# PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3

## PSEB 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 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 vertices of the ∆ABC are 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 = 1/2 [x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})

= 1/2 [2 × (0 + 4) – 1 × (- 4 – 3) + 2 × (3 – 0)]

= 1/2 [8 + 7 + 6] = 21/2

= 10.5 sq units.

(ii) Let vertices of the ∆ABC are 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 = 1/2 [x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})]

= 1/2 [- 5 (- 5 – 2) + 3 (2 + 1) + 5 (- 1 + 5)]

= 1/2 [35 + 9 + 20]

= 1/2 × 64 = 32 sq units.

Question 2.

In each of the following find the value of ‘k’ for which the points are coimear.

(i) (7, – 2); (5, 1); (3, k)

(ii) (8, 1); (k, – 4); (2, – 5)

Solution:

(i) Let given points be A (7, – 2); B (5, 1) and C (5, k)

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

y_{1} = – 2, y_{2} = 1 y_{3} = k

Three points are collinear iff

1/2 [x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})] = 0

or 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 = −8/−2 = 4 .

Hence k = 4.

ii) Let 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 = – 4, y = – 5

Three points are collinear iff

1/2 [x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})] = 0

or 1/2 [8 (- 4 + 5) + k (- 5 – 1) + 2 (1 + 4) = 0]

or 8 – 6k + 10 = 0

or – 6k = – 18 .

or k = −18/−6 = 3.

Hence k = 3.

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 the area of the triangle formed to the area of the given triangle.

Solution:

Let vertices of given triangle ABC are A(0, – 1); B (2, 1) and C (0, 3).

Also, D, E, F be the mid points of AB, BC, CA respectively.

Using mid point formula,

Question 4.

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

Solution:

Let co-ordinates of the given quadrilateral ABCD are A(- 4, – 2); B(-3, – 5); C(3, – 2) and D (2, 3).

Join AC then Quad. ABCD divides in two triangles

i.e. ∆ABC and ∆CDA

Question 5.

You have studied in Class IX, (Chapter 9, Q. 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:

Given that coordinates of the vertices of ∆ABC are A(4, – 6); B (3, – 2) and C (5, 2)

Let CD is the median i.e. D is the mid point of AB which divides AABC into two pails i.e.

In ∆CDB

x = 5, x = 35, x = 3

y = 2, y = – 4, y = – 2

Area of ∆CDB = 1/2 [x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})]

= 1/2 [5 (- 4 + 2) + 3.5 (- 2 – 2) + 3 (2 + 4)]

= 1/2 [- 10 – 14 + 18]

= 1/2 × – 6 = – 3

= 3 sq. units(∵ area cannot be negalive)

From above discussion it is clear that area of ∆ADC = area of ∆CDB = 3 sq. units

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

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