# Gujarat Board Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3

Gujarat Board Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3

## Gujarat Board Textbook Solutions Class 10 Maths 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 the given point be A(2, 3), B(-1, 0) and C(2, -4).

Here, we have

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

x_{2} = -1, y_{2} = 0

and x_{3} = 2, y_{3} = -4

Now, area of ΔABC

= 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 +(-1)(-4 -3) + 2(3 – 0)1

= 1/2(8 + 7 + 6) = 1/2 x 21

= 21/2 sq. units

(ii) Let the given points be A(-5, -1), B(3, -5) and C(5, 2).

Here, we have

x_{1} = -5, y_{1} = -1

x_{2} = 3, y_{2} = -5

and x_{3} = 5, y_{3} = 2

Now, area of ΔABC

= 1/2[x_{1}(y_{1} – y_{1}) + 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]

= 32 sq.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, we have,

x_{1} = 7 y_{1} = -2

x_{2} = 5, y_{2} = 1

and x_{3} = 3, y_{3} = k

Now, area of ΔABC

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

= 1/2 7[(1 -k)] + 5[k -(-2)] + 31(-2 – 1)]

= 1/2 [7 – 7k + 5k + 10 – 9]

= 1/2 [8 – 2k] = 4 – k

If the points are collinear, then area of the triangle = 0

⇒ 4 – k = 0

⇒ k = 4.

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

Here, we have

x_{1} = 8, y_{1} = 1

x_{2} = k, y_{2} = -4

and x_{3} = 2, y_{3} = -5

Now, area of ΔABC

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

= 1/2 [8 – 6k + 10]

= [18 – 6k] = 9 – 3k

If the points are collinear, then area of the triangle = 0

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 area of the given triangle.

Solution:

Let A(0, -1), B(2, 1) and C(0, 3) be the vertices of ABC. Let D, E, F be the mid-points of sides BC, CA and AB respectively.

Then, the coordinates of D, E and F are (1, 2), (0, 1) and (1, 0) respectively.

Now, Area of ΔABC

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

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

= 1/2 [0 + 8 + 0] = 4 sq. units

Area of ΔDEF

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

Area of ΔDEF

= [1(1 – 0) + 0(0 – 2) + [(2 – 1)]

= Area of ADEF = 1/2 [1 + 1]

= 1 sq. unit

∴ Area of DEF : Area of ABC = 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). [National Olympiad]

Solution:

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

Area of ΔABC

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

Here, we have

x_{1} = -4, y_{1} = -2

x_{2}= -3, y_{2} = -5

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

Now, area of ΔABC

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

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

= 1/2 [12 + 0 + 9] = 1/2 x 21

= 21/2 sq. units,

We know that, area of ΔACD

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

Here, we have

x_{1} = -4, y_{2} = -2

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

and x_{3} = 2, y_{3}= 3

Question 5.

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

Solution:

According to the question, AD is the median of ΔABC, therefore D is the mid-point of BC.

For ΔABD,

[Let, x_{1} = 4, y_{1} = -6, x_{2} = 3, y_{2} = -2, x_{3} = 4, y_{3} = 0]

= [-8 + 18 – 16] = (-6) = -3

= 3 sq. units [∴ Area of triangle is positive.]

Area of ADC = Area of AED

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

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