Question

Find (a) the equation of a line that is parallel to the given line and includes the given point, and (b) the equation of a line that is perpendicular to the given line through the given point. Write both answers in slope-intercept form. (c) Graph both of these lines on the same axes. Please show all of your work.

y = 2x + 6, (-1, -2)

Answer 1

y = 2x + 6, (-1, -2)
here the slope m = 2

Answer 2

next use the formula y-y1=m(x-x1)

Answer 3

for parallel line use m=m but for perpendicukar line of it use m=-1/m

Answer 4

for this point (-1, -2)
the parallel line is
y-y1=m(x-x1)
y-(-2)=m(x-(-1))
y+2=2(x+1)

Answer 5

for the perpendicular line of y=2x+6 we use m=-1/2
we have
y-y1=m(x-x1)
y-(-2)=m(x-(-1))
y+2=(-1/2)(x+1)

Answer 6

now mult and add
y+2=2(x+1)
y=-2+2x+2
=2x ans....this is the parallel line of y=2x+6

Answer 7

for the perpendicular line of y=2x+6 at point(-1,-2)
we use m=-1/2 we have
y-y1=m(x-x1)
y-(-2)=m(x-(-1)) here use m=-1/2
y+2=(-1/2)(x+1)

Answer 8

now mult and add y+2=-1/2 (x+1)
y=-2-1/2x+2
=-1/2x ans....this is the perpendicular line of y=2x+6 at point(-1.-2)

Answer 9

http://www.wolframalpha.com/input/?i=graph+y+%3D+2x+%2B+6%2C+from+%28-1%2C+-2%29+

Answer 10

Thank you so much Mark O.