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# 10.6: Formulas for Justifying Translations

Difficulty Level: At Grade Created by: CK-12

Objectives

The lesson objectives for justifying translations are the following formulas:

• Mid-Point Formula
• Distance Formula
• Slope Formula

## Mid-Point Formula

Concept Content

In the first lessons on transformations, you learned about translations, reflections, rotations, and dilations. The last concept in each of these lessons talked about the properties of the transformation type. One additional property is common to all of these transformations and that is mid-point. The mid-point is the average of the two endpoints in a segment. Mid-point has the symbol $M$ and the formula:

$M= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right)$

Midpoints can be used in proving figures are reflections, for example. Look at the equilateral triangle in the diagram below.

In an equilateral triangle there are three lines of symmetry. A line of symmetry is the same as a line of reflection. What is important for a line of symmetry is that each of the halves that result from drawing a line of symmetry is congruent or is the same size and shape. In the section Reflections of Geometric Shapes you learned about the line of symmetry, which is also called the mirror line.

If you were to draw the lines of symmetry in the triangle above, you would draw a line from each vertex to the midpoint on the opposite side.

$C$ is the mid-point of $AB, G$ is the midpoint of $BF$, and $H$ is the midpoint of $AF$. The lines $AG, FC,$ and $BH$ are all lines of symmetry or lines of reflection. You can now say the $\Delta ABH$ is reflected about the line $BH$ to form the $\Delta BHF$.

Guidance

Find the midpoints for the digram below in order to draw the lines of reflection (or the lines of symmetry).

$M_{AB}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) && M_{AD}= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{AB}&= \left( \frac{10+10}{2}, \frac{1+5}{2} \right) && M_{AD}= \left( \frac{10+1}{2}, \frac{5+5}{2} \right) \\M_{AB}&= \left( \frac{20}{2}, \frac{6}{2} \right) && M_{AD}= \left( \frac{11}{2}, \frac{10}{2} \right) \\M_{AB}&=(10,3) && M_{AD}=(5.5,5)$

$M_{BC}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) && M_{CD}= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{BC}&= \left( \frac{10+1}{2}, \frac{1+1}{2} \right) && M_{CD}= \left( \frac{1+1}{2}, \frac{1+5}{2} \right) \\M_{BC}&= \left( \frac{11}{2}, \frac{2}{2} \right) && M_{CD}= \left( \frac{2}{2}, \frac{6}{2} \right) \\M_{BC}&=(5.5,1) && M_{CD}=(1,3)$

As seen in the graph above, a rectangle has two lines of symmetry.

Examples

Example A

In the diagram below, $C$ is the midpoint between $A(-9, -1)$ and $B(-3, 7)$. Find the coordinates of $C$.

$M_{AB}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{AB}&= \left( \frac{-9+-3}{2}, \frac{-1+7}{2} \right) \\M_{AB}&= \left( \frac{-12}{2}, \frac{6}{2} \right) \\M_{AB}&=(-6,3)$

Example B

Find the coordinates of point $T$ on the line $ST$ knowing that $S$ has coordinates (–3, 8) and the midpoint is (12, 1).

Look at the midpoint formula.

$M= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right)$

For this problem, if you let point $T$ have the coordinates $x_1$ and $y_1$, then you need to find $x_1$ and $y_1$ using the midpoint formula.

$M_{ST}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\(12,1)&= \left( \frac{-3+x_1}{2}, \frac{8+y_1}{2} \right)$

Next you need to separate the $x$-coordinate formula and the $y$-coordinate formula to solve for your unknowns.

$12= \frac{-3+x_1}{2} \quad 1= \frac{8+y_1}{2}$

Now multiply each of the equations by 2 in order to get rid of the fraction.

$24=-3+x_1 \quad 2=8+y_1$

Now you can solve for $x_1$ and $y_1$.

$27=x_1 \quad -6=y_1$

Therefore the point $T$ in the line $ST$ has coordinates (27, –6).

Example C

Find the midpoints for the digram below in order to draw the lines of reflection (or the lines of symmetry).

$M_{IL}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) && M_{IJ}= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{IL}&= \left( \frac{2+-1}{2}, \frac{-2+1}{2} \right) && M_{IJ}= \left( \frac{2+5}{2}, \frac{-2+1}{2}\right) \\M_{IL}&= \left( \frac{1}{2}, \frac{-1}{2} \right) && M_{IJ}= \left( \frac{7}{2}, \frac{-1}{2}\right) \\& && M_{IJ}=(3.5,-0.5)$

$M_{JK}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{JK}&= \left( \frac{5+2}{2}, \frac{4+1}{2} \right) \\M_{JK}&= \left( \frac{7}{2}, \frac{5}{2} \right) \\M_{JK}&=(3.5,2.5)$

$M_{KL}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{KL}&= \left( \frac{2+-1}{2}, \frac{1+4}{2} \right) \\M_{KL}&= \left( \frac{1}{2}, \frac{5}{2} \right) \\M_{KL}&=(0.5,2.5)$

As seen in the graph above, a square has two lines of symmetry drawn from the mid-points of the opposite sides. A square actually has two more lines of symmetry that are the diagonals of the square.

