Recall that the point-slope form of the line through (h,k) with slope m is
y-k = m(x-h)
So,
(-4,0) parallel to y = ¾x+1
the line has slope 3/4, so your line, with the same slope is
y-0 = 3/4 (x+3)
Perpendicular lines have slopes whose product is -1
(3,-2) perpendicular to y =- 3x + 2
so you want a line with slope -1/(-3) = 1/3
y+2 = 1/3 (x-3)
Do the others in like wise
Write the equation of the PARALLEL line:
2)(-4,0) parallel to y = ¾x+1
3)(5,-2) parallel to y=-7/5x-1
4)(-4,0) parallel to y = -6/5x – 3
5)(-2,1) perpendicular to y = ⅓x-1
6)(3,3) perpendicular to y= -7/3x-2
7)(-4,-3) perpendicular to y = 4x - 2
8)(3,-2) perpendicular to y =- 3x + 2
I'm really confused on how to do this. If someone could help or explain on how to do it, I'd greatly appreciate it. Thanks!
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