Haryana State Board HBSE 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 Textbook Exercise Questions and Answers.

## Haryana Board 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 A (2, 3), B (- 1, 0) and C(2, – 4) be the vertices of the given triangle.

Here, x = 2, y = 3, x = -1, y = 0, x = 2, y = – 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]

= \(\frac{1}{2}\) × 21 = 10.52

Hence, Area of ∆ABC = 10.5 square units.

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

Here x_{1} = – 5, y_{1} = – 1, x_{2} = 3, y_{2} = 5, x_{3} = 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)]

= \(\frac{1}{2}\) [- 5 × – 7 + 3 × 3 + 5 × 4]

= \(\frac{1}{2}\) (35 + 9 + 20)

= \(\frac{1}{2}\) (64) = 32

Hence, Area of ∆ABC = 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, y_{1} = – 2, x_{2} = 5, y_{2} = 1, x_{3} = 3, y_{3} = k

Since, the given points are collinear.

Area of ∆ABC = 0

⇒ \(\frac{1}{2}\) [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})] = 0

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

⇒ \(\frac{1}{2}\) [7 – 7k + 5k + 10 + 3 × (- 3)] = 0

⇒ \(\frac{1}{2}\) (- 2k + 17 – 9)= 0

⇒ \(\frac{1}{2}\) (-2k + 8) = 0

⇒ – 2k + 8 = 0

⇒ – 2k = – 8

⇒ k = \(\frac{-8}{-2}\)

⇒ k = 4

⇒ Hence, k = 4.

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

Here, x_{1} = 8, y_{1} = 1, x_{2} = k, y_{2} = – 4, x_{3} = 2, y_{3} = – 5

Since, the given points are collinear.

Area of ∆ABC = 0

⇒ \(\frac{1}{2}\) [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})] = 0

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

⇒ \(\frac{1}{2}\) [8 × 1 + k × (- 6) + 2 × 5] = 0

⇒ \(\frac{1}{2}\) [8 – 6k + 10] = 0

⇒ (- 6k + 18) = 0

⇒ – 6k + 18 = 0

⇒ 6k = 18

⇒ k = \(\frac{18}{6}\)

⇒ k = 3

Hence, k = 3.

Question 3.

Find the area of the triangle formed by the 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 A (0, – 1), B (2, 1) and C (0, 3) be the vertices of given triangle.

Let D, E and F are the mid points of sides of the triangle AB, BC and CA respectively. Then,

The co-ordinates of D are \(\left(\frac{0+2}{2}, \frac{-1+1}{2}\right)\) i.e., (1, 0)

The co-ordinates of E are \(\left(\frac{2+0}{2}, \frac{1+3}{2}\right)\) i.e., (1, 2)

The co-ordinates of F are \(\left(\frac{0+0}{2}, \frac{-1+3}{2}\right)\) i.e., (0, 1)

Thus, co-ordinates of the mid-points are D(1, 0), E(1, 2), F(0, 1)

∴ Area of ∆DEF = \(\frac{1}{2}\) [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 + 1 × 1 + 0)

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

= \(\frac{1}{2}\) × 2

= 1 square unit.

Now, 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(- 2) + 2 × 4 + 0 (- 2)]

= \(\frac{1}{2}\) (0 + 8 + 0)

= \(\frac{1}{2}\) × 8

= 4 square units.

Area of ∆DEF : Area of ∆ABC = 1 : 4

Hence, area of ∆DEF = 1 square unit and

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).

Solution :

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

Area of quadrilateral ABCD = Area(∆ABC) + Area(∆ACD)

Now, 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)]

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

= \(\frac{1}{2}\) (12 – 0 + 9)

= \(\frac{21}{2}\) square units.

and, Area of ∆ACD = \(\frac{1}{2}\) [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

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

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

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

= \(\frac{1}{2}\) × 35

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

Hence, Area of quadrilateral ABCD = \(\frac{21}{2}\) + \(\frac{35}{2}\)

= \(\frac{56}{2}\) = 28 square 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 ∆ABC whose vertices are A(4, – 6), B(3, – 2) and C(5, 2).

Solution :

Since AD is the median

∴ D is the mid point of BC. Then co-ordinates of D are \(\left(\frac{3+5}{2}, \frac{-2+2}{2}\right)\) i.e., (4, 0)

Now Area of ∆ABD = \(\frac{1}{2}\) [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

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

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

= \(\frac{1}{2}\) [- 8 + 18 – 16]

= \(\frac{1}{2}\) (- 24 + 18)

= \(\frac{1}{2}\) × – 6 = – 3

Since, area cannot be negative, we will take numerical value of – 3 i.e. 3 square units and

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(0 – 2) + 4(2 + 6) + 5(- 6 – 0)]

= \(\frac{1}{2}\) [4 × (- 2) + 4 × 8 + 5 (- 6)]

= \(\frac{1}{2}\) (- 8 + 32 – 30)

= \(\frac{1}{2}\) (- 6)

= – 3 i.e., 3 square units.

Hence Area of ∆ABD = Area of ∆ADC.