Equations of lines, passing though (perpendicular & parallel to) (-1,5) 2x+4y=1?

Question

Is this correct...

y-5=1/2(x-(-1))
y=1/2x+2
y=1/2x+3/2 (parallel)

y-5=-2(x-5)
y=-2x+5 (perpendicular)

Thank you!

Answer 1

First we put the equation
2x+4y=1
into slope-intercept form.

4y = -2x + 1
y = -1/2x + 1/4
The slope is -1/2.

Then using point-slope form, we find parallel equation through (-1,5): x1 is -1 and y1 is 5
(y-y1) = m(x-x1)
y-5 = -1/2(x--1)
y-5 = -1/2(x+1)
y-5 = -1/2x - 1/2
y-10/2 = -1/2x - 1/2
y = -1/2x + 9/2

Then using point-slope form, we find perpendicular equation through (-1,5) with slope 2. To get the perpendicular slope, you take the negative reciprocal. Again, x1 is -1 and y1 is 5.
(y-y1) = m(x-x1)
y-5 = 2(x--1)
y-5 = 2(x+1)
y-5 = 2x + 2
y = 2x + 7

Answer 2

The original line has the equation:
y= -1/2X+1

so a parallel line will have slope -1/2

and a perpindicular line will have slope 2

Passing through the point -1,5

y-5=-1/2(x+1)
y=-1/2X+9/2

Use the same technique for the perpendicular line:

y-5=2(x+1)
y=2x+7

j

Answer 3

exactly