Finding coordinates of 4th point in a quadilateral

Question

A(1,5), B(4,0), C(-3,-5) are three vertices of a parallelogram ABCD.

Find the coordinates of D, the fourth vertex of the parallelogram

*What I've Done:*I've created the structure on a number plane. I was wondering do you have to find the intersection of AC (These are the points that join up to D)

Thanks!!

Answer 1

Hint

Write $\vec{AB}=\vec{DC}$ and solve two equations knowing that $\vec{MN}=(x_N-x_M,y_N-y_M)$.

Answer 2

Hint

Distance between two points $P(p_1,p_2)$ and $Q(q_1,q_2)$ in plane is $\sqrt{(q_1-p_1)^2+(q_2-p_2)^2}$

The distance between $D(x,y)$ and $C(-3,-5)$ is equal to the distance between $A(1,5)$ and $B(4,0)$. You can find $x,y$ by the above formula.

Answer 3

To get from A to B, you have to add $(3,-5)$; that is, $$(1,5)+(3,-5)=(4,0)$$ If you add $(3,-5)$ to C, you get one correct value for D.

You get other correct answers if you see what you have to add to A to get to C, and then add that to B, or if you see what you have to add to B to get to C, and then add that to A.