Write the slope intercept form of the equation that passes through the point (-2,3) and is parallel to the line y= -6x -9 y=-6 - 9 y=-6 - 1 y=6x - 9 y=1/6x - 1

y = -6x -9

Step-by-step explanation:

Slope intercept form: y=mx + b

Point slope form: y-y1 = m(x-x1)

The slope is -6

So, plug in the slope for m and (-2,3) for x and y.

y-3 = -6(x-(-2)

or

y - 3 = -6(x + 2)

Now distribute,

y - 3 = -6x - 12

Simplify by isolating y and adding 3 to both sides of the equation

y - 3 = -6x - 12

  +3          + 3

y = -6x - 9

This is now in slope intercept form y = mx + b

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'.