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.
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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 (m1) = -3/b
And,
The slope of line ax – 5y + 7 = 0 is
5y = ax + 7
y = (a/5) x + 7/5
So, slope (m2) = a/5
As the lines are perpendicular, we have
m1 x m2 = -1
-3/b x a/5 = -1
-3a/5b = -1
-3a = – 5b
3a = 5b
Hence, the relation connecting a and b is 3a = 5b.