Vocabulary

Line of Symmetry
The line of symmetry (or the line of reflection or the mirror line) is the line drawn so that each of the halves that result from drawing the line is congruent or is the same size and shape.
Mid-Point
The mid-point is the average of the two endpoints in a segment. Mid-point has the symbol $M$ and the formula:

$M= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right)$

Guided Practice

1. In the diagram below, $Z$ is the midpoint between $X (-5, 6)$ and $Y (3, -4)$. Find the coordinates of $Z$.

2. Find the coordinates of point $K$ on the line $JK$ knowing that $J$ has coordinates (–2, 5) and the midpoint is (10, 1).

3. A diameter is drawn in the circle as shown in the diagram below. What are the coordinates for the center of the circle, $O$?

1. $M_{XY}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{XY}&= \left( \frac{-5+3}{2}, \frac{-4+6}{2} \right) \\M_{XY}&= \left( \frac{-2}{2}, \frac{2}{2} \right) \\M_{XY}&=(-1,1)$

2. Let point $K$ have the coordinates $x_1$ and $y_1$, then find $x_1$ and $y_1$ using the midpoint formula.

$M_{JK}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\(10,1)&= \left( \frac{-2+x_1}{2}, \frac{5+y_1}{2} \right)$

Next you need to separate the $x$-coordinate formula and the $y$-coordinate formula to solve for your unknowns.

$10= \frac{-2+x_1}{2} \quad 1= \frac{5+y_1}{2}$

Now multiply each of the equations by 2 in order to get rid of the fraction.

$20=-2+x_1 \quad 2=5+y_1$

Now you can solve for $x_1$ and $y_1$.

$22=x_1 \quad -3=y_1$

Therefore the point $K$ in the line $JK$ has coordinates (22, –3).

3. $M_{AB}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{AB}&= \left( \frac{1+-4}{2}, \frac{2+7}{2} \right) \\M_{AB}&= \left( \frac{-3}{2}, \frac{9}{2} \right) \\M_{AB}&=(-1.5,4.5)$

Summary

In this concept you looked at the mid-point formula. The mid-point formula has the symbol $M$ and the formula:

$M= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right)$

Mid-point is a property of all transformations. It is particularly useful in determining the lines of symmetry or the lines of reflection in figures.

Problem Set

Find the mid-point for each line below given the endpoints:

1. Line $AB$ given $A(5, 7)$ and $B(3, 9)$.
2. Line $BC$ given $B(3, 8)$ and $C(5, 2)$.
3. Line $CD$ given $C(4, 6)$ and $D(3, 5)$.
4. Line $DE$ given $D(9, 11)$ and $E(2, 2)$.
5. Line $EF$ given $E(1, 1)$ and $F(8, 7)$.

For the following lines, one endpoint is given and then the mid-point. Find the other endpoint.

1. Line $AB$ given $A(-3, 5)$ and $M_{AB}(7, 7)$.
2. Line $BC$ given $B(2, 4)$ and $M_{BC}(4, 9)$.
3. Line $CD$ given $C(-2, 6)$ and $M_{CD}(1, 1)$.
4. Line $DE$ given $D(2, 9)$ and $M_{DE}(8, 2)$.
5. Line $EF$ given $E(-6, -5)$ and $M_{EF}(-2, 6)$.

For each of the diagrams below, find the midpoints.

## Distance Formula

Concept Content

In this second concept of lesson Formulas for Justifying Translations, you will learn how the distance formula is used in transformations. All of the transformations except dilations have the common property of side lengths being the same. Dynamic geometry software can calculate the distance formula but you can calculate it as well. In order to calculate the distance of side lengths, you use the distance formula. The distance formula has the symbol $d$ and is calculated using the formula:

$d= \sqrt{ \left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2}$

Guidance

Triangle $ABC$ has vertices $A(-5, 7), B(-8, 6)$ and $C(-3, 3)$. The triangle is reflected about the $y$-axis to form triangle $A^\prime B^\prime C^\prime$. Assuming that $\angle A= \angle A^\prime, \angle B= \angle B^\prime$,and $\angle C= \angle C^\prime$ prove the two triangles are congruent.

To prove congruence, prove that $m \overline{AB}=m \overline{A^\prime B^\prime},m \overline{AC}=m \overline{A^\prime C^\prime},$ and $m \overline{BC}=m \overline{B^\prime C^\prime}$.

