Write the equation of a secant line? screenshot of the question is attached

y = 12x - 13

Step-by-step explanation:

The two points you want the line through are ...

... (-4, h(-4)) = (-4, -61)

and

... (2, h(2)) = (2, 11)

The 2-point form of the equation of a line is applicable. For (x1, y1) and (x2, y2), the equation is

... y = (y2-y1)/(x2-x1)·(x-x1) +y1

... y = (-61-11)/(-4-2)·(x -2) + 11 . . . . putting in the point values

... y = 12(x -2) + 11 . . . . . . . . . . . . . simplify

... y = 12x -13 . . . . . . . . . . . . . . . . . eliminate parentheses, collect terms

1.0

step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think 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 because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.