Solve (-36 4/9)- (-10 2/9)- (18 2/9)

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like,   however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

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'.

y + 11 = (1/4)(x + 16)

step-by-step explanation: