We know the of the of PQR. We are first asked to graph â–³ PQR. We can graph â–³ PQR by plotting the vertices P(0,0), Q(2,0), and R(0,2) and connecting them.
Next we are asked to graph the of the triangle after a over the x-axis followed by a by a of 2.
Graphing the Image
Let's start with the reflection of â–³ PQR over the x-axis. When we reflect â–³ PQR, the orientation of the figures will be reversed.
Now we can dilate the image of the reflection by a scale factor of 2. To do this, we will multiply the coordinates of the vertices of the reflected triangle by the scale factor 2.
| Vertices of Image of Reflection
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Vertices of Image of Dilation
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| P_\text{r}(0,0)
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P_\text{d}({\color{#FD9000}{2}}(0),{\color{#FD9000}{2}}(0))=P_\text{d}(0,0)
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| Q_\text{r}(2,0)
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Q_\text{d}({\color{#FD9000}{2}}(2),{\color{#FD9000}{2}}(0))=Q_\text{d}(4,0)
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| R_\text{r}(0,\text{-} 2)
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R_\text{d}({\color{#FD9000}{2}}(0),{\color{#FD9000}{2}}(\text{-}4))=R_\text{d}(0,\text{-}4)
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Next, we will graph the image of the dilation.
Now let's look at the and the image of the series of the transformations.
Finding the Side Lengths
We want to find the side lengths of the preimage and the image. Notice that QPR and angle R_\text{d}P_\text{d}Q_\text{d} are . This means that both triangles are .
We can find the lengths of the of the triangles by using the grids of the graph. Then we can use the to find the lengths of the . Let's start by finding the side lengths of â–³ PQR. The lengths of RP and PQ are 2 units by counting the grids.
Let's substitute these values into the formula.
RP^2+PQ^2=RQ^2
2^2+ 2^2=RQ^2
4+4=RQ^2
8=RQ^2
sqrt(8)=sqrt(RQ^2)
± sqrt(8)=RQ
± sqrt(2*2*2)=RQ
± sqrt(2^2*2)=RQ
± sqrt(2^2)*sqrt(2)=RQ
± 2*sqrt(2)=RQ
± 2sqrt(