Question

Three vertices of parallelogram ABCD are A(2,-6),B(-1,2),C(5,3).find the coordinates of vertex D

Answer 1

Check the first picture below.

for a parallelogram, it has to have two pairs of parallel sides, so BC || AD and AB || CD.

if two lines are parallel, they have the same slope, namely the same rise and run.

notice, the line BC has a run of 6 and a rise of 1, arrows in red, therefore, AD must also have a run of 6 and a rise of 1, so if we move from A 6 units over and 1 up, we'd end up at point D.

now, for part atop

Check the 2nd picture, in a parallelogram, opposite sides are parallel and opposite angles are equal, so if WLP is 144°, then PNW is also 144°.

in a parallelogram, diagonals bisect each other, so each cut the other in halves, so if PM is 9 then MW is also 9 and thus PW is just 9+9=18.

now, about QRST.

notice what we said about a parallelogram above, thus

$\bf 8x+13=11x-23\implies 8x+36=11x\implies 36=3x \\\\\\ \cfrac{36}{3}=x\implies \boxed{12=x} \\\\[-0.35em] ~\dotfill\\\\ 2y+12=4y-4\implies 2y+16=4y\implies 16=2y \\\\\\ \cfrac{16}{2}=y\implies \boxed{8=y}$

so then TQ ==> 2(8) + 12 ==> 16 + 12 ==> 28.

∡Q ==> 8(12) + 13 ==> 96 + 13 ==> 109°.

and since in a parallelogram, adjacent interior angles add up to 180°, then ∡Q + ∡T = 180°.

∡T ==> 180 - 109 ==> 71°.