Question

The three vertices of a parallelogram taken in order are $(-1,0),(3,1)$ and $(2,2)$ respectively. Find the coordinates of the fourth vertex.

• A. $(-2,-1)$
• B. $(-2,1)$
• C. $(2,1)$
• D. $(2,-1)$

Answer 1

Answer: (B)

$(-2,1)$

If $ABCD$ is a parallelogram with vertices $A(-1,0)$, $B(3,1)$, $C(2,2)$ and let $D(x,y)$

We use the principal that diagonals of a parallelogram bisect each other.

Therefore, mid point of $AC$ = mid point of $BD$

Now midpoint of a line with points $(x_1,y_1)$ and $(x_2,y_2)$:

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

We find midpoint of $AC$ as follows:

$(\frac{-1+2}{2},\frac{0+2}{2})=(\frac{1}{2},\frac{2}{2})=(\frac{1}{2},1)$

And now midpoint of $BD$ is:

$(\frac{3+x}{2},\frac{1+y}{2})$

Now equate the midpoints to find the coordinates of $D$:

$(\frac{3+x}{2},\frac{1+y}{2})=(\frac{1}{2},1)$

$\Rightarrow \frac{3+x}{2}=\frac{1}{2},\frac{1+y}{2}=1$

$\Rightarrow 3+x=1,1+y=2$

$\Rightarrow x=1-3,y=2-1$

$\Rightarrow x=-2,y=1$

Hence, the coordinates of the fourth vertex of parallelogram is $(-2,1)$.