# 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:**

Reasoning: Three or more points are said to be collinear if they lie on a single straight line.

(i) Let A(x_{1}, y_{1}) = (7, - 2), B(x_{2}, y_{2}) = (5 , 1) and C(x_{3}, y_{3}) = (3, k)

For three points to be collinear, the area of triangle must be equal to zero.

1/2 (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3 }- y_{1}) + x_{3} (y_{1} - y_{2})) = 0

By substituting the values of vertices, A, B, C in the above equation

1/2 [7{1 - k} + 5{k - ( - 2)} + 3{( - 2) - 1}] = 0

7 - 7k + 5k + 10 - 9 = 0

- 2k + 8 = 0

k = 4

Hence, the given points are collinear for k = 4

(ii) Let A(x_{1}, y_{1}) = (8, 1), B(x_{2}, y_{2}) = (k , - 4) and C(x_{3}, y_{3}) = (2, - 5)

For three points to be collinear, the area of triangle must be equal to zero.

1/2 (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3 }- y_{1}) + x_{3} (y_{1} - y_{2})) = 0

By substituting the values of vertices, A, B, C in the above equation.

1/2 [8{ - 4 - ( - 5)} + k{( - 5) - (1)} + 2{1 - ( - 4)}] = 0

8 - 6k + 10 = 0 (By Transposing)

6k = 18

k = 3

Hence, the given points are collinear for k = 3

**Video Solution:**

## 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)

### NCERT Class 10 Maths Solutions - Chapter 7 Exercise 7.3 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)

The value of ‘k’, for which the points are collinear , for points (7, - 2), (5, 1), (3, k) is k = 4 and for points (8, 1), (k, - 4), (2, - 5) is k = 3