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# Find the equation of the line passing through (- 3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6)

**Solution:**

The slope of the line joining the points (2, 5) and (-3, 6) is

m = (6 - 5)/(- 3 - 2) = - 1/5

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, the slope of the line perpendicular to the line through the points (2, 5) and (-3, 6) is

- 1/m = - 1/(- 1/5) = 5

Now, the equation of the line passing through point (-3, 5) , whose slope is 5, is

(y - 5) = 5(x + 3)

y - 5 = 5x + 15

5x - y + 20 = 0

Hence, the equation of the line is 5x - y + 20 = 0

NCERT Solutions Class 11 Maths Chapter 10 Exercise 10.2 Question 10

## Find the equation of the line passing through (- 3, 5) and perpendicular to the line through the points (2, 5) and (- 3, 6)

**Summary:**

The equation of the line passing through (- 3, 5) and perpendicular to the line through the points (2, 5) and (- 3, 6) is 5x - y + 20 = 0

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