$d_{AB}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{AB}&= \sqrt{\left(-5- \left(-8 \right) \right)^2+ \left(7-6 \right)^2} && d_{A^\prime B^\prime}= \sqrt{ \left(5-8 \right)^2+ \left(7-6 \right)^2} \\d_{AB}&= \sqrt{\left(3 \right)^2+ \left(1 \right)^2} && d_{A^\prime B^\prime}= \sqrt{ \left(-3 \right)^2+ \left(1 \right)^2} \\d_{AB}&= \sqrt{9+1} && d_{A^\prime B^\prime}= \sqrt{9+1} \\d_{AB}&= \sqrt{10} && d_{A^\prime B^\prime}= \sqrt{10} \\d_{AB}&=3.16 \ cm && d_{A^\prime B^\prime}=3.16 \ cm$

$d_{AC}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{A^\prime C^\prime}= \sqrt{ \left(x_2-x_1 \right)^2+ \left(y_2-y_1\right)^2} \\d_{AC}&= \sqrt{\left(-5- \left(-3 \right) \right)^2+ \left(7-3 \right)^2} && d_{A^\prime C^\prime}= \sqrt{\left(5-3 \right)^2+ \left(7-3 \right)^2} \\d_{AC}&= \sqrt{\left(-2 \right)^2+ \left(4 \right)^2} && d_{A^\prime C^\prime}= \sqrt{\left(2 \right)^2+ \left(4 \right)^2} \\d_{AC}&= \sqrt{4+16} && d_{A^\prime C^\prime}= \sqrt{4+16} \\d_{AC}&= \sqrt{20} && d_{A^\prime C^\prime}= \sqrt{20} \\d_{AC}&=4.47 \ cm && d_{A^\prime C^\prime}=4.72 \ cm$

$d_{BC}&= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} && d_{A^\prime C^\prime}= \sqrt{ \left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} \\d_{BC}&= \sqrt{\left(-8- \left(-3\right)\right)^2+ \left(6-3\right)^2} && d_{A^\prime C^\prime}= \sqrt{\left(8-3\right)^2+ \left(6-3\right)^2} \\d_{BC}&= \sqrt{\left(-5\right)^2+ \left(3\right)^2} && d_{A^\prime C^\prime}= \sqrt{\left(5\right)^2+ \left(3\right)^2} \\d_{BC}&= \sqrt{25+9} && d_{A^\prime C^\prime}= \sqrt{25+9} \\d_{BC}&= \sqrt{34} && d_{A^\prime C^\prime}= \sqrt{34} \\d_{BC}&=5.83 \ cm && d_{A^\prime C^\prime}=5.83 \ cm$

It is given that $\angle A= \angle A^\prime, \angle B= \angle B^\prime,$ and $\angle C= \angle C^\prime$, and the distance formula proved that $m \overline{AB}=m \overline{A^\prime B^\prime},m \overline{AC}=m \overline{A^\prime C^\prime},$ and $m \overline{BC}=m \overline{B^\prime C^\prime}$. Therefore the two triangles are congruent.

Examples

Example A

Line $AB$ is translated 5 units to the right and 6 units down to produce line $A^\prime B^\prime$. The diagram below shows the endpoints of lines $AB$ and $A^\prime B^\prime$. Prove the two lines are congruent.

$d_{AB}&= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} \\d_{AB}&= \sqrt{\left(-4-3\right)^2+ \left(2-2\right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(1-8\right)^2+ \left(-4- \left(-4\right)\right)^2} \\d_{AB}&= \sqrt{\left(-7\right)^2+ \left(0\right)^2} && d_{A^\prime B^\prime}= \sqrt{ \left(-7\right)^2+ \left(0\right)^2} \\d_{AB}&= \sqrt{49+0} && d_{A^\prime B^\prime}= \sqrt{49+0} \\d_{AB}&= \sqrt{49} && d_{A^\prime B^\prime}= \sqrt{49} \\d_{AB}&=7 \ cm && d_{A^\prime B^\prime}=7 \ cm$

Example B

Line $AB$ has been rotated about the origin $90^\circ$CCW to produce $A^\prime B^\prime$. The diagram below shows the lines $AB$ and $A^\prime B^\prime$. Prove the two lines are congruent.

$d_{AB}&= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} \\d_{AB}&= \sqrt{\left(-4-3\right)^2+ \left(2-2\right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(-2- \left(-2\right)\right)^2+ \left(-4-3\right)^2} \\d_{AB}&= \sqrt{\left(-7\right)^2+ \left(0\right)^2} && d_{A^\prime B^\prime}= \sqrt{ \left(0\right)^2+ \left(-7\right)^2} \\d_{AB}&= \sqrt{49+0} && d_{A^\prime B^\prime}= \sqrt{0+49} \\d_{AB}&= \sqrt{49} && d_{A^\prime B^\prime}= \sqrt{49} \\d_{AB}&=7 \ cm && d_{A^\prime B^\prime}=7 \ cm$

Example C

The diamond $ABCD$ has been reflected about the line $y = x$ to produce $A^\prime B^\prime C^\prime D^\prime$ as shown in the diagram below. Prove the two are congruent.

Since the figures are diamonds, you can conclude that all angles are the same and equal to $90^\circ$. So you have to prove that $m \overline{AB}=m \overline{A^\prime B^\prime},m \overline{AD}=m \overline{A^\prime D^\prime},m \overline{BC}=m \overline{B^\prime C^\prime},$ and $m \overline{CD}=m \overline{C^\prime D^\prime}$.

