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3 years ago

Find the equation of the line through ( - 10, - 8) that is perpendicular to the line through (10,6), (5,5).

Mathematics

1 answer:

Furkat [3]3 years ago

4 0

Answer:

y=-5x-58

Step-by-step explanation:

The equation of a line in slope-intercept form is y=mx+b where m is the slope and b is the y-intercept.

Perpendicular lines have opposite reciprocal slopes.

Anyways we need to find the slope of the line going through (10,6) and (5,5).

To find the slope, we are going to line up the points vertically and subtract vertically, then put 2nd difference over 1st difference. Like so:

( 10 , 6)

- ( 5 , 5)

------------------

5 1

So the slope of the line through (10,6) and (5,5) is 1/5.

The slope of a line that is perpendicular will be the opposite reciprocal of 1/5.

The opposite reciprocal of 1/5 is -5.

The line we are looking for is y=-5x+b where we need to find the y-intercept b.

y=-5x+b goes through (-10,-8)

So we can use (x,y)=(-10,-8) to find b in y=-5x+b.

y=-5x+b with (x,y)=(-10,-8)

-8=-5(-10)+b

-8=50+b

Subtract 50 on both sides:

-8-50=b

-58=b

So the equation is y=-5x-58

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[ $\frac{-7}{3\sqrt{6} }$ , $\frac{-2}{3\sqrt{6} }$ , $\frac{1}{3\sqrt{6} }$ ]

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