Plz, . will give ! 40 find the equation of a line through (4, -3) that is parallel to the line that passes through these points (-4,7) and (2,4). write your answer in standard form (ax + by = c). show all your work.

x + 2y = -2

Step-by-step explanation:

First, find the slope of the line that passes through (-4,7) and (2,4).  As we move to the right from -4 to 2, x increases by 6, and y decreases by 3.

Thus, the slope of this given line is m = rise / run = -3/6, or -1/2.

Then a new line which is parallel to the given one must have the same slope:  -1/2.

Starting with the slope-intercept form y = mx + b, we find the equation of the line through (4, -3) that is parallel to the line we just studied (above).

y = mx + b becomes -3 = (-1/2)(4) + b,  or  -3 = -2 + b.  Then b = -1:  

y = (-1/2)x - 1.  We must put this into the form (Ax + By = C).

Multiply all terms of y = (-1/2)x - 1 by 2 to eliminate the fractional coefficient:

2y = -x - 2.  Moving -x to the left side, we get:

x + 2y = -2

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i will pick c put check it over to be sure

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'.