Can someone help?
Given the point (-3, 6) and a line y=2x-8, write an equation of a line through the point and parallel to the line.

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Explanation:

The line y = 2x-8 is in y = mx+b form with m = 2 as the slope. Anything parallel to this will have the same slope m = 2. Parallel lines always have equal slopes, but different y intercepts.

We will use m = 2 and the given point (x,y) = (-3,6) to find the y intercept b

y = mx+b

6 = 2(-3)+b

6 = -6+b

6+6 = -6+b+6

12 = b

b = 12

So y = mx+b turns into y = 2x+12

As a way to check, plug in x = -3 and we should get y = 6 as an output

y = 2x+12

y = 2*(-3)+12

y = -6+12

y = 6

So the answer is confirmed.

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You could also use the point-slope formula

m = 2 is the slope;   is the point the parallel line goes through.

So,

and we get the same answer.

step-by-step explanation:

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)