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Expert-verified Found in: Page 171 ### Geometry

Book edition Student Edition
Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Pages 227 pages
ISBN 9780395977279 # The coordinates of the three vertices of a parallelogram are given. Find all the possibilities you can for the coordinates of the fourth vertex.$\left(-1,0\right),\left(2,-2\right),\left(2,2\right)$

All possibilities of the fourth vertex are $\left(-1,4\right)$, $\left(-1,-4\right)$ and $\left(5,0\right)$.

See the step by step solution

## Step 1. Description of step.

Consider the parallelogram be ABCD.

Consider the points $\left(-1,0\right),\left(2,-2\right),\left(2,2\right)$ and $\left(x,y\right)$ as A, B, C and D respectively.

In the parallelogram, the diagonals bisect each other.

In the parallelogram ABCD, the diagonals are AC and BD.

Let O be the point where the diagonals AC and BD intersect each other.

Therefore, the midpoint of AC will be the midpoint of BD that is O is midpoint of both AC and BD.

The midpoint Y of the line segment joining the point $X\left({x}_{1},{y}_{1}\right)$ and $Z\left({x}_{2},{y}_{2}\right)$ is given by:

$Y\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)$.

Therefore, the midpoint of the line segment AC is given by:

$\begin{array}{c}O\left(\frac{-1+2}{2},\frac{0+2}{2}\right)=O\left(\frac{1}{2},\frac{2}{2}\right)\\ =O\left(\frac{1}{2},1\right)\end{array}$

Therefore, the midpoint of AC is $O\left(\frac{1}{2},1\right)$.

As, O is the midpoint of BD, therefore it can be obtained that:

$\begin{array}{c}\left(\frac{1}{2},1\right)=\left(\frac{2+x}{2},\frac{-2+y}{2}\right)\\ ⇒\frac{2+x}{2}=\frac{1}{2}\text{and}\frac{-2+y}{2}=1\end{array}$

Therefore, it can be obtained that:

$\begin{array}{c}\frac{2+x}{2}=\frac{1}{2}\\ 2+x=1\\ x=1-2\\ x=-1\end{array}$

and

$\begin{array}{c}\frac{-2+y}{2}=1\\ -2+y=2\\ y=2+2\\ y=4\end{array}$

Therefore, $x=-1$ and $y=4$.

Therefore, the coordinate of point D is $\left(-1,4\right)$.

Therefore, the coordinate of the fourth vertex is $\left(-1,4\right)$.

## Step 2. Description of step.

Consider the parallelogram be ABCD.

Consider the points $\left(-1,0\right),\left(2,-2\right),\left(2,2\right)$ and $\left(x,y\right)$ as A, B, C and D respectively.

In the parallelogram, the diagonals bisect each other.

In the parallelogram ABCD, the diagonals are AC and BD.

Let O be the point where the diagonals AC and BD intersect each other.

Therefore, the midpoint of AC will be the midpoint of BD that is O is the midpoint of both AC and BD.

The midpoint Y of the line segment joining the point $X\left({x}_{1},{y}_{1}\right)$ and $Z\left({x}_{2},{y}_{2}\right)$ is given by:

$Y\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)$.

Therefore, the midpoint of the line segment AC is given by:

$\begin{array}{c}O\left(\frac{-1+2}{2},\frac{0-2}{2}\right)=O\left(\frac{1}{2},\frac{-2}{2}\right)\\ =O\left(\frac{1}{2},-1\right)\end{array}$

Therefore, the midpoint of AC is $O\left(\frac{1}{2},-1\right)$.

As, O is the midpoint of BD, therefore it can be obtained that:

$\begin{array}{c}\left(\frac{1}{2},-1\right)=\left(\frac{2+x}{2},\frac{2+y}{2}\right)\\ ⇒\frac{2+x}{2}=\frac{1}{2}\text{and}\frac{2+y}{2}=-1\end{array}$

Therefore, it can be obtained that:

$\begin{array}{c}\frac{2+x}{2}=\frac{1}{2}\\ 2+x=1\\ x=1-2\\ x=-1\end{array}$

and

$\begin{array}{c}\frac{2+y}{2}=-1\\ 2+y=-2\\ y=-2-2\\ y=-4\end{array}$

Therefore, $x=-1$ and $y=-4$.

Therefore, the coordinate of the point D is $\left(-1,-4\right)$.

Therefore, the coordinate of the fourth vertex is $\left(-1,-4\right)$.

## Step 3. Description of step.

Consider the parallelogram be ABCD.

Consider the points $\left(-1,0\right),\left(2,2\right),\left(x,y\right)$ and as A, B, C and D respectively.

In the parallelogram, the diagonals bisect each other.

In the parallelogram ABCD, the diagonals are AC and BD.

Let O be the point where the diagonals AC and BD intersect each other.

Therefore, the midpoint of AC will be the midpoint of BD that is O is the midpoint of both AC and BD.

The midpoint Y of the line segment joining the point $X\left({x}_{1},{y}_{1}\right)$ and $Z\left({x}_{2},{y}_{2}\right)$ is given by:

$Y\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)$.

Therefore, the midpoint of the line segment BD is given by:

$\begin{array}{c}O\left(\frac{2+2}{2},\frac{2-2}{2}\right)=O\left(\frac{4}{2},\frac{0}{2}\right)\\ =O\left(2,0\right)\end{array}$

Therefore, the midpoint of BD is $O\left(2,0\right)$.

As, O is the midpoint of AC, therefore it can be obtained that:

$\begin{array}{c}\left(2,0\right)=\left(\frac{-1+x}{2},\frac{0+y}{2}\right)\\ ⇒\frac{-1+x}{2}=2\text{and}\frac{0+y}{2}=0\end{array}$

Therefore, it can be obtained that:

$\begin{array}{c}\frac{-1+x}{2}=2\\ -1+x=4\\ x=4+1\\ x=5\end{array}$

and

$\begin{array}{c}\frac{0+y}{2}=0\\ 0+y=0\\ y=0\end{array}$

Therefore, $x=5$ and $y=0$.

Therefore, the coordinate of the point C is $\left(5,0\right)$.

Therefore, the coordinate of the fourth vertex is $\left(5,0\right)$.

## Step 4. Description of step.

Therefore, all possibilities of the fourth vertex are $\left(-1,4\right),\left(-1,-4\right)$ and $\left(5,0\right)$. ### Want to see more solutions like these? 