The angle between $3 x + 4 y + 5 = 0, 4 x - 3 y + 6 = 0$ is

π / 2

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π / 4

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3 π / 8

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π / 8

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Solution

The correct option is B $π / 2$
Given lines are $3 x + 4 y + 5 = 0 4 x - 3 y + 6 = 0$
Angle between them is given by $cos θ = \frac{| a_{1} a_{2} + b_{1} b_{2} |}{\sqrt{a_{1}^{2} + b_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2}}} cos θ = \frac{3 (4) - 4 (3)}{\sqrt{3^{2} + 4^{2}} \sqrt{3^{2} + 4^{2}}} c o s θ = 0 θ = π / 2$

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