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

1 Answer

Jul 27, 2016

$x - 4 y = 20$ $x-4y=20$

Explanation:

For the general case:
$XXX A x + B y = C$ $color(white)("XXX")Ax+By=C$ has a slope of $m = - \frac{A}{B}$ $m=-A/B$

So $4 x + 1 = 0$ $4x+1=0$ has a slope of $m = - 4$ $m=-4$

In general if a line has a slope of $m$ $m$ all lines perpendicular to it will have a slope of $- \frac{1}{m}$ $-1/m$

There any line perpendicular to $4 x + y = 0$ $4x+y=0$ will have a slope of $\frac{1}{4}$ $1/4$ .

If such a perpendicular line passes through $(0, - 5)$ $(0,-5)$
then using the slope-point form:
$XXX y - (- 5) = \frac{1}{4} (x - 0)$ $color(white)("XXX")y-(-5)=1/4(x-0)$
or
$XXX x - 4 y = 20$ $color(white)("XXX")x-4y=20$

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