Finding the coordinates of the last point to form a parallelogram

Question

Points: $$A(-2,6),\quad B(-5,0),\quad C(1,0).$$

Find the coordinates of $D$ such that $ABCD$ is a parallelogram.

My workings:

• Midpoint of $AD$ = Midpoint of $AC$
• Letting $D$ be $(X,Y)$ $$\left( \frac{2x+x}{2} , 6+\frac{y}{2} \right) = \left( -5+\frac{1}{2} , 0+\frac{0}{2} \right)$$ $$2x+x = -5+1$$ $$X= -2 $$ $$6+y= 0$$ $$Y= -6$$

Therefore, $D(-2,-6)$.

My Friend said that my answer is wrong, may I get some help as to why my Ans is wrong or is there any faster way to do it? Thanks for the help ..

Answer 1

The way I would go about finding $D$ is to find the pair of lines that are parallel to the unknown sides and find their intersection. We know that each pair of sides is parallel to each other.

The slope of $AB = (6-0)/(-2- -5)= 2$. The side parallel to $AB$ is $CD$ which passes through point $C (1, 0)$. The equation of the line that contains $CD$ is therefore $y = 2 x - 2$.

Now we need to do the same thing with $BC$. $BC$ is a horizontal line which means that it is defined by the equation $y = 0$. $AD$ is parallel to $BC$ and is defined by the line $y = 6$.

We now have two equations and we can solve them to find point $D$. When you solve it you will find that point is $(4,6)$.

Answer 2

Since you need a paralellogram, sieds $AB$ and $CD$ have to be parallel, as well as $BC$ and $DA$.

So, IMHO the easiest way is to take vector $\vec{BA}$ and translate $C$ with it.

In other words, $D = C + \vec{BA} = (1,0) + (-2-(-5),6-0) = (4,6)$.

If you still wan to use your midpoint trick, note that you need to compare midpoints of $BD$ (not $AD$) and $AC$. So, given $D(X,Y)$, you have to solve $$ \left( \frac{-5 + X}{2} , \frac{0 + Y}{2} \right) = \left( \frac{-2 + 1}{2} , \frac{6 + 0}{2} \right). $$