# Find the ratio in which the line segment joining the points (- 3, 10) and (6, - 8) is divided by (- 1, 6)

**Solution:**

The coordinates of the point P(x, y) which divides the line segment joining the points A(x_{1}, y_{1}) and B(x_{2}, y_{2}), internally, in the ratio m_{1}: m_{2} is given by the Section Formula: P(x, y) = [(mx_{2} + nx_{1 }/ m + n)] , [(my_{2} + ny_{1 }/ m + n)]

Let the ratio in which the line segment joining A(- 3, 10) and B(6, - 8) is divided by point P(- 1, 6) be k : 1.

By Section formula, P(x, y) = [(mx_{2} + nx_{1 }/ m + n)] , [(my_{2} + ny_{1 }/ m + n)]

Therefore,

- 1 = (6k - 3) / (k + 1)

- k - 1 = 6k - 3

7k = 2 (By Cross Multiplying & Transposing)

k = 2 / 7

Hence, the point P divides AB in the ratio 2 : 7

**Video Solution:**

## Find the ratio in which the line segment joining the points (- 3, 10) and (6, - 8) is divided by (- 1, 6)

### NCERT Class 10 Maths Solutions - Chapter 7 Exercise 7.2 Question 4:

Find the ratio in which the line segment joining the points (- 3, 10) and (6, - 8) is divided by (- 1, 6)

The ratio in which the line segment joining the points (- 3, 10) and (6, - 8) is divided by (- 1, 6) is 2 : 7