$d_{AB}&= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} \\d_{AB}&= \sqrt{\left(-6.1- \left(-3\right)\right)^2+ \left(9.3-4.9\right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(9.3- 4.9\right)^2+ \left(-6.1- \left(-3\right)\right)^2} \\d_{AB}&= \sqrt{\left(-3.1\right)^2+ \left(4.4\right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(4.4\right)^2+ \left(-3.1\right)^2} \\d_{AB}&= \sqrt{9.61+19.36} && d_{A^\prime B^\prime}= \sqrt{19.36+9.61} \\d_{AB}&= \sqrt{28.97} && d_{A^\prime B^\prime}= \sqrt{28.97} \\d_{AB}&=5.38 \ cm && d_{A^\prime B^\prime}=5.38 \ cm$

$d_{AD}&= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} && d_{A^\prime D^\prime}= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} \\d_{AD}&= \sqrt{\left(-6.1- \left(-10.5\right)\right)^2+ \left(9.3-6.2\right)^2} && d_{A^\prime D^\prime}= \sqrt{\left(9.3- 6.2\right)^2+ \left(-6.1- \left(-10.5\right)\right)^2} \\d_{AD}&= \sqrt{\left(4.4\right)^2+ \left(3.1\right)^2} && d_{A^\prime D^\prime}= \sqrt{\left(3.1\right)^2+ \left(4.4\right)^2} \\d_{AD}&= \sqrt{19.36+9.61} && d_{A^\prime D^\prime}= \sqrt{9.61+19.36} \\d_{AD}&= \sqrt{28.97} && d_{A^\prime D^\prime}= \sqrt{28.97} \\d_{AD}&=5.38 \ cm && d_{A^\prime D^\prime}=5.38 \ cm$

$d_{BC}&= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} && d_{B^\prime C^\prime}= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} \\d_{BC}&= \sqrt{\left(-3- \left(-7.4\right)\right)^2+ \left(4.9-1.8\right)^2} && d_{B^\prime C^\prime}= \sqrt{\left(4.9- 1.8\right)^2+ \left(-3- \left(-7.4\right)\right)^2} \\d_{BC}&= \sqrt{\left(4.4\right)^2+ \left(3.1\right)^2} && d_{B^\prime C^\prime}= \sqrt{\left(3.1\right)^2+ \left(4.4\right)^2} \\d_{BC}&= \sqrt{19.36+9.61} && d_{B^\prime C^\prime}= \sqrt{9.61+19.36} \\d_{BC}&= \sqrt{28.97} && d_{B^\prime C^\prime}= \sqrt{28.97} \\d_{BC}&=5.38 \ cm && d_{B^\prime C^\prime}=5.38 \ cm$

$d_{CD}&= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} && d_{C^\prime D^\prime}= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} \\d_{CD}&= \sqrt{\left(-7.4- \left(-10.5\right)\right)^2+ \left(1.8-6.2\right)^2} && d_{C^\prime D^\prime}= \sqrt{\left(1.8- 6.2\right)^2+ \left(-7.4- \left(-10.5\right)\right)^2} \\d_{CD}&= \sqrt{\left(3.1\right)^2+ \left(4.4\right)^2} && d_{C^\prime D^\prime}= \sqrt{\left(-4.4\right)^2+ \left(3.1\right)^2} \\d_{CD}&= \sqrt{9.61+19.36} && d_{C^\prime D^\prime}= \sqrt{19.36+9.61} \\d_{CD}&= \sqrt{28.97} && d_{C^\prime D^\prime}= \sqrt{28.97} \\d_{CD}&=5.38 \ cm && d_{C^\prime D^\prime}=5.38 \ cm$

Since all of the side measures are the same and $m \overline{AB}=m \overline{A^\prime B^\prime},m \overline{AD}=m \overline{A^\prime D^\prime},m \overline{BC}=m \overline{B^\prime C^\prime},$ and $m \overline{CD}=m \overline{C^\prime D^\prime}$, the two diamonds are congruent figures.

Vocabulary

Distance Formula
The distance formula has the symbol $d$ and is calculated using the formula $d= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2}$

Guided Practice

1. Line $\overline{ST}$ drawn from $S(-3, 4)$ to $T(-4, 8)$ has undergone a reflection in the $y$-axis to produce Line $S^\prime T^\prime$ drawn from $S^\prime (3, 4)$ to $T^\prime (4, 8)$. Draw the preimage and image and prove the two lines are congruent.

2. The triangle below has undergone a rotation of $90^\circ$CW about the origin. Given that all of the angles are equal, draw the translated image and prove the two figures are congruent.

3. The polygon below has undergone a translation of 7 units to the left and 1 unit up. Given that all of the angles are equal, draw the translated image and prove the two figures are congruent.

1.

