Line has a slope of -3 and passes through, (0,9)
y = -3x+9
Identify a point on the line and the slope of the line in the equation:
y - 6 = 2(x + 4)
Point: (-4,6) Slope: 2
(12,8) and (4,3)
5/8
y-2=8(x-3)
8x-y=22
Rewrite in Slope-Int Form y-11=1/2(x+5)
y = 1/2x+27/2
Slope = undefined
Point = (-6,2)
x=-6
The slope of a line perpendicular to 5x+3y=10
3/5
y+17=-3(x-1)
3x+y=-14
Line through (9,4) and (0,10)
y = -2/3x+10
(9,7) and (4,5)
y - 7 = 2/5 (x - 9) or
y - 5 = 2/5(x - 4)
(-2,7) and (4,7)
0
x-intercept (3,0) and y-intercept (0,-9)
3x-y=9
Line has a slope of 0 and passes through; (8,-5)
y = -5
Perpendicular to
y-2=1/3(x+5) through (-1,8)
y-8 = -3(x+1)
(0,4) and (0,7)
Parallel to y+2=1/6(x-3) through (-2,5)
x-6y=-32
Line parallel to y-3=-5(x-2) and through the point (-6,1).
y = -5x-29
y-5=5/4(x-0) or y-5=5/4x
y-0=5/4(x+4) or y=5/4x+5
4x+7y=-10
4/7
Perpendicular to 5x+3y=12 through the point (-1,10)
3x-5y=-53