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Q36

Expert-verifiedFound in: Page 171

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.

$(-1,0),(2,-2),(2,2)$

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

Consider the parallelogram be *ABCD*.

Consider the points $(-1,0),(2,-2),(2,2)$ 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:

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

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 $\left(-1,4\right)$.

Consider the parallelogram be *ABCD*.

Consider the points $(-1,0),(2,-2),(2,2)$ 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:

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

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 $\left(-1,-4\right)$.

Consider the parallelogram be *ABCD*.

Consider the points $(-1,0),(2,2),(x,y)$ 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:

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

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 $\left(5,0\right)$.

Therefore, all possibilities of the fourth vertex are $(-1,4),(-1,-4)$ and $(5,0)$.

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