Enter the equation of the line in slope-intercept form.

Slope is -2/5, and (-9,2) is on the line.

y = (-2/5)x - 8/5

Step-by-step explanation:

Start with the slope-intercept form y = mx + b.

Replace m with -2/5, y with 2 and x with -9:

2 = (-2/5)(-9) + b

Simplifying this, we get:

2 = 18/5 + b, or 10/5 = 18/5 + b

Subtracting 18/5 from both sides yields  b = -8/5, and  so

the desired equation is

y = (-2/5)x - 8/5

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'.