒ 49 = 44x – 28
⇒ 44x = 77
$\Rightarrow x=\frac{77}{44}$
Hence, $x=\frac{7}{4}$
Question 20
If the area of the quadrilateral whose angular points taken in order are (1, 2), (-5, 6), (7, -4) and (h,
-2) be zero, show that h=3.
Sol :Given: vertices of the quadrilateral be A(1, 2), B(-5, 6), C(7, -4) and D(h, -2).
Let join AC to form two triangles,
Now, We know that
Area of triangle = $\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of triangle ABC = $\frac{1}{2}[1(6+4)-5(-4-2)+7(2-6)]$
= $\frac{1}{2}[10+30-28]$
=$\frac{12}{2}$
= 6 sq units
Now, Area of triangle ADC = $\frac{1}{2}[1(-2+4)+\mathrm{h}(-4-2)+7(2+2)]$
=$\frac{1}{2}[2-6 h+28]$
=$\frac{1}{2}[-6 h+30]$
= 3h - 15Area of quadrilateral ABCD = Area of triangle ABC+ Area of triangle ADC
= 3h – 15 + 6
= 3h = 9
= h = 3
Hence, h is 3
Question 21
Find the area of the triangle whose vertices A, B, C are (3, 4) (-4, 3), (8, 6) respectively and hence find
the length of the perpendicular from A to BC.
Sol :Given: A triangle whose vertices A (3, 4) B(-4, 3), C(8, 6)
To find: Find the area of Triangle and length of AD
Area of triangle = $\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of triangle ABC = $\frac{1}{2}|3(3-6)-4(6-4)+8(4-3)|$
= $\frac{1}{2}|-9-8+8|$
= $=\frac{8}{2}$
= 4 square units
We need to find the length of AD on BC
Hence, We need to find the slope first,
The slope $=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Now the slope of BC $=\frac{8+2}{-1+3}=\frac{10}{2}=5$
If AD perpendicular BC then the slope of AD is $=-\frac{1}{5}$
Therefore, The equation of is (y - y1) = m(x - x1)
$(\mathrm{y}+1)=\frac{-\frac{1}{5}}{5}(\mathrm{x}-5)$
5y + 5 = - x + 5
Given: A(3, 1) and B(1, -3)
To find: Find the coordinate of the third vertex C.
Let Assume C = (a, b)
Centroid on C $=\left[\frac{3+1+a}{3}, \frac{1-3+b}{3}\right]$
$=\left[\frac{4+a}{3}, \frac{-2+b}{3}\right]$
Therefore, G(1, 0) as it lies on x-axis
⇒ 4 + a = 3
⇒ a = -1
⇒ - 2 + b = 0
⇒ b = 2
Hence, C is (-1, 2)
Given: Coordinates of Triangle are A(3, 1) and B(1, -3)
Centroid of triangle lies on x- axis.
Let the third coordinate be C(x, y)
Centroid of the triangle with vertices (x1, y1), (x2, y2), (x3, y3) is given by,
C(X, Y) $=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$
The y coordinate of centroid will be 0, as it lies opn x – axis.
Therefore,
y1 + y2 + y3 = 0
1 – 3 + y3 = 0
y3 = 2
So, C(x, y) becomes (x, 2)
We are given that the area of triangle = 3 square units.
We know that,
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Putting the values we get,
5y + 5 = - x + 5
Question 22
The coordinates of the centroid of a triangle and those of two of its vertices are A(3, 1), B(1, -3) and the
centroid of the triangle lies on the x-axis. Find the coordinates of the third vertex C.
Sol :Given: A(3, 1) and B(1, -3)
To find: Find the coordinate of the third vertex C.
Let Assume C = (a, b)
Centroid on C $=\left[\frac{3+1+a}{3}, \frac{1-3+b}{3}\right]$
$=\left[\frac{4+a}{3}, \frac{-2+b}{3}\right]$
Therefore, G(1, 0) as it lies on x-axis
⇒ 4 + a = 3
⇒ a = -1
⇒ - 2 + b = 0
⇒ b = 2
Hence, C is (-1, 2)
Question 23
The area of a triangle is 3 square units. Two of its vertices are A(3,1), B(1,-3) and the centroid of the
triangle lies on x-axis. Find the coordinates of the third vertex C.
Sol :Given: Coordinates of Triangle are A(3, 1) and B(1, -3)
Centroid of triangle lies on x- axis.
