Question

Write the equation of a line that is perpendicular to the given line and that passes through the given point.

Answer 1

Greetings and Happy Holidays!

1) Perpendicular to $-x+5y=14$
In order for lines to be perpendicular, their slopes must be negative reciprocals.
Example of slopes with negative reciprocals: 5 and $\frac{-1}{5}$

First, rearrange the equation into slope y-intercept form:
$-x+5y=14$

$5y=x+14$

$\frac{5y}{5}=\frac{x+14}{5}$

$y=\frac{1}{5}x+\frac{14}{5}$

The slope of the equation is: \frac{1}{5}

The negative reciprocal formula: $(m_{1})(m_{2})=-1$

Solve for the negative reciprocal: 
$\frac{1}{5}m_{2}=-1$

Divide both sides by $\frac{1}{5}$
$\frac{\frac{1}{5}m_{2} }{ \frac{1}{5}} = \frac{-1}{ \frac{1}{5}}$

$m_{2}=(-1)(\frac{5}{1})$

$m_{2}=(\frac{-5}{1})$

$m_{2}=-5$

The slope of the new line is: -5

2) Passes through (-5,-2)

Create an equation with the slope discovered in slope y-intercept form.
$y=-5x+b$

Input the point the line passes through.
$(-2)=-5(-5)+b$

Solve for b (the y-intercept).
$-2=-5(-5)+b$

Multiply.
$-2=25+b$

Add -25 to both sides.
$(-2)+(-25)=b$

$-27=b$

The y-intercept is equal to -27

The Equation of the line is:
$y=-5x-27$

I hope this helped!
-Benjamin