The midpoint of UV is (5, -11). The coordinates of one endpoint are U(3, 5). Find the coordinates of endpoint V. I don't understand so can someone please help me and explain the process?

Question

polaris_s0i · Answer

You just need to find the equation of the line and then use symmetry to figure this out.

$$y = mx + b$$
$$y = \frac{\Delta y}{\Delta x} x + b$$
$$y = \frac{y_2 - y_1}{x_2 - x_1} x + b$$
$$y = \frac{(-11) - 5}{5 - 3} x + b$$
$$y = \frac{-16}{2} x + b$$
$$y = -8x + b$$

substitute in a point to figure out b (I'll put U in).
$$5 = (-8)(3) + b$$
$$5 = -24 + b$$
$$b = 29$$

So the equation of you're line is:
$$ y = -8x + 29 $$

Now the symmetry part:
The midpoint is +2 in the x  direction, so add another 2 to the midpoint to get V's x coordinate.

$$y_3 = -8x_3 + 29$$
$$y_3 = -8(7) + 29$$
$$y_3 = -56 + 29$$
$$y_3 = -27$$

So V should be (7, -27)

to check, calculate the distance from U to the midpoint.

$$s = \sqrt{(\Delta y)^2 + (\Delta x)^2}$$
$$s = \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}$$
$$s = \sqrt{(-11 - 5)^2 + (5 - 3)^2}$$
$$s = \sqrt{(-16)^2 + (2)^2}$$
$$s = \sqrt{256 + 4}$$
$$s = \sqrt{260}$$

then do the same from the midpoint to V

$$s = \sqrt{(\Delta y)^2 + (\Delta x)^2}$$
$$s = \sqrt{(y_3 - y_2)^2 + (x_3 - x_2)^2}$$
$$s = \sqrt{((-27) - (-11))^2 + (7 - 5)^2}$$
$$s = \sqrt{(-16)^2 + (2)^2}$$
$$s = \sqrt{256 + 4}$$
$$s = \sqrt{260}$$

The lengths are the same, so we are correct.