## EXERCISE 4.3

#### Page No 122:

#### Question 1:

Find area of the triangle with vertices at the point given in each of the following:

(i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8)

(iii) (−2, −3), (3, 2), (−1, −8)

#### Answer:

(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,

(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,

(iii) The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8)

is given by the relation,

Hence, the area of the triangle is.

#### Page No 123:

#### Question 2:

Show that points

are collinear

#### Answer:

Area of ΔABC is given by the relation,

Thus, the area of the triangle formed by points A, B, and C is zero.

Hence, the points A, B, and C are collinear.

#### Question 3:

Find values of *k* if area of triangle is 4 square units and vertices are

(i) (*k*, 0), (4, 0), (0, 2) (ii) (−2, 0), (0, 4), (0, *k*)

#### Answer:

We know that the area of a triangle whose vertices are (*x*_{1}, *y*_{1}), (*x*_{2}, *y*_{2}), and

(*x*_{3}, *y*_{3}) is the absolute value of the determinant (Δ), where

It is given that the area of triangle is 4 square units.

∴Δ = ± 4.

(i) The area of the triangle with vertices (*k*, 0), (4, 0), (0, 2) is given by the relation,

Δ =

**∴**−*k* + 4 = ± 4

When −*k* + 4 = − 4, *k* = 8.

When −*k* + 4 = 4, *k* = 0.

Hence, *k* = 0, 8.

(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0, *k*) is given by the relation,

Δ =

∴*k* − 4 = ± 4

When *k* − 4 = − 4, *k* = 0.

When *k* − 4 = 4, *k* = 8.

Hence, *k* = 0, 8.

#### Question 4:

(i) Find equation of line joining (1, 2) and (3, 6) using determinants

(ii) Find equation of line joining (3, 1) and (9, 3) using determinants

#### Answer:

(i) Let P (*x*, *y*) be any point on the line joining points A (1, 2) and B (3, 6). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.

Hence, the equation of the line joining the given points is *y* = 2*x*.

(ii) Let P (*x*, *y*) be any point on the line joining points A (3, 1) and

B (9, 3). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.

Hence, the equation of the line joining the given points is *x* − 3*y* = 0.

#### Question 5:

If area of triangle is 35 square units with vertices (2, −6), (5, 4), and (*k*, 4). Then *k* is

**A.** 12 **B.** −2 **C.** −12, −2 **D.** 12, −2

#### Answer:

**Answer: D**

The area of the triangle with vertices (2, −6), (5, 4), and (*k*, 4) is given by the relation,

It is given that the area of the triangle is ±35.

Therefore, we have:

When 5 − *k* = −7, *k* = 5 + 7 = 12.

When 5 − *k* = 7, *k* = 5 − 7 = −2.

Hence, *k* = 12, −2.

The correct answer is D.