Log In Start studying!

Select your language

Suggested languages for you:

Americas

English (US)

Europe

Q36

Expert-verified

Geometry

Found in: Page 171

Book edition Student Edition

Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

Pages 227 pages

ISBN 9780395977279

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later

Short Answer

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 $(- 1, 4)$ , $(- 1, - 4)$ and $(5, 0)$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Description of step.

Consider the parallelogram be ABCD.

Consider the points $(- 1, 0), (2, - 2), (2, 2)$ and $(x, y)$ 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 $X (x_{1}, y_{1})$ and $Z (x_{2}, y_{2})$ is given by:

$Y (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .

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

$\begin{matrix} O (\frac{- 1 + 2}{2}, \frac{0 + 2}{2}) = O (\frac{1}{2}, \frac{2}{2}) \\ = O (\frac{1}{2}, 1) \end{matrix}$

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

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

$\begin{matrix} (\frac{1}{2}, 1) = (\frac{2 + x}{2}, \frac{- 2 + y}{2}) \\ \Rightarrow \frac{2 + x}{2} = \frac{1}{2} and \frac{- 2 + y}{2} = 1 \end{matrix}$

Therefore, it can be obtained that:

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

and

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

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

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

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

Step 2. Description of step.

Consider the parallelogram be ABCD.

Consider the points $(- 1, 0), (2, - 2), (2, 2)$ and $(x, y)$ 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 $X (x_{1}, y_{1})$ and $Z (x_{2}, y_{2})$ is given by:

$Y (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .

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

$\begin{matrix} O (\frac{- 1 + 2}{2}, \frac{0 - 2}{2}) = O (\frac{1}{2}, \frac{- 2}{2}) \\ = O (\frac{1}{2}, - 1) \end{matrix}$

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

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

$\begin{matrix} (\frac{1}{2}, - 1) = (\frac{2 + x}{2}, \frac{2 + y}{2}) \\ \Rightarrow \frac{2 + x}{2} = \frac{1}{2} and \frac{2 + y}{2} = - 1 \end{matrix}$

Therefore, it can be obtained that:

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

and

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

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

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

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

Step 3. Description of step.

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 $X (x_{1}, y_{1})$ and $Z (x_{2}, y_{2})$ is given by:

$Y (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .

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

$\begin{matrix} O (\frac{2 + 2}{2}, \frac{2 - 2}{2}) = O (\frac{4}{2}, \frac{0}{2}) \\ = O (2, 0) \end{matrix}$

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

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

$\begin{matrix} (2, 0) = (\frac{- 1 + x}{2}, \frac{0 + y}{2}) \\ \Rightarrow \frac{- 1 + x}{2} = 2 and \frac{0 + y}{2} = 0 \end{matrix}$

Therefore, it can be obtained that:

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

and

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

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

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

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

Step 4. Description of step.

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

Most popular questions for Math Textbooks

Prove Theorem 5-3.

State the principal definition or theorem that enables you to deduce, from the information given, that quadrilateral SACK is a parallelogram.

$∠ S K C ≅ ∠ C A S$ ; $∠ K C A ≅ ∠ A S K$

Quadrilateral ABCD is a parallelogram. Name the principal theorem or definition that justifies the statement.

$\bar{A D} ≅ \bar{B C}$

Find something interesting to prove. Then prove it. Answers may vary.

Given: $▱ EFIH; ▱ EGHJ; ∠ 1 ≅ ∠ 2$

Given: $Plane X ∥ Plane Y; \bar{LM} ≅ \bar{ON}$

Prove: LMNO is a parallelogram.

Want to see more solutions like these?

Sign up for free to discover our expert answers

Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free