If the lines $(p + 1) x + y = 3$ and $3 y - (p - 1) x = 4$ are perpendicular to each other, find $p$ .

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Solution

The correct options are
B
2

D
-2

When we convert the given lines into $y = m x + c$ form ,we get

$y = - (p + 1) x + 3 a n d y = \frac{(p - 1)}{3} x + \frac{4}{3}$
Here $m_{1} = - (p + 1) a n d m_{2} = \frac{(p - 1)}{3}$
$∵$ Lines are perpendicular to each other
$m_{1} m_{2} = - 1$
$\Rightarrow - (p + 1) \frac{(p - 1)}{3} = - 1$
$\Rightarrow p^{2} - 1 = 3$

$\Rightarrow p^{2} = 4$

$\Rightarrow p = \pm 2$

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Similar questions

If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.

Find the value of p if the lines, whose equations are 2x - y + 5 = 0 and px + 3y = 4 are perpendicular to each other.

(p + 1) x + y = 3

3 y - (p - 1) x = 4

p

\frac{2 x - 1}{2} = \frac{3 - y}{1} = \frac{z - 1}{3}

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2 x + 3 y - 7 = 0

4 y - p x - 12 = 0

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