Answer:
Step-by-step explanation:
19)
P1=(2,3) and parallel to y = x - 4
the slope is 1 so any formula with a slope of 1 will be parallel
use the point-slope formula
y-y1 = m(x-x1)
y-3 = 1(x-2)
y-3 = x-2
y = x +1
20)
P1=(-6,-1) parallel to 2x + 3y = 3
first solve the above equation into slope-intercept form
3y = -2x +3
y = (-2/3)x +3/3
y = -2/3)x + 1
now make an equation in point-slope form with the same slope as above
y- ( -1) = (-2/3)(x- ( -6))
y+1 = (-2/3)(x +6)
y+1 = -2/3x - 4
y = -2/3x -5
y = (-2/3)x -5
21)
P1 = (-2,8) perpendicular to y = 2x + 5
when a line is perpendicular the slope is a reciprocal and the sign is switched
so we need - 1/2 to be perpendicular to the above equation
use the point-slope again
y - 8 = (-1/2)(x-(-2))
y - 8 = (-1/2)(x+2)
y - 8 = -1/2x - 1
y = -1/2x +7
y = (-1/2)x + 7 ( just making it clear that the x is not in the denominator)
22)
(-2,-11) perpendicular to x + 4y = 8
move into slope-intercept form
4y = -x +8
y = (-1/4)x +2
find the reciprocal and then change the sign so +4 is the slope
use the point-slope again
y- ( -11) = 4(x- ( -2))
y+11 = 4(x+2)
y+11 = 4x +8
y = 4x -3
there you go :)
