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3 years ago

Find an equation of a line through (-2, 4) and perpendicular to -x+8y=16

Mathematics

2 answers:

erica [24]3 years ago

6 0

Answer: $y=-8x-12$

Step-by-step explanation:

The equation of the line has the form:

$y=mx+b$

Where m is the slope and b is the y-intercept.

Let's find the slope of the given line. Solve for y:

$-x+8y=16\\8y=x+16\\y=(1/8)x+2$

Then m=1/8

By definition the slopes of perpendicular lines are negative reciprocals, therefore the slope of the other line is:

m=-8

Substitute the slope and the point given into the equation of the line to calculate b:

$4=(-8)(-2)+b\\4=16+b\\b=-12$

Then the equation is:

$y=-8x-12$

Lena [83]3 years ago

3 0

Answer:

y = -8x-12

Step-by-step explanation:

We have given an equation of line and a point.

-x+8y=16 and (x,y) = (-2, 4)

We have to find the equation of line that is perpendicular -x+8y=16.

y = mx+c is equation of line where m is slope and c is y-intercept.

8y = x+16

y = 1/8x+2

Hence, slope of perpendicular line is -8.

y = -8x+c is equation of perpendicular line to -x+8y=16.

putting the value of point in above equation , we have

4 = -8(-2)+c

4 = 16+c

c = 4-16

c = -12

hence, y = -8x-12 is equation of line passes through (-2, 4) and perpendicular to -x+8y=16.

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