Question

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)

Answer 1

The point-slope form:

$y-y_1=m(x-x_1)$

m - slope

We have the line

$y-3=-\dfrac{1}{5}(x+2)\to m=-\dfrac{1}{5}$

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

$m_1m_2=-1\to m_2=-\dfrac{1}{m_1}$

Therefore the slope of ouer line is

$m=-\dfrac{1}{-\frac{1}{5}}=5$

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

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