ICSE Board Question Based on Equation of a Straight Line Chapter of M.L Aggarwal for class10

In this question given lines are perpendicular to each other so you have to find the relation connecting a and b.

This is the Question Number 07, Exercise 12.2 of M.L Aggarwal.

Deepak BoraNewbie

# If the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other, find the relation connecting a and b.

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Given that the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other

Then the product of their slopes must be -1.

Slope of line 3x + by + 5 = 0 is,

by = -3x – 5

y = (-3/b) – 5/b

So, slope (m

_{1}) = -3/bAnd,

The slope of line ax – 5y + 7 = 0 is

5y = ax + 7

y = (a/5) x + 7/5

So, slope (m

_{2}) = a/5As the lines are perpendicular, we have

m

_{1}x m_{2}= -1-3/b x a/5 = -1

-3a/5b = -1

-3a = – 5b

3a = 5b

Hence, the relation connecting a and b is 3a = 5b.