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)?

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The equation of a line:
[tex]y=mx+b[/tex]
m - the slope, b - the y-intercept

The y-intercept of the first line is -6.
[tex]y=m_1x-6[/tex]

It passes through (x,y)=(-2,4). Plug the values into the equation and calculate m:
[tex]4=m_1 \times (-2)-6 \\ 4+6=-2m_1 \\ 10=-2m_1 \\ \frac{10}{-2}=m_1 \\ m_1=-5[/tex]

The slope of the first line is -5.
The product of the slopes of two perpendicular lines is -1.
 [tex]m_1 \times m_2=-1 \\ -5 \times m_2=-1 \\ m_2=\frac{-1}{-5} \\ m_2=\frac{1}{5}[/tex]

The slope of the second line is 1/5.
[tex]y=\frac{1}{5}x+b[/tex]

It passes through (x,y)=(5,-4). Plug the values into the equation and calculate b:
[tex]-4=\frac{1}{5} \times 5+b \\ -4=1+b \\ -4-1=b \\ b=-5 \\ \\ \boxed{y=\frac{1}{5}x-5}[/tex]