Question

Can someone please answer 22-26 for me?

Answer 1

22. D
j(-5, 6) & slope (m) = -4

`y-y_1=m(x-x_1) \\y-6=-4(x-(-5)) \\y-6=-4(x+5) \\y-6=-4x-20 \\y=-4x-20+6 \\y=-4x-14`

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23. D
(4, -4) (1, 2)
First, we have to find the slope


`m= (y_2-y_1)/(x_2-x_1) \\\\m= (2 - (-4))/(1-4) \\\\m=(2+4)/(-3) \\\\m=(6)/(-3) \\\\m=-2`


`y-y_1=m(x-x_1) \\ y-(-4)=-2(x-4) \\y+4=-2(x-4)`

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24. D
First, we have to find the slope
y = mx + b
where m is the slope and b is the y intercept.
y = -7x + 15 ---> m = -7

a parallel line has an identical slope, then the parallel line through (9, -6) will have slope m = -7

(9, -6) & m = -7

`y - y_1=m(x-x_1) \\y-(-6)=-7(x-9) \\y+6=-7(x-9)`

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25. I cant do it bcz the point "S" doesnt appear in the image

"The slopes of two perpendicular lines are negative reciprocals."

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26. C
E(8, 1)

Vertical lines always have the formula "x = (x coordinate)" and horizontal lines always have the formula "y = (y coordinate)"

x = 8

Answer 2

2y is c. And for 2t if you find both slopes just see if they are reciprocal which means flipping denominators in the fractions. Sorry I can't answer them all.