What is the slope of a line perpendicular to Ax+By+C=0?

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Answer 1 · 2016-07-31T05:16:31

$B x - A y + C = 0$ $Bx-Ay+C=0$

Given -

$A x + B y + C = 0$ $Ax+By+C=0$

In such case you have to interchange the co-efficients of $x$ $x$ and $y$ $y$ and then change the sign of the coefficient of $y$ $y$

The equation of the perpendicular line is

$B x - A y + C = 0$ $Bx-Ay+C=0$

You can check it by using the formula

$m_{1} \times m_{2} = - 1$ $m_1xxm_2=-1$

Slope of the first line -

$m_{1} = - \frac{a}{b} = - \frac{A}{B}$ $m_1=-a/b=-A/B$

Slope of the second line -

$m_{2} = - \frac{a}{b} = - \frac{B}{- A} = \frac{B}{A}$ $m_2=-a/b=-B/(-A)=B/A$

Substituting these values in the formula $m_{1} \times m_{2} = - 1$ $m_1xxm_2=-1$

We get -

$-cancelA/cancelB xx cancelB/cancelA=-1$