Let the third coordinate be C(x, y)
Centroid of the triangle with vertices (x1, y1), (x2, y2), (x3, y3) is given by,
C(X, Y) $=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$
The y coordinate of centroid will be 0, as it lies opn x – axis.
Therefore,
y1 + y2 + y3 = 0
1 – 3 + y3 = 0
y3 = 2
So, C(x, y) becomes (x, 2)
We are given that the area of triangle = 3 square units.
We know that,
Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Putting the values we get,
$3=\frac{1}{2}[3(-3-2)+1(2-3)+x(1+3)]$
-15 – 1 + 4x = 6
4x = 22
$x=\frac{22}{4}$
Hence, coordinates of the third vertex are $C\left(\frac{22}{4}, 2\right)$
Given: The area of a parallelogram is 12. A(-1, 3) and B(-2, 4)
To find: Find the other two vertices of the parallelogram.
Let C is (x, y) and A(-1, 3)
Since, AC is bisected at P, y coordinate ( when p = 0)
Then, $\frac{y+3}{2}=0$
y = -3
So, Coordinate of C is (x, -3)
Now, Area of parallelogram ABCD = area of triangle ABC + Area of triangle BAD
Since, 2(Area of triangles) = area of parallelogram
We have, A(-1, 3) B(-2, 4) and C(x, -3)
Now, Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of triangle ABC $=\frac{1}{2}[-1(4+3)-2(-3-3)+x(3-4)]$
-15 – 1 + 4x = 6
4x = 22
$x=\frac{22}{4}$
Hence, coordinates of the third vertex are $C\left(\frac{22}{4}, 2\right)$
Question 24
The area of a parallelogram is 12 square units. Two of its vertices are the points A(-1, 3) and B (-2, 4).
Find the other two vertices of the parallelogram, if the point of intersection of diagonals lies on x-axis on its
positive side.
Sol :Given: The area of a parallelogram is 12. A(-1, 3) and B(-2, 4)
To find: Find the other two vertices of the parallelogram.
Let C is (x, y) and A(-1, 3)
Since, AC is bisected at P, y coordinate ( when p = 0)
Then, $\frac{y+3}{2}=0$
y = -3
So, Coordinate of C is (x, -3)
Now, Area of parallelogram ABCD = area of triangle ABC + Area of triangle BAD
Since, 2(Area of triangles) = area of parallelogram
We have, A(-1, 3) B(-2, 4) and C(x, -3)
Now, Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of triangle ABC $=\frac{1}{2}[-1(4+3)-2(-3-3)+x(3-4)]$
$6=\frac{1}{2}[-7+12-\mathrm{x}]$
$6=\frac{1}{2}[5-\mathrm{x}]$
12 = 5 - x
So, x = -7
Hence, Coordinate of C is (-7, -3)
In the same we will calculate for D
Let D is (x, y) and A(-2, 4)
Since, BD is bisected at Q, y coordinate ( when Q = 0)
Then, $\frac{y+4}{2}=0$
y = -4
So, Coordinate of C is (x, -4)
We have, A(-1, 3) B(-2, 4) and C(x, -4)
Now, Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of triangle ABC $=\frac{1}{2}[-1(4+4)-2(-4-3)+x(3-4)]$
${6}=\frac{1}{2}[-8+14-\mathrm{x}]$
${6}=\frac{1}{2}[6-\mathrm{x}]$
12 = 6 - x
So, x = -6
Hence, Coordinate of D is (-6, -4)
Hence, C (-7, -3) and D(-6, -4)
Given: Let ABCD is a quadrilateral whose vertices A(-2, 5), B(4, -1), C(9, 1) and D(3, 7).
To prove: ABCD is a parallelogram .
We have to find |AD|, |AB|, |BC|, |DC|
The distance between two sides $=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}$
|AD|$=\sqrt{(3+2)^{2}+(7-5)^{2}}$
= √29
|AB| $=\sqrt{(4+2)^{2}+(1+5)^{2}}$
= √72
|DC| = $=\sqrt{(9-3)^{2}+(1-7)^{2}}$
= √72
|BC| = $\sqrt{(9-4)^{2}+(1+1)^{2}}$
= √29
Therefore, AB = DC and AD = BC
Hence, ABCD is a parallelogram
Now, The Area of ABCD is = |a×b| =$\left|\begin{array}{ccc}1 & j & k \\ 6 & -6 & 0 \\ 5 & 2 & 0\end{array}\right|$
$=\left|\begin{array}{cc}-6 & 0 \\ 2 & 0\end{array}\right| \mathrm{i}-\left|\begin{array}{cc}6 & 0 \\ 5 & 0\end{array}\right| \mathrm{j}+\left|\begin{array}{cc}6 & -6 \\ 5 & 2\end{array}\right| \mathrm{k}$
= 0i – 0j+ 42 k
|a×b| = 42
Hence The area of parallelgram is 42
Given: Points of the quadrilatreral A(-3, -1), B(2, -1), C(1, 1), and D(-2, 1).
