Find the slope‐intercept equation of the line with the following properties:Perpendicular to the line x-4y=2; containing the point (5,2).

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The slope-intercept form of a line is written as `y=mx+b` where m is the slope and b is the y-intercept. To find this equation, we need the slope of the line and any point on the line. We have the point, it is (5,2). The slope is perpendicular to the line:

`x-4y=2` rearrange

`-4y=-x+2` divide by -4

`y=1/4 x-2/4`

which has slope `1/4` . A line perpendicular to this one has a slope that is the negative reciprocal, which has slope -4.

This means the line we are looking for has slope -4 and goes through the point (5,2).

Sub into the equation of the line:

`y=mx+b`

`2=-4(5)+b` solve for b

`2+20=b`

`b=22`

**This means the equation of the line is `y=-4x+22` .**

Thanks so very much I understand

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