Write equation of a line in slope-intercept form that is perpendicular to y-3=-4(x+7) and passes through (-8,4)

Answer 1

Hi...

First things first, you need to rewrite the equation of the 1st line in the y=mx + b form to determine the slope. Remember that PERPENDICULAR lines has slopes whose product is -1 OR the slope of one line is the NEGATIVE RECIPROCAL of the other....m₂ = -¹/m₁

y - 3 = -4(x + 7)...distribute the -4 to all terms in the parenthesis (Distributive Property of Multiplication)

y - 3 = -4x - 28

y = -4x - 28 + 3 ...Isolate the y term by +3 on both sides

y = -4x - 25

So the slope of this 1st line, m₁ = -4 OR -⁴/₁

That means the slope of the 2nd line (which is PERPENDICULAR), m₂ = ¹/₄

With the point given (-8, 4), you can use the POINT-SLOPE formula to develop the equation of the 2nd line

with x₁ = -8 and y₁ = 4 and m = ¹/₄

y - y₁ = m (x - x₁)

y - 4 = ¹/₄(x - (-8))

y - 4 = ¹/₄(x + 8)

y - 4 = ¹/₄(x) + ¹/₄(8)

y - 4 = ¹/₄(x) + 2

y = ¹/₄(x) + 2 + 4

y = ¹/₄(x) + 6 Final Answer

Hope this helps...

Answer 2

1/4 = (y-4)/(x+8)

You should be able to "massage" this around to slope-intercept form.

Note: the slope of a line perpendicular to a given line is the negative reciprocal of the given line.

Answer 3

Hey Anastasia,

Do the same thing for this problem as you did for the Parallel one. The only difference is that a perpendicular slope is the opposite reciprocal.

y-3 = -4(x+7)

y-3 = -4x-28

y = -4x-25

The slope of this line is -4, so its opposite reciprocal is +1/4 and the point is (-8,4)

m=1/4

x1 = -8

y1 = 4

The point slope formula is

y-y1=m*(x-x1)

y-4=1/4(x-(-8))

y-4 = 1/4x + 2

y = 1/4x + 6

Answer 4

to be perpendicular, the slope is the negative inverse of -4 or 1/4

y=(1/4)x + b where b is the y intercept. Plug in the point to calculate b

4 = (1/4)(-8) + b

b = -4+2 = 6

y= (1/4)x +6