To Prove: ABCD is a trapezium
Proof:
For proving ABCD to be a trapezium, we need to prove that two of the sides are parallel.
Therefore, AB and CD are parallel.
For proving ABCD a trapezium,
Slope of AB = Slope of CD
Slope of AB $=\frac{-1+1}{2+5}=0$
Slope of CD $=\frac{1-1}{-2-1}=0$
Hence, the quadrilateral is a trapezium.
$6=\frac{1}{2}[5-\mathrm{x}]$
12 = 5 - x
So, x = -7
Hence, Coordinate of C is (-7, -3)
In the same we will calculate for D
Let D is (x, y) and A(-2, 4)
Since, BD is bisected at Q, y coordinate ( when Q = 0)
Then, $\frac{y+4}{2}=0$
y = -4
So, Coordinate of C is (x, -4)
We have, A(-1, 3) B(-2, 4) and C(x, -4)
Now, Area of triangle $=\frac{1}{2}\left\{\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right\}$
Then,
Area of triangle ABC $=\frac{1}{2}[-1(4+4)-2(-4-3)+x(3-4)]$
${6}=\frac{1}{2}[-8+14-\mathrm{x}]$
${6}=\frac{1}{2}[6-\mathrm{x}]$
12 = 6 - x
So, x = -6
Hence, Coordinate of D is (-6, -4)
Hence, C (-7, -3) and D(-6, -4)
Question 25
Prove that the quadrilateral whose vertices are A(-2, 5), B(4, -1), C(9, 1) and D(3, 7) is a parallelogram
and find its area. If E divides AC in the ratio 2:1, prove that D, E and the middle point F of BC are
collinear.
Sol :Given: Let ABCD is a quadrilateral whose vertices A(-2, 5), B(4, -1), C(9, 1) and D(3, 7).
To prove: ABCD is a parallelogram .
We have to find |AD|, |AB|, |BC|, |DC|
The distance between two sides $=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}$
|AD|$=\sqrt{(3+2)^{2}+(7-5)^{2}}$
= √29
|AB| $=\sqrt{(4+2)^{2}+(1+5)^{2}}$
= √72
|DC| = $=\sqrt{(9-3)^{2}+(1-7)^{2}}$
= √72
|BC| = $\sqrt{(9-4)^{2}+(1+1)^{2}}$
= √29
Therefore, AB = DC and AD = BC
Hence, ABCD is a parallelogram
Now, The Area of ABCD is = |a×b| =$\left|\begin{array}{ccc}1 & j & k \\ 6 & -6 & 0 \\ 5 & 2 & 0\end{array}\right|$
$=\left|\begin{array}{cc}-6 & 0 \\ 2 & 0\end{array}\right| \mathrm{i}-\left|\begin{array}{cc}6 & 0 \\ 5 & 0\end{array}\right| \mathrm{j}+\left|\begin{array}{cc}6 & -6 \\ 5 & 2\end{array}\right| \mathrm{k}$
= 0i – 0j+ 42 k
|a×b| = 42
Hence The area of parallelgram is 42
Question 26
Prove that points (-3, -1), (2, -1), (1, 1) and (-2, 1) taken in order are the vertices of a
trapezium.
Sol :Given: Points of the quadrilatreral A(-3, -1), B(2, -1), C(1, 1), and D(-2, 1).
To Prove: ABCD is a trapezium
Proof:
For proving ABCD to be a trapezium, we need to prove that two of the sides are parallel.
Therefore, AB and CD are parallel.
For proving ABCD a trapezium,
Slope of AB = Slope of CD
Slope of AB $=\frac{-1+1}{2+5}=0$
Slope of CD $=\frac{1-1}{-2-1}=0$
Hence, the quadrilateral is a trapezium.
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