KC Sinha Mathematics Solution Class 10 Chapter 10 Coordinates Geometry Exercise 10.4

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 :
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 units

Question 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

$\left.=\frac{1}{2}[\{-1.5(-2-(4))\}+(6)\{(4-3)\}-3(3+2)\}\right]$

$=\frac{1}{2}\{+9+6-15\}$

$=\frac{0}{2}$

= 0 square units

Question 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

$\left.=\frac{1}{2}[\{-5(-5-2)\}+3\{(2+1)\}+5(-1+5)\}\right]$

$=\frac{1}{2}\{35+9+20\}$

$=\frac{64}{2}$

= 32 square units

Question 1 D

Find the area of the triangle whose vertices are
(5, 2), (4, 7), (7, -4)

Sol :
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

$=\frac{1}{2}[\{5(7-(-4))\}+4\{(-4-2)+7(2-7)\}]$

$=\frac{1}{2}\{55-24-35\}$

$=\frac{4}{2}$

= 2 square units

Question 1 E

Find the area of the triangle whose vertices are
(2, 3), (-1, 0), (2, -4)

Sol :
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

$=\frac{1}{2}[\{2(0-(-4))\}+(-1)\{(-4-3)+2(3-0)\}]$

$=\frac{1}{2}\{8+7+6\}$

$=\frac{21}{2}$ Square units

Question 1 F

Find the area of the triangle whose vertices are
(1, -1), (-4, 6), (-3, -5)

Sol :
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

$=\frac{1}{2}[\{1(6-(-5))\}+(-4)\{(-5+1)-3(-1-6)\}]$

$=\frac{1}{2}\{30+16+21\}$

$=\frac{67}{2}$ 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 :
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

$=\frac{1}{2}\left[\left\{\mathrm{at}_{1}^{2}\left(2 \mathrm{at}_{2}-2 \mathrm{at}_{3}\right)\right\}+\mathrm{at}_{2}^{2}\left\{\left(2 \mathrm{at}_{3}-2 \mathrm{at}_{1}\right)-\mathrm{at}_{3}^{2}\left(2 \mathrm{at}_{1}-2 \mathrm{at}_{2}\right)\right\}\right]$

$=\frac{1}{2}\{30+16+21\}$

$=\frac{67}{2}$ Square units

Question 1 H

Find the area of the triangle whose vertices are
(-5, 7), (-4, -5), (4, 5)

Sol :
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

$=\frac{1}{2}[\{-5(-5-(5))\}+(-4)\{(5-7)+4(7+5)\}]$

$=\frac{1}{2}\{50+8+48\}$

$=\frac{106}{2}$

= 53 square units

Question 2 A

Find the area of the quadrilateral whose vertices are
(1, 1), (7, -3), (12, 2) and (7, 21)

Sol :
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 units

Area 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 :
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

$=\frac{1}{2}|-4(7+5)+0(-5-5)+5(5-7)|$

$=\frac{1}{2}[-48+0-10]$

$=\frac{58}{2}$

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 units

Area 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 :

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

$=\frac{1}{2}|-5(-5+6)-4(-6-7)-1(7+5)|$

$=\frac{1}{2}|-5(1)-4(-13)-1(12)|$

$=\frac{1}{2}|-5+52-12|$

$=\frac{35}{2}$

Area of given triangle ABC

$=\frac{1}{2}|-5(-6-5)-1(5-7)+4(7+6)|$

$=\frac{1}{2}|-5(-11)-1(-2)+4(13)|$

$=\frac{1}{2}|55+2+52|$

$=\frac{112}{2}$

= 56 square units

Area 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 :

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

$=\frac{1}{2}|0(0-3)+6(3-0)+4(0-0)|$

$=\frac{1}{2}|0+18+0|$

$=\frac{1}{2}|18|$

= 9 Square units

Area 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 units

Area 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 :

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

$=\frac{1}{2}|1(3-7)+5(7-0)+2(0-3)|$

$=\frac{1}{2}|-4+35-6|$

$=\frac{1}{2}|25|$

$=\frac{25}{2}$ Square units

Area 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 units

Area 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

$\left.=\frac{1}{2} \mid(-4)\{(-5+2)\}+(-3)\{((-2)+2)\}+3\{((-2)+2)\}\right] \mid$

$=\frac{1}{2}[12+0+9]$

$=\frac{21}{2}$ 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 :
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

