Question

A line passes through the point (–2, 4), and its y-intercept is (0, –6). What is the equation of the line that is perpendicular to the first line and passes through the point (5, –4)?

Answer 1

The equation of a line:

`y=mx+b`
m - the slope, b - the y-intercept

The y-intercept of the first line is -6.

`y=m_1x-6`

It passes through (x,y)=(-2,4). Plug the values into the equation and calculate m:

`4=m_1 * (-2)-6 \\ 4+6=-2m_1 \\ 10=-2m_1 \\ (10)/(-2)=m_1 \\ m_1=-5`

The slope of the first line is -5.
The product of the slopes of two perpendicular lines is -1.

`m_1 * m_2=-1 \\ -5 * m_2=-1 \\ m_2=(-1)/(-5) \\ m_2=(1)/(5)`

The slope of the second line is 1/5.

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

It passes through (x,y)=(5,-4). Plug the values into the equation and calculate b:

`-4=(1)/(5) * 5+b \\ -4=1+b \\ -4-1=b \\ b=-5 \\ \\ \boxed{y=(1)/(5)x-5}`