# Find the equation of the line perpendicular to the line -4x+y=-23; containing the point (8,8) . Express the equation in general form or slope-intercept form The equation of the line is?...

Find the equation of the line perpendicular to the line -4x+y=-23; containing the point (8,8) . Express the equation in general form or slope-intercept form

The equation of the line is?

(simplify your answer. Type in general form or slope intercept form)

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### 1 Answer

We want the equation of the line perpendicular to the line -4x+y=-23 containing the point (8,8).

If two lines are perpendicular, then their slopes are opposite reciprocals (or the product of their slopes will be -1). The slope of the given line is 4. (We can rewrite in slope-intercept form: y=4x-23 or you might know that if you are given a line in standard form Ax+By=C, then the slope is `-A/B` )

Thus the slope of the required line will be `m=-1/4` .

We now have a point and the slope, so we can use the point-slope form: `y-y_1=m(x-x_1)` where the given point is `(x_1,y_1)`

So `y-8=-1/4(x-8)==>y-8=-1/4x+2 ==> y=-1/4x+10`

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The equation we seek is `y=-1/4x+10` or in standard form x+4y=10.

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