$=\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1} \mathrm{y}_{2}}{2}\right]$

$=\left[\frac{2+3}{2}, \frac{5+1}{2}\right]$

$=\left[\frac{5}{2}, \frac{6}{2}\right]$

$=\left[\frac{5}{2}, 3\right]$

Area of triangle

$=\frac{1}{2}\left|\left[1(5-3)+2(3-2)+\frac{5}{2}(2-5)\right]\right|$

$=\frac{1}{2}\left[2+2-\frac{15}{2}\right]$

$=\frac{1}{2}\left|-\frac{7}{2}\right|$

$=\frac{7}{4}$ sq units

Area 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 units

Hence, ∆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

$=\frac{1}{2}[4(6-3)+2(3-4)+1(4-6)]$

$=\frac{1}{2}[12-2-2]$

$=\frac{1}{2}[8]$

= 4

Therefore 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

$=\frac{1}{2}|(1)\{(6-4)\}-3(4-2)+5(2-6)|$

$=\frac{1}{2}|2-6-20|$

$=\frac{-24}{2}$

= 12

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

$=\frac{1}{2}|1(3-4)+3(4-5)-1(5-3)|$

$=\frac{1}{2}[-1-3-2]$

$=\frac{1}{2}|-6|$

= 3

Therefore 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

$=\frac{1}{2}|0(1-3)+2(3+1)+0(-1-1)|$

$=\frac{1}{2}[8]$
= 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

$=\frac{1}{2}|1(1-0)+0(0-2)+1(2-1)|$

$=\frac{1}{2}[1+1]$

$=\frac{1}{2}[2]$

= 1

Therefore 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

$=\frac{1}{2}|-1(1-5)+6(5+2)+3(-2-1)|$

$=\frac{1}{2}[4+42-9]$

$=\frac{1}{2}[37]$

$=\frac{37}{2}$ square units

Hence 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

$=\frac{1}{2}|3(6-9)+0(9-0)+6(0-6)|$

$=\frac{1}{2}[-9-36]$

$=\frac{1}{2}[-45]$

$=\frac{45}{2}$ square units

Now, According to the question,

DE internally divides AB in the ratio 1:2 hence

Coordinates of D

$=\left[\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{m} 1+\mathrm{m} 2}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{m} 1+\mathrm{m} 2}\right]$

$=\left[\frac{1 \times 0+2 \times 3}{2}, \frac{1 \times 6+2 \times 0}{1+2}\right]$

$=\left[\frac{6}{3}, \frac{6}{3}\right]$

= (2, 2)

E internally divides AC in the ratio 1:2 hence

Coordinates of D

$=\left[\frac{1 \times 6+2 \times 3}{2+1}, \frac{1 \times 9+2 \times 0}{1+2}\right]$

$=\left[\frac{12}{3}, \frac{9}{3}\right]$

= (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 units

Therefore, 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

$=\frac{1}{2}[\mathrm{a}\{(\mathrm{c}+\mathrm{a})-(\mathrm{a}+\mathrm{b})\}+\mathrm{b}\{(\mathrm{a}+\mathrm{b})-(\mathrm{b}+\mathrm{c})\}+\mathrm{c}\{(\mathrm{b}+\mathrm{c})-(\mathrm{c}+\mathrm{a})\}]$

$=\frac{1}{2}[\mathrm{a}\{\mathrm{c}+\mathrm{a}-\mathrm{a}-\mathrm{b}\}+\mathrm{b}\{\mathrm{a}+\mathrm{b}-\mathrm{b}-\mathrm{c}\}+\mathrm{c}\{\mathrm{b}+\mathrm{c}-\mathrm{c}-\mathrm{a}\}$

$=\frac{1}{2}[\mathrm{a}\{\mathrm{c}-\mathrm{b}\}+\mathrm{b}\{\mathrm{a}-\mathrm{c}\}+\mathrm{c}\{\mathrm{b}-\mathrm{a}\}]$

= 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

$=\frac{1}{2}\left\{\mathrm{x}_{1} \mathrm{y}_{2}-\mathrm{x}_{1} \mathrm{y}_{3}+\mathrm{x}_{2} \mathrm{y}_{3}-\mathrm{x}_{2} \mathrm{y}_{1}+\mathrm{x}_{3} \mathrm{y}_{1}-\mathrm{x}_{3} \mathrm{y}_{2}\right\}=0$

