# How do you write an equation of a line passing through (5, 0), perpendicular to 4x + y = -1?

Sep 14, 2016

$x - 4 y - 5 = 0$.

#### Explanation:

A Useful Result : The Eqn. of the line passing through $\left({x}_{0} , {y}_{0}\right)$

and $\bot$ to line$: a x + b y + c = 0$ is $b \left(x - {x}_{0}\right) = a \left(y - {y}_{0}\right)$.

Using this Result, we can immediately write the reqd. eqn. of line as,

$1 \left(x - 5\right) = 4 \left(y - 0\right) , i . e . , x - 4 y - 5 = 0$

Alternatively, rewriting the given eqn. as $: y = - 4 x - 1$, we get

the slope of this line is $- 4$.

Therefore, the slope of the reqd. $\bot$ line must be $\frac{1}{4}$, which,

passes through a pt. $\left(5 , 0\right)$.

By the Slope-Point Form, the reqd. eqn. is,

$y - 0 = \frac{1}{4} \left(x - 5\right) , \mathmr{and} , x - 4 y - 5 = 0$, as we had it earlier!.

Enjoy Maths.!