Unit 2 Dilations Similarity And Introducing Slope — Unit Plan

In this diagram, the larger rectangle is a scaled copy of the smaller one, and the scale factor is $\frac32$ because $4 \boldcdot \frac32=6$ and $2 \boldcdot \frac32=3$. Scaled copies of rectangles have another interesting property: the diagonal of the large rectangle contains the diagonal of the smaller rectangle. This is the case for any two scaled copies of a rectangle if we line them up as shown. If two rectangles are not scaled copies of one another, then their diagonals would not match up.

A small rectangle inside a larger rectangle with same lower left point and a dashed diagonal line from that lower left point through both upper points. The small rectangle has length 4 and height 2. The large rectangle has length 6 and height 3.

Square grids can be useful for showing dilations, especially when the center of dilation and the point(s) being dilated lie at grid points. Rather than using a ruler to measure the distance between the points, we can count grid units.

For example, the dilation of point $Q$ with center of dilation $P$ and scale factor $\frac32$ will be 6 grid squares to the left and 3 grid squares down from $P$, since $Q$ is 4 grid squares to the left and 2 grid squares down from $P$ . The dilated image is marked as $Q’$.

Sometimes the square grid comes with coordinates, giving us a convenient way to name points. Sometimes the coordinates of the image can be found just using arithmetic, without having to measure.

For example, to perform a dilation with center of dilation at $(0,0)$ and scale factor 2 on the triangle with coordinates $(\text-1, \text-2)$, $(3,1)$, and $(2, \text-1)$, we can just double the coordinates to get $(\text-2, \text-4)$, $(6,2)$, and $(4, \text-2)$.

Two triangles on a coordinate plane, origin O. Horizontal axis scale negative 7 to 7 by 1’s. Vertical axis scale negative 5 to 5 by 1’s. The coordinates of the triangle are (negative 1 comma negative 2), (3 comma 1), (2 comma negative 1 ). The coordinates of the image are(negative 2 comma negative 4), (6 comma 2), (4 comma negative 2).

One important use of coordinates is to communicate geometric information precisely. Like an address in a city, they tell you exactly where to go. Because the plane is laid out in a grid, these “addresses” are simple, consisting of 2 signed numbers. 

Consider a quadrilateral $ABCD$ in the coordinate plane. Performing a dilation of $ABCD$ requires 3 vital pieces of information:

1. The coordinates of $A$, $B$, $C$, and $D$

2. The coordinates of the center of dilation

3. The scale factor 

With this information, we can dilate each of the vertices $A$, $B$, $C$, and $D$ and then draw the corresponding segments to find the dilation of $ABCD$. Without coordinates, describing the location of the new points would likely require sharing a picture of the polygon and the center of dilation.

Let’s show that triangle $ABC$ is similar to triangle $DEF$:

Two figures are similar if one figure can be transformed into the other by a sequence of translations, rotations, reflections, and dilations. There are many correct sequences of transformations, but we only need to describe one to show that two figures are similar.

One way to get from triangle $ABC$ to triangle $DEF$ follows these steps:

• Reflect triangle $ABC$ across line $f$
• Rotate $90^\circ$ counterclockwise around $D$
• Dilate with center $D$ and scale factor 2

Another way to show that triangle $ABC$ is similar to triangle $DEF$ would be to dilate triangle $DEF$ by a scale factor of $\frac12$ with center of dilation at $D$, then translate $D$ to $A$, then rotate it $90^\circ$ clockwise around $D$, and finally reflect it across the vertical line containing $DF$ so it matches up with triangle $ABC$.

Two polygons are similar when there is a sequence of translations, rotations, reflections, and dilations taking one polygon to the other. When the polygons are triangles, we only need to check that both triangles have two corresponding angles to show they are similar.

For example, triangle $ABC$ and triangle $DEF$ both have a 30-degree angle and a 45-degree angle.

We can translate $A$ to $D$ and then rotate around point $D$ so that the two 30-degree angles are aligned, giving this picture:

If 2 polygons are similar, then the side lengths in one polygon are multiplied by the same scale factor to give the corresponding side lengths in the other polygon.

For these triangles the scale factor is 2:

Here is a table that shows relationships between the lengths of the short and medium sides of the 2 triangles.

The lengths of the medium side and the short side are in a ratio of $4:3$. This means that the medium side in each triangle is $\frac43$ as long as the short side. This is true for all similar polygons: the ratio between 2 sides in one polygon is the same as the ratio of the corresponding sides in a similar polygon.

We can use these facts to calculate missing lengths in similar polygons. For example, triangles $ABC$ and  $A’B’C’$ are similar. 

Since side $BC$ is twice as long as side $AB$, side $B’C’$ must be twice as long as side $A’B’$. Since $A’B’$ is 1.2 units long and $2\boldcdot1.2=2.4$, the length of side $B’C’$ is 2.4 units.

Here is a line drawn on a grid. There are also four right triangles drawn.

<p>Four right triangles each with hypotenuse on the same line. First horizontal side 6, vertical side 4. Second horizontal side 3, vertical side 2. Third horizontal side 1, vertical side fraction 2 over 3. Fourth horizontal side 6, vertical side 4.</p>  

These four triangles are all examples of slope triangles. The longest side of a slope triangle is on the line, one side is vertical, and another side is horizontal. The slope of the line is the quotient of the vertical length and the horizontal length of the slope triangle. This number is the same for all slope triangles for the same line because all slope triangles for the same line are similar.

In this example, the slope of the line is $\frac{2}{3}$. Here is how the slope is calculated using the slope triangles:

• Points $A$ and $B$ give $2\div 3=\frac23$.
• Points $D$ and $B$ give $4\div 6=\frac23$.
• Points $A$ and $C$ give $4\div 6=\frac23$.
• Points $A$ and $E$ give $\frac23 \div 1=\frac23$.

Here is a line with a few of the points labeled.

We can use what we know about slope to decide if a point lies on a line.

First, use points and slope triangles to write an equation for the line.

• The slope triangle with vertices $(0,1)$ and $(2,5)$ gives a slope of $\frac{5-1}{2-0} =2$.
• The slope triangle with vertices $(0,1)$ and $(x,y)$ gives a slope of $\frac{y-1}{x}$.
• Since these slopes are the same, $\frac{y-1}{x} = 2$ is an equation for the line.

To check whether or not the point $(11,23)$ lies on this line, we can check that $\frac{23-1}{11} =2$. Since $(11,23)$ is a solution to the equation, it's on the line!