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

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Answer 1 · 2016-09-14T14:55:34

$x-4y-5=0$ .

A Useful Result : The Eqn. of the line passing through $(x_0,y_0)$

and $bot$ to line $: ax+by+c=0$ is $b(x-x_0)=a(y-y_0)$ .

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

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

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

the slope of this line is $-4$ .

Therefore, the slope of the reqd. $bot$ line must be $1/4$ , which,

passes through a pt. $(5,0)$ .

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

$y-0=1/4(x-5), or, x-4y-5=0$ , as we had it earlier!.

Enjoy Maths.!

How do you write an equation of a line passing through (5, 0), perpendicular to 4x + y = -14x + y = -1?