Find the equation of the line that passes through (1,4) and is parallel to 3x+y+2=0. Leave your answer in the form y=mx+c.

y = -3x + 7.

Step-by-step explanation:

First find the slope of the given line by converting to intercept form:

3x + y + 2 = 0

y = -3x - 2

So the slope is -3.

Then the line we want has a slope of -3 also (as it is parallel).

y - y1 = m(x - x1)   where m is the slope and (x1, y1) is a point on the line:

m = -3,  x1 = 1 and y1 = 4, so we have:

y - 4 = -3(x - 1)

y = -3x + 3 + 4

y = -3x + 7.

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'.