$d_{ST}&= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} && d_{S^\prime T^\prime}= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} \\d_{ST}&= \sqrt{\left(-3- \left(-4\right)\right)^2+ \left(4-8\right)^2} && d_{S^\prime T^\prime}= \sqrt{\left(3-4\right)^2+ \left(4-8\right)^2} \\d_{ST}&= \sqrt{\left(1\right)^2+ \left(-4\right)^2} && d_{S^\prime T^\prime}= \sqrt{\left(-1\right)^2+ \left(-4\right)^2} \\d_{ST}&= \sqrt{1+16} && d_{S^\prime T^\prime}= \sqrt{1+16} \\d_{ST}&= \sqrt{17} && d_{S^\prime T^\prime}= \sqrt{17} \\d_{ST}&=4.12 \ cm && d_{S^\prime T^\prime}=4.12 \ cm$

2.

$d_{AB}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{AB}&= \sqrt{\left(2-7 \right)^2+ \left(2-3 \right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(2-3 \right)^2+ \left(-2- \left(-7 \right)\right)^2} \\d_{AB}&= \sqrt{\left(-5 \right)^2+ \left(-1 \right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(-1 \right)^2+ \left (5 \right)^2} \\d_{AB}&= \sqrt{25+1} && d_{A^\prime B^\prime}= \sqrt{1+25} \\d_{AB}&= \sqrt{26} && d_{A^\prime B^\prime}= \sqrt{26} \\d_{AB}&=5.10 \ cm && d_{A^\prime B^\prime}=5.10 \ cm$

$d_{AC}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{A^\prime C^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{AC}&= \sqrt{\left(2-4 \right)^2+ \left(2-6 \right)^2} && d_{A^\prime C^\prime}= \sqrt{\left(2-6 \right)^2+ \left(-2- \left(-4 \right)\right)^2} \\d_{AC}&= \sqrt{\left(-2 \right)^2+ \left(-4 \right)^2} && d_{A^\prime C^\prime}= \sqrt{\left(-4 \right)^2+ \left (2 \right)^2} \\d_{AC}&= \sqrt{4+16} && d_{A^\prime C^\prime}= \sqrt{16+4} \\d_{AC}&= \sqrt{20} && d_{A^\prime C^\prime}= \sqrt{20} \\d_{AC}&=4.47 \ cm && d_{A^\prime C^\prime}=4.72 \ cm$

$d_{BC}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{B^\prime C^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{BC}&= \sqrt{\left(7-4 \right)^2+ \left(3-6 \right)^2} && d_{B^\prime C^\prime}= \sqrt{\left(3-6 \right)^2+ \left(-7- \left(-4 \right)\right)^2} \\d_{BC}&= \sqrt{\left(3 \right)^2+ \left(-3 \right)^2} && d_{B^\prime C^\prime}= \sqrt{\left(-3 \right)^2+ \left(-3 \right)^2} \\d_{BC}&= \sqrt{9+9} && d_{B^\prime C^\prime}= \sqrt{9+9} \\d_{BC}&= \sqrt{18} && d_{B^\prime C^\prime}= \sqrt{18} \\d_{BC}&=4.24 \ cm && d_{B^\prime C^\prime}=4.24 \ cm$

3.

$d_{DE}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{D^\prime E^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{DE}&= \sqrt{\left(3.5-6 \right)^2+ \left(1-3 \right)^2} && d_{D^\prime E^\prime}= \sqrt{\left(-3.5- \left(-1 \right) \right)^2+ \left(2-4 \right)^2} \\d_{DE}&= \sqrt{\left(-2.5 \right)^2+ \left(-2 \right)^2} && d_{D^\prime E^\prime}= \sqrt{\left(-2.5 \right)^2+ \left(-2 \right)^2} \\d_{DE}&= \sqrt{6.25+4} && d_{D^\prime E^\prime}= \sqrt{6.25+4} \\d_{DE}&= \sqrt{10.25} && d_{D^\prime E^\prime}= \sqrt{10.25} \\d_{DE}&=3.20 \ cm && d_{D^\prime E^\prime}=3.20 \ cm$

$d_{EF}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{E^\prime F^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{EF}&= \sqrt{\left(6-5 \right)^2+ \left(3-6 \right)^2} && d_{E^\prime F^\prime}= \sqrt{\left(-1- \left(-2 \right) \right)^2+ \left(4-7 \right)^2} \\d_{EF}&= \sqrt{\left(1 \right)^2+ \left(-3 \right)^2} && d_{E^\prime F^\prime}= \sqrt{\left(1 \right)^2+ \left(-3 \right)^2} \\d_{EF}&= \sqrt{1+9} && d_{E^\prime F^\prime}= \sqrt{1+9} \\d_{EF}&= \sqrt{10} && d_{E^\prime F^\prime}= \sqrt{10} \\d_{EF}&=3.16 \ cm && d_{E^\prime F^\prime}=3.16 \ cm$

