Indicate in standard form the equation of the line passing through the given points. r(3, 3), s(-6, -6)

The slope of this line is:

3-(-6)

m = = 1

3-(-3)

and so, using the point-slope formula for the equation of a str. line,

y-3 = 1(x-3), or y-3 = x - 3, or 1x - 3 - y + 3 = 0. Then 1x - 1y = 0 (in standard form). Alternatively, y = x (slope-intercept form).

1. d.90

2. e. none of these (80)

3. a. 0

4. b. 6

5. e. none of these (10)

6. a. 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'.