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

$=\frac{1}{2}\left\{\mathrm{a}\left(\mathrm{b}_{1}-\left(\mathrm{b}-\mathrm{b}_{1}\right)+\mathrm{a}_{1}\left\{\left(\mathrm{b}-\mathrm{b}_{1}\right)-\mathrm{b}\right\}+\left(\mathrm{a}-\mathrm{a}_{1}\right)\left(\mathrm{b}-\mathrm{b}_{1}\right\}=0\right.\right.$

$\frac{1}{2}\left\{a b_{1}-\left(a b-a b_{1}\right)+a_{1} b-a_{1} b_{1}-a_{1} b\right\}+\left(a b-a b_{1}-a_{1} b+a_{1} b_{1}\right\}=0$

{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

$=\frac{1}{2}\{a(b-1)+0\{1-0\}+1(0-b)\}=0$

$\frac{1}{2}[\mathrm{ab}-\mathrm{a}+0+0-\mathrm{b}]=0$

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

$=\frac{1}{2}|2 x(2 x+1-0)+3(0-2 x)+1(2 x-2 x-1)|=0$

$\frac{1}{2}\left[4 \mathrm{x}^{2}+2 \mathrm{x}-6 \mathrm{x}-1\right]=0$

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

$\frac{1}{2}[\{2(\mathrm{k}+3)\}+\{4(-3-3)\}+\{6(3-\mathrm{k})\}]=0$

= 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

$\frac{1}{2}[\{7(1-\mathrm{k})\}+\{5(\mathrm{k}+2)\}+\{3(-2-1)\}]=0$

= 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

$\frac{1}{2}[\{8(-4+5)\}+\{\mathrm{k}(-5-1)\}+\{2(-1+4)\}]=0$

= 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

$\frac{1}{2}[\{2(-1-3)\}+\{p(3-1)\}+\{-1(1+1)\}]=0$

= 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 = x

Therefore, 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

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 :
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

$\frac{1}{2}[(a+1)(a+2)\{(a+3)-(a+4)\}+(a+2)(a+3)\{(a+4)-(a+2)\}+(a+3)(a+4)\{(a+2)-(a+3)\}]$

$=\frac{1}{2}\left[\left(a^{2}+3 a+2\right)(-1)+\left(a^{2}+5 a+6\right)(2)+\left(a^{2}+7 a+12\right)(-1)\right]$

$=\frac{1}{2}\left[-\mathrm{a}^{2}-3 \mathrm{a}-2+2 \mathrm{a}^{2}+10 \mathrm{a}+12-\mathrm{a}^{2}+7 \mathrm{a}+12\right]$

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

$=\left[\frac{3 \mathrm{k}-5}{\mathrm{k}+1}, \frac{5 \mathrm{k}+1}{\mathrm{k}+1}\right]$

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

$=\frac{1}{2}[x(5+2)-3(-2-3 x)+4(3 x-5)$

$=\frac{1}{2}[5 x+2 x+6+3 x+12 x-20]$

$=\frac{1}{2}[22 \mathrm{x}-14]$

= 11x – 7 sq units

Now, The area of ∆DBC

$=\frac{1}{2}[6(5+2)-3(-2-3)+4(3-5)$

$=\frac{1}{2}[42+15-8]$

$=\frac{1}{2}[49]$

$=\frac{49}{2}$

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 - 15

Area 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 =

$\frac{1}{2}|3(3-6)-4(6-4)+8(4-3)|$

=

$\frac{1}{2}|-9-8+8|$

=

$=\frac{8}{2}$

= 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

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

$3=\frac{1}{2}[3(-3-2)+1(2-3)+x(1+3)]$

-15 – 1 + 4x = 6

4x = 22

$x=\frac{22}{4}$

Hence, coordinates of the third vertex are

$C\left(\frac{22}{4}, 2\right)$

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

$=\frac{1}{2}[-1(4+3)-2(-3-3)+x(3-4)]$

$6=\frac{1}{2}[-7+12-\mathrm{x}]$

$6=\frac{1}{2}[5-\mathrm{x}]$

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

$=\frac{1}{2}[-1(4+4)-2(-4-3)+x(3-4)]$

${6}=\frac{1}{2}[-8+14-\mathrm{x}]$

${6}=\frac{1}{2}[6-\mathrm{x}]$

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.

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