$d_{FG}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{F^\prime G^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{FG}&= \sqrt{\left(5-2 \right)^2+ \left(6-6 \right)^2} && d_{F^\prime G^\prime}= \sqrt{\left(-2- \left(-5 \right) \right)^2+ \left(7-7 \right)^2} \\d_{FG}&= \sqrt{\left(3 \right)^2+ \left(0 \right)^2} && d_{F^\prime G^\prime}= \sqrt{\left(3 \right)^2+ \left(0 \right)^2} \\d_{FG}&= \sqrt{9+0} && d_{F^\prime G^\prime}= \sqrt{9+0} \\d_{FG}&= \sqrt{9} && d_{F^\prime G^\prime}= \sqrt{9} \\d_{FG}&=3.00 \ cm && d_{F^\prime G^\prime}=3.00 \ cm$

$d_{GH}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{G^\prime H^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{GH}&= \sqrt{\left(2-1 \right)^2+ \left(6-3 \right)^2} && d_{G^\prime H^\prime}= \sqrt{\left(-5- \left(-6 \right) \right)^2+ \left(7-4 \right)^2} \\d_{GH}&= \sqrt{\left(1 \right)^2+ \left(3 \right)^2} && d_{G^\prime H^\prime}= \sqrt{\left(1 \right)^2+ \left(3 \right)^2} \\d_{GH}&= \sqrt{1+9} && d_{G^\prime H^\prime}= \sqrt{1+9} \\d_{GH}&= \sqrt{10} && d_{G^\prime H^\prime}= \sqrt{10} \\d_{GH}&=3.16 \ cm && d_{G^\prime H^\prime}=3.16 \ cm$

$d_{HD}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{H^\prime D^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{HD}&= \sqrt{\left(1-3.5 \right)^2+ \left(3-1 \right)^2} && d_{H^\prime D^\prime}= \sqrt{\left(-6- \left(-3.5 \right) \right)^2+ \left(4-2 \right)^2} \\d_{HD}&= \sqrt{\left(-2.5 \right)^2+ \left(2 \right)^2} && d_{H^\prime D^\prime}= \sqrt{\left(-2.5 \right)^2+ \left(2 \right)^2} \\d_{HD}&= \sqrt{6.25+4} && d_{H^\prime D^\prime}= \sqrt{6.25+4} \\d_{HD}&= \sqrt{10.25} && d_{H^\prime D^\prime}= \sqrt{10.25} \\d_{HD}&=3.20 \ cm && d_{H^\prime D^\prime}=3.20 \ cm$

Summary

In this concept of Chapter Is it a Slide, a Flip, or a Turn?, lesson Formulas for Justifying Translations, you were introduced to the distance formula. The distance formula has the symbol $d$ and is calculated using the formula:

$d= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2}$

The distance formula is used to help justify congruence by proving that the sides of a preimage have the same length as the sides of the transformed image.

Problem Set

Find the distance for each line below given the endpoints:

1. Line $AB$ given $A(5, 7)$ and $B(3, 9)$.
2. Line $BC$ given $B(3, 8)$ and $C(5, 2)$.
3. Line $CD$ given $C(4, 6)$ and $D(3, 5)$.
4. Line $DE$ given $D(9, 11)$ and $E(2, 2)$.
5. Line $EF$ given $E(1, 1)$ and $F(8, 7)$.

For each of the diagrams below, find the distances to prove congruence knowing the angles are congruent.

## Slope Formula

Concept Content

For transformations and rotations, the slope of a line can also be found to determine if the lines or figures are congruent. In this last lesson of the chapter, you will use the slope formula to calculate the slope of lines in preimages that have been translated or rotated. The slope formula has the symbol $m$ and can be calculated using the formula:

$m= \frac{y_2-y_1}{x_2-x_1}$

Lines that are parallel have the same slope. In contrast lines that are perpendicular have negative reciprocal slopes. For transformations, the following table shows how the slope formula can be used to help prove congruence.

Transformation Type Line type What happens to the slope?
Translations Parallel Slopes will remain the same
Rotation of $90^\circ$ Perpendicular Slopes will be negative reciprocals
Rotation of $180^\circ$ Parallel Slopes will be the same
Dilations Parallel Slopes will be the same

So you can use slope as another justification for transformations including translations, rotations, and dilations.

Guidance

The figure below shows a dilation of two trapezoids. Show the dilation lines have the same slope.

$m_{BC}&= \frac{6-6}{-5- \left(-9 \right)} && m_{B^\prime C^\prime}= \frac{3-3}{-2.5- \left(-4.5 \right)} \\m_{BC}&= \frac{0}{4} && m_{B^\prime C^\prime}= \frac{0}{2} \\m_{BC}&=0 && m_{B^\prime C^\prime}=0$

$m_{BC}$ and $m_{B^\prime C^\prime}$ have the same slopes.

$m_{BE}&= \frac{-3-6}{-10- \left(-9 \right)} && m_{B^\prime E^\prime}= \frac{-1.5-3}{-5- \left(-4.5 \right)} \\m_{BE}&= \frac{-9}{-1} && m_{B^\prime E^\prime}= \frac{-4.5}{-0.5} \\m_{BE}&=9 && m_{B^\prime E^\prime}=9$

$m_{BE}$ and $m_{B^\prime E^\prime}$ have the same slopes.

