Question

Need help with these 4 problems will give brainliest to the person who answers them all

Answer 1

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 :)