Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3

Question 1.

Find the area of the triangle whose vertices are :

(i) (2, 3), (- 1, 0), (2, – 4)

(ii) (- 5, – 1), (3, – 5), (5, 2)

Solution:

(i) Let the vertices of △ABC be

A(2, 3), B(-1, 0) and C(2, – 4)

Here

x_{1} = 2, x_{2} = -1, x_{3} = 2

y_{1} = 3, y_{2} = 0, y_{3} = – 4

∴ Area of △ABC

= \(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

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

= \(\frac{1}{2}\)[8 + 7 + 6] = \(\frac{21}{2}\)

= 10.5 square units

(ii) Let the vertices of △ABC be A(- 5, – 1), B(3, – 5) and C(5, 2).

Here

x_{1} = – 5, x_{2} = 3, x_{3} = 5

y_{1} = -1, y_{2} = – 5, y_{3} = 2

∴ Area of △ABC

= \(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

= \(\frac{1}{2}\)[(- 5) (-5 – 2) + 3 (2 + 1) + 5 (-1 + 5)]

= \(\frac{1}{2}\)[35 + 9 + 20] = \(\frac{1}{2}\) × 64

= 32 square units

Question 2.

In each of the following find the value of ‘k’, for which the points are collinear :

(i) (7, – 2), (5, 1), (3, k)

(ii) (8, 1), (k, – 4), (2, – 5)

Solution:

(i) Let the given points be A(7, -2); B(5, 1) and C(3, k).

Here

x_{1} = 7, x_{2} = 5, x_{3} = 3

y_{1} = -2 ,y_{2} = 1, y_{3} = k

If the three points are collinear then the area of the triangle must be zero.

i.e., \(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})] = 0

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

or 7 – 7k + 5k + 10 – 9 = 0

or – 2k + 8 = 0 or – 2k = – 8

or k = \(\frac{-8}{-2}\) = 4

Hence for k = 4 the three points are collinear.

(ii) Let the given points be A(8, 1), B(k, – 4) and C(2, – 5).

Here

x_{1} = 8, x_{2} = k, x_{3} = 2

y_{1} = 1, y_{2} = -4, y_{3} = 5

If the three points are collinear then the area of the triangle must be zero.

i.e.,\(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{3}) = o

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

or 8 – 6k + 10 = 0

or – 6k = – 18 or k = \(\frac{-18}{-6}\) = 3

Hence for k = 3 the three points are collinear.

Question 3.

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, – 1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Solution:

Let the vertices of the given triangle ABC be A(0, – 1), B(2, 1) and C(0, 3) and D, E, F are the mid-points of AB, BC, CA respectively. Hence by mid-point formula

∴ Coordinates of the vertices of △DEF are D(1, 0), E(1, 2) and F(0, 1).

Here x_{1} = 1, x_{2} = 1, x_{3} = 0

y_{1} = 0, y_{2} = 2, y_{3} = 1

Area of △DEF

= [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

= \(\frac{1}{2}\)[1(2 – 1) + 1(1 – 0) + 0(0 – 2)]

= \(\frac{1}{2}\)[1 + 1 + 0] = \(\frac{2}{2}\)

= 1 square unit Ans.

In △ABC,

x_{1} = 0, x_{2} = 2, x_{3} = 0

y_{1} = -1, y_{2} = 1, y_{3} = 3

Area of △ABC

= \(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

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

= \(\frac{1}{2}\)[0 + 8 + 0] = \(\frac{8}{2}\) = 4 square units

Required ratio = \(\frac{Area of △DEF}{Area of △ABC}\)

= \(\frac{1}{4}\) i e., 1 : 4

Hence the ratio of the areas of both triangles = 1 : 4

Question 4.

Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).

Solution:

Let the coordinates of the vertices of given quadrilateral ABCD be A(-4, -2); B(- 3, -5); C(3, -2) and D(2, 3).

Join AC, then quadrilateral ABCD is divided into two triangles, i.e., △ABC and △CDA

In △ABC,

Here x_{1} = – 4, x_{2} = – 3, x_{3} = 3

y_{1} = -2, y_{2} = – 5, y_{3} = – 2

Area of △ABC

= \(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

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

= \(\frac{1}{2}\)[12 + 0 + 9] = \(\frac{21}{2}\) square units

In △CDA,

x_{1} = -3, x_{2} = 2, x_{3} = – 4

y_{1} = – 2, y_{2} = 3, y_{3} = – 2

Area of △CDA,

= \(\frac{1}{2}\)[x_{1}(y_{2} – y_{3} + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

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

= \(\frac{1}{2}\)[15 + 0 + 20] = \(\frac{35}{2}\) square units

Now, area of the quadrilateral ABCD

= (Area of △ABC) + (Area of △ACD)

= \(\frac{21}{2}\) + \(\frac{35}{2}\) = \(\frac{21+35}{2}\)

= \(\frac{56}{2}\) = 28 square units

Hence the area of the required quadrilateral

= 28 square units

Question 5.

You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for △ABC whose vertices are A(4, – 6), B(3, – 2) and C(5, 2).

Solution:

According to the question, the coordinates of the vertices of △ABC are A(4, – 6), B(3, – 2) and C(5, 2).

Let CD be a median, i.e., D is the mid-point of AB which divides △ABC into two parts i.e., △ADC and △CDB.

= (3.5, – 4)

In △ADC,

x_{1} = 4, x_{2} = 3.5, x_{3} = 5

y_{1} = – 6, y_{2} = – 4, y_{3} = 2

Area of △ADC

= \(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

= \(\frac{1}{2}\)[4(-4 – 2) + 3.5(2 + 6) + 5(-6 + 4)]

= \(\frac{1}{2}\)[-24 + 28 – 10]

= \(\frac{1}{2}\)[-34 + 25] = \(\frac{1}{2}\)[-6] = -3

= 3 square units [∵ Area cannot be negative]

In △CDB,

x_{1} = 5, x_{2} = 3.5, x_{3} = 3

y_{1} = 2, y_{2} = – 4, y_{3} = – 2

Area of △CDB

= \(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

= \(\frac{1}{2}\)[5 (- 4 + 2) + 3.5 (- 2 – 2) + 3(2 + 4)]

= \(\frac{1}{2}\)[- 10 – 14 + 18] = \(\frac{1}{2}\)

[-24 + 18] = \(\frac{1}{2}\)[- 6] = -3

= 3 square units [∵ Area cannot be negative]

Hence it is clear that Area of △ADC= Area of △CDB = 3 square units

Hence a median of a triangle divides it into two triangles of equal areas.

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