$m_{ED}&= \frac{-3- \left(-1 \right)}{-10- \left(-5 \right)} && m_{E^\prime D^\prime}= \frac{-1.5- \left(-0.5 \right)}{-5- \left(-2.5 \right)} \\m_{ED}&= \frac{-2}{-5} && m_{E^\prime D^\prime}= \frac{-1}{-2.5} \\m_{ED}&= \frac{2}{5} && m_{E^\prime D^\prime}= \frac{2}{5}$

$m_{ED}$ and $m_{E^\prime D^\prime}$ have the same slopes.

$m_{DC}&= \frac{-1-6}{-5- \left(-5 \right)} && m_{D^\prime C^\prime}= \frac{-0.5-3}{-2.5- \left(-2.5 \right)} \\m_{DC}&= \frac{-7}{0} && m_{D^\prime C^\prime}= \frac{3.5}{0} \\m_{DC}&=undefined && m_{D^\prime C^\prime}=undefined$

$m_{DC}$ and $m_{D^\prime C^\prime}$ have the same slopes.

The lines in Image A that correspond to the dilation lines in Image B have the same slope so are parallel.

Examples

Example A

Find the slope of the following line.

$m_{AB}&= \frac{-3-5}{-13- \left(-3 \right)} \\m_{AB}&= \frac{-8}{-10} \\m_{AB}&= \frac{4}{5}$

Example B

The figure below shows a rotation of two quadrilaterals $90^\circ$CW about the origin. Show the slopes of reflected lines have negative reciprocal slopes as one justification of $90^\circ$ rotations.

$m_{AH}&= \frac{6.1-2.5}{2.3-4} && m_{A^\prime H^\prime}= \frac{-2.3- \left(-4 \right)}{6.1-2.5} \\m_{AH}&= \frac{3.6}{-1.7} && m_{A^\prime H^\prime}= \frac{1.7}{3.6} \\m_{AH}&=-2.12 && m_{A^\prime H^\prime}=0.472$

$m_{HI}&= \frac{3.2-2.5}{8-4} && m_{H^\prime I^\prime}= \frac{-8- \left(-4 \right)}{3.2-2.5} \\m_{HI}&= \frac{0.7}{4} && m_{H^\prime I^\prime}= \frac{-4}{0.7} \\m_{HI}&=0.175 && m_{H^\prime I^\prime}=-5.71$

The lines in preimage that correspond to the rotated lines in the final image have negative reciprocal slopes so they are perpendicular.

Example C

The figure below shows a translation of line $VW$ over 4 units to the right and down 3 units. Show the slopes of the translated lines are the same as one justification of translations.

$m_{VW}&= \frac{3- \left(-2 \right)}{6- \left(-1 \right)} && m_{V^\prime W^\prime}= \frac{-5-0}{3-10} \\m_{VW}&= \frac{-5}{-7} && m_{V^\prime W^\prime}= \frac{5}{7}\\m_{VW}&= \frac{5}{7} && m_{V^\prime W^\prime}= \frac{5}{7}$

The lines $VW$ and $V^\prime W^\prime$ for the translation are parallel.

Vocabulary

Slope
The slope formula has the symbol $m$ and can be calculated using the formula:

$m= \frac{y_2-y_1}{x_2-x_1}$

Guided Practice

1. A line passes through the points (4, 25) and (10, 40). What is the slope of the line?

2. The figure below shows a translation of line $QT$ over 3 units to the left and up 6 units. Show the slopes of the translated lines are the same as one justification of translations.

3. The figure below shows a rotation of $180^\circ$ about the origin of Image A to produce Image B. Show the slopes of the rotated lines are the same as one justification of this type of rotation.

1. $m&= \frac{40-25}{10-4} \\m&= \frac{15}{6} \\m&=2.5$

2. $m_{QT}&= \frac{-6-1}{3- \left(-2 \right)} && m_{Q^\prime T^\prime}= \frac{0-7}{0- \left(-5 \right)} \\m_{QT}&= \frac{-7}{5} && m_{Q^\prime T^\prime}= \frac{-7}{5} \\m_{QT}&=-1.4 && m_{Q^\prime T^\prime}=-1.4$

The lines $QT$ and $Q^\prime T^\prime$ for the translation are parallel.

3. $m_{AB}&= \frac{1-2}{-3- \left(-8 \right)} && m_{A^\prime B^\prime}= \frac{-1- \left(-2 \right)}{3-8} \\m_{AB}&= \frac{-1}{5} && m_{A^\prime B^\prime}= \frac{1}{-5} \\m_{AB}&=-0.2 && m_{A^\prime B^\prime}=-0.2$

$m_{AC}&= \frac{1-4}{-3- \left(-5 \right)} && m_{A^\prime C^\prime}= \frac{-1- \left(-4 \right)}{3-5} \\m_{AC}&= \frac{-3}{2} && m_{A^\prime C^\prime}= \frac{3}{-2} \\m_{AC}&=-1.5 && m_{A^\prime C^\prime}=-1.5$

$m_{BC}&= \frac{2-4}{-8- \left(-5 \right)} && m_{B^\prime C^\prime}= \frac{-2- \left(-4 \right)}{8-5} \\m_{BC}&= \frac{-2}{-3} && m_{B^\prime C^\prime}= \frac{2}{3} \\m_{BC}&=0.67 && m_{B^\prime C^\prime}=0.67$

The lines in Image A that correspond to the rotated lines in Image B have the same slope so they are parallel.

Summary

In this last concept of lesson Formulas for Justifying Translations, and the chapter Is It a Slide, a Flip, or a Turn?, you learned that the slope can be used as a justification for translations, rotations, and dilations. With reflections, it depends on the type of reflection and the slope cannot always be used as a justification.

The slope formula has the symbol $m$ and can be calculated using the formula:

$m= \frac{y_2-y_1}{x_2-x_1}$

Lines that are parallel have the same slope. In contrast lines that are perpendicular have negative reciprocal slopes. For transformations, the following table shows how the slope formula can be used to help justify congruence.

Transformation Type Line type What happens to the slope?
Translations Parallel Slopes will remain the same
Rotation of $90^\circ$ Perpendicular Slopes will be negative reciprocals
Rotation of $180^\circ$ Parallel Slopes will be the same
Dilations Parallel Slopes will be the same

Therefore you can use slope as another justification for transformations including translations, rotations, and dilations.

Problem Set

Find the slope for each line below given the endpoints:

1. Line $AB$ given $A(5, 7)$ and $B(3, 9)$.
2. Line $BC$ given $B(3, 8)$ and $C(5, 2)$.
3. Line $CD$ given $C(4, 6)$ and $D(3, 5)$.
4. Line $DE$ given $D(9, 11)$ and $E(2, 2)$.
5. Line $EF$ given $E(1, 1)$ and $F(8, 7)$.

Justify the following transformations of the preimage A to the transformed images in the diagrams below using the slope formula.

## Summary

In this sixth lesson of Chapter Is it a Slide, a Flip, or a Turn? you learned some of the formulas for justifying transformations. You learned first about the mid-point formula that is used to help justify transformations. It is a property common to all transformations. The mid-point formula has the symbol $M$ and is found using the formula:

$M= \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)$

You next learned about the distance formula. You can use dynamic geometry software to determine distance but you could also calculate the distance given two points. The distance formula has the symbol $d$ and is found using the formula:

$d= \sqrt{ \left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2}$

Finally, you learned in concept 3 the slope formula. You could use the slope formula to justify the property of parallelism if it exists between two or more figures. Remember that parallel lines have the same slope. The slope formula has the symbol $m$ and is found using the formula:

$m= \frac{y_2-y_1}{x_2-x_1}$

Keep in mind that reflections can result in perpendicular lines not parallel ones. Perpendicular lines have negative reciprocal slopes. You see perpendicular lines all the time, such as the lines on a tennis court. (Tennis courts have parallel lines as well!)

The graph above is a sketch of the lines on a tennis court. The green lines are an example of lines that are parallel. The red lines are an example of lines that are perpendicular. You also learned that the parallel and perpendicular slopes depend on the type of transformation. The table below shows the relationship between slope and transformation type.

Transformation Type Line type What happens to the slope?
Translations Parallel Slopes will remain the same
Rotation of $90^\circ$ Perpendicular Slopes will be negative reciprocals
Rotation of $180^\circ$ Parallel Slopes will be the same
Dilations Parallel Slopes will be the same

Mid-point, distance, and slope can be used for justification of transformations.

For chapter Is it a Slide, a Flip, or a Turn? you have worked with the four types of transformations, graphed them, wrote notations, worked with the properties of transformations, and then justified them. The table below is a summary of the concepts learned.

Transformation Notation Mapping Rule Properties
Translation $T_{a,b}(x,y)=(x+a,y+b)$ $(x,y) \rightarrow (x+a,y+b)$

1. distance

2. angle measure

3. parallelism

4. colinearity

5. Mid-point

Reflection

$r_{x-axis}(x,y)=(x,-y)$

$r_{y-axis}(x,y)=(-x,y)$

$r_{y=x}(x,y)=(y,x)$

$r_{y=-x}(x,y)=(-y,-x)$

$r_{origin}(x,y)=(-x,-y)$

$(x,y) \rightarrow (x,-y)$

$(x,y) \rightarrow (-x,y)$

$(x,y) \rightarrow (y,x)$

$(x,y) \rightarrow (-y,-x)$

$(x,y) \rightarrow (-x,-y)$

Rotation

$R_{90^\circ}(x,y)=(-y,x)$

$R_{180^\circ}(x,y)=(-x,-y)$

$R_{270^\circ}(x,y)=(y,-x)$

$(x,y) \rightarrow (-y,x)$

$(x,y) \rightarrow (-x,-y)$

$(x,y) \rightarrow (y,-x)$

Dilation $D_k(x,y)=(kx,ky)$ $(x,y) \rightarrow (kx,ky)$

1. angle measure

2. parallelism

3. colinearity

4. midpoint

## Date Created:

May 28, 2014

Jan 14, 2015
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