# At what coordinates should the fourth vertex be located?

A parallelogram is drawn on a coordinate grid so that three vertices are located at $(3, 4)$, $(-2, 4)$ and $(-4, 1)$. At what coordinates should the fourth vertex be located?

-
@HarishChandraRajpoot: What is the point of this edit to a question from 3 years ago? Did you not read this metathread and other related ones. Remember that an edit bumps the post to the front page, where it disturbs all the users. Only essential edits to old posts. –  Jyrki Lahtonen 2 days ago

Hint: Have you drawn a picture? There are three answers, one seeming more obvious to me than the others. You can connect the points with two line segments to make two sides of the parallelogram in three ways. Then through the two points that have only one side drawn so far, draw a line parallel to the other side. Where the two new lines intersect is the point you want.

Added: another way to think of it is that a parallelogram can be divided into two congruent triangles by either diagonal. One of the triangles is upside down relative to the other. Your three points form a triangle, which you can reflect in any of the three sides.

-

Let fourth point be $D=(x,y)$.

$A=(3,4),B=(-2,4),C=(-4,1)$

As, Diagonals of $||_{gm}$ bisect each other $\implies$ mid-point of $AC$ is same as the mid-point of $BD$

Now, coordinates of mid-point of $AC=(\frac{3+(-4)}{2},\frac{4+1}{2})$ and coordinates of mid-point of $BD=(\frac{x+(-2)}{2},\frac{y+4}{2})$

Thus, equating $x$-coordinate and $y$-coordinate gives $x=1$ and $y=1$.

Thus, point $D$ is $(1,1)$

-

Keep the points as A,B,C. Find mid point(A1)of B and C, Find A2 as A.A2:A2.A1 = -2:1 (external). A2 will be the 4th point opposite to A. IIIy find B2, C2.

-
What does $A.A2$ mean? What does IIIy mean? –  Gerry Myerson Sep 18 '12 at 8:59

Let the vertices of parallelogram be $A(3, 4)$, $B(-2, 4)$, $C(-4, 1)$ & $D(a, b)$ Then the lines $AD$ & $BC$ will be parallel to each other thus we have $$\text{slope of line}\space AD=\text{slope of line}\space BC$$ $$\implies \frac{b-4}{a-3}=\frac{1-4}{-4-(-2)}$$ $$\implies -2b+8=-3a+9$$ $$\implies b=\frac{3a-1}{2}\tag 1$$ Now, we have $$AD=BC$$ $$\implies \sqrt{(a-3)^2+(b-4)^2}=\sqrt{(-4-(-2))^2+(1-4)^2}$$ $$\implies (a-3)^2+\left(\frac{3a-1}{2}-4\right)^2=13$$ $$\implies (a-3)^2+\frac{9}{4}\left(a-3\right)^2=13$$ $$\implies \frac{13}{4}\left(a-3\right)^2=13$$ $$\implies \left(a-3\right)^2=4$$ $$\implies a-3=\pm 2 \quad\text{or}\quad a=3\pm 2$$ $$\text{if}\quad \color{blue}{a=5}\implies \color{blue}{b}=\frac{3(5)-1}{2}=\color{blue}{7}$$ $$\text{if}\quad \color{blue}{a=1}\implies \color{blue}{b}=\frac{3(1)-1}{2}=\color{blue}{1}$$ It can be easily checked that both the above coordinates satisfies all the conditions of a parallelogram. Hence, the fourth vertex $D$ of the parallelogram may be $\color{blue}{(1, 1)}$ or $\color{blue}{(5, 7)}$

-