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

User Sherica
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2 Answers

4 votes

Answer:

y-7=1/4(x+5)

Explanation:

User Sysix
by
5.1k points
6 votes

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)}

User Sji
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5.4k points