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If the lines 2x + 3ay – 1 = 0 and 3x + 4y + 1 = 0 are mutually perpendicular, then the value of a will be2None of these

Question

If the lines 2x + 3ay – 1 = 0 and 3x + 4y + 1 = 0 are mutually perpendicular, then the value of a will be

2

None of these

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Solution

The condition for two lines Ax + By + C = 0 and Px + Qy + R = 0 to be perpendicular is AP + BQ = 0.

Given the two lines are 2x + 3ay - 1 = 0 and 3x + 4y + 1 = 0.

Comparing these with the general equation of a line, we get:

A1 = 2, B1 = 3a for the first line and A2 = 3, B2 = 4 for the second line.

Substituting these values into the condition for perpendicularity:

A1 * A2 + B1 * B2 = 0

2 * 3 + 3a * 4 = 0

6 + 12a = 0

12a = -6

a = -6 / 12

a = -0.5

So, the value of a is -0.5.

This problem has been solved

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