Gujarat Board GSEB Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3 Textbook Questions and Answers.

## 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 Î”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 +(-1)(-4 -3) + 2(3 – 0)1

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

= \(\frac {21}{2}\) 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 Î”ABC

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

= 32 sq.units

Question 2.

In each of the following find the value of ‘kâ€™, 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 Î”ABC

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

= \(\frac {1}{2}\) 7[(1 -k)] + 5[k -(-2)] + 31(-2 – 1)]

= \(\frac {1}{2}\) [7 – 7k + 5k + 10 – 9]

= \(\frac {1}{2}\) [8 – 2k] = 4 – k

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

â‡’ 4 – k = 0

â‡’ 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 Î”ABC

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

= \(\frac {1}{2}\) [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 Î”ABC

= \(\frac {1}{2}\)[x_{1}(y_{1} – 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] = 4 sq. units

Area of Î”DEF

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

Area of Î”DEF

= [1(1 – 0) + 0(0 – 2) + [(2 – 1)]

= Area of ADEF = \(\frac {1}{2}\) [1 + 1]

= 1 sq. unit

âˆ´ 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 Î”ABC

= \(\frac {1}{2}\) [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 Î”ABC

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

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

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

= \(\frac {21}{2}\) sq. units,

We know that, area of Î”ACD

= [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

Now, area of Î”ACD

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

= \(\frac {1}{2}\) [(-4 x -5) + (3 x 5) + (2 x 0)]

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

= \(\frac {1}{2}\) x 35 = \(\frac {35}{2}\)

Hence, the area of quadrilateral ABCD

= ar(Î”ABC) + ar(Î”ACD)

= \(\frac {21}{2}\) + \(\frac {35}{2}\) = \(\frac {1}{2}\)(21 + 35)

= \(\frac {1}{2}\) x 56 = 28 sq. units.

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 Î”ABC whose vertices are A(4, -6), B(3, -2) and C(5, 2).

Solution:

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

âˆ´ Coordinates of D are \(\left(\frac{3+5}{2}, \frac{-2+2}{2}\right)\), i.e., (4, 0).

Here, for Î”ADC,

x_{1} = 4, y_{1} = 0,

x_{2}= 4, y_{2} = 0,

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

Area of Î”ABD

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

= \(\frac {1}{2}\)[4(0 – 2) + 4 (2 + 6) + 5(-6 – 0)]

= \(\frac {1}{2}\)[(-8 + 32 – 30)] = \(\frac {1}{2}\) [-6]

= -3 = 3 sq. units

[ âˆ´ Area of triangle is positive and Area of Î”ABD]

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

For Î”ABD,

[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 [âˆ´ 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.