What is the equation of a line that is perpendicular to y-3=-4(x+2) and passes through the point (-5,7)

Question

asked May 15, 2020 145k views

2 Answers

Sji · Answer 1 · 2020-05-22T19:29:45+0000

The point-slope form:

y-y_1=m(x-x_1)

m - slope

Let
k:y=m_1x+b_1 and
l:y=m_2x+b_2

$l\ \perp\ k\iff m_1m_2=-1\to m_2=-(1)/(m_1)$

We have

$y-3=-4(x+2)\to m_1=-4$

Therefore

m_2=-(1)/(-4)=(1)/(4)

We have the slope and the point (-5, 7). Substitute to the point-slope formula:

$y-7=(1)/(4)(x-(-5))\\\\\boxed{y-7=(1)/(4)(x+5)}$