What is the slope intercept form equation of the line that passes through (3, 4) and (5, 16)?

It is convenient to start with the 2-point form of the equation for a line.

... y - y1 = (y2 - y1)/(x2 - x1)×(x - x1)

Either point can be (x1, y1), and the other can be (x2, y2). If we take them in order, we get

... y - 4 = (16 - 4)/(5 - 3)×(x - 3) . . . . . fill in the two points

... y = 12/2(x -3) +4 . . . . . . . . . . . . . . add 4, simpliffy a bit

... y = 6x -18 +4 . . . . . . . . . . . . . . . . . eliminate parentheses

... y = 6x -14 . . . . . . . . . . . . . . . . . . . . put in slope-intercept form

77

step-by-step explanation:

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)