Find the equation of the line that is perpendicular to y = –2x + 5 and contains the point
(2,–1).

The given equation is:

To find the line perpendicular to it, we interchange coefficients and switch the signs of one coefficient.

The equation to a line perpendicular to it is:

$ 2y-x=c$

where, $c$ is some constant we have determine using the condition given.

It passes through $(2,-1)$

Put the point in our equation:

$2(-1)-(2)=c$

$c=-2-2$

$c=-4$

The final equation is:

$\boxed{ 2y-x=-4}$

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'.

r = 78539458390583095803458309458039485093485 or just 3

step-by-step explanation: