If the two lines $(a + b) x - 4 y + 5 = 0$ and $x + (a - b) y + 10 = 0$ are perpendicular to each other then

a = b

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2 a + b = 0

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3 a - 5 b = 0

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5 a - 2 b = 0

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Solution

The correct option is C $3 a - 5 b = 0$
Slope of $(a + b) x - 4 y + 5 = 0$ is
$m_{1} = - \frac{a + b}{- 4} = \frac{a + b}{4}$
Slope of $x + (a - b) y + 10 = 0$
$m_{1} = - \frac{1}{(a - b)}$
Since lines are perpendicular
$∴ m_{1} . m_{2} = - 1$
$\Rightarrow \frac{a + b}{4} (- \frac{1}{a - b}) = - 1$
$\Rightarrow a + b = 4 a - 4 b$
$\Rightarrow 3 a - 5 b = 0$

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Similar questions

(a + b) x - 4 y + 5 = 0

x + (a - b) y + 10 = 0

(a + b) x - 4 y + 5 = 0

x + (a - b) y + 10 = 0

a x^{3} - 9 y x^{2} - y^{2} x + 4 y^{3} = 0

a x^{3} - 9 y x^{2} - y^{2} x + 4 y^{3} = 0

a

a x^{3} - 9 x^{2} y - x y^{2} + 4 y^{3} = 0

(a > 0)

a

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