Question

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

–x + 5y = 14; (–5, –2)

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
`(-1)/(5)`

First, rearrange the equation into slope y-intercept form:

`-x+5y=14`


`5y=x+14`


`(5y)/(5)=(x+14)/(5)`


`y=(1)/(5)x+(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:

`(1)/(5)m_(2)=-1`

Divide both sides by
`(1)/(5)`

`((1)/(5)m_(2) )/( (1)/(5)) = (-1)/( (1)/(5))`


`m_(2)=(-1)((5)/(1))`


`m_(2)=((-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