Write the equation of a line that is perpendicular to the given line and that passes through the given point. y-3=-1/5(x+2);(-2,7)

Question

asked Jul 22, 2019 189k views

1 Answer

Dan Carlson · Answer 1 · 2019-07-28T23:33:59+0000

The point-slope form:

y-y_1=m(x-x_1)

m - slope

We have the line

$y-3=-(1)/(5)(x+2)\to m=-(1)/(5)$

If m₁ and m₂ are the slopes of the perpendicular lines, then

$m_1m_2=-1\to m_2=-(1)/(m_1)$

Therefore the slope of ouer line is

m=-(1)/(-(1)/(5))=5

We have the point (-2, 7). Put the coordinates to the equation of a line in the point-slope form:

y-7=5(x-(-2))

y-7=5(x+2) use the distributive property

y-7=(5)(x)+(5)(2)

y-7=5x+10 add 7 to both sides

$\boxed{y=5x+17}$