Solve 4x- c= k for x.
O A.
x-**
k+c
4
O B. x = 4(k-c)
O c.
X =
K-C
4
O D. x = 4(k+ c)

y-3=-3/2(x+4)

ur very much welcome

question says to graph a triangle (xyz) and reflect it over the line y=x to create triangle x'y'z'.

we know that line y=x has both coordinates equal like (1,1), (5,5) etc so we can easily graph that reflecting line.

description of transformation:

we can see that x and y coordinate gets switched after transformation like (2,8) changed to (8,2)

what do you notice about the two line segments you drew:

we see that line segments xp and x'p are of same length and perpendicular to the reflecting line y=x.

do you think you would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line? :

yes, i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line.

how do you know? :

by measuring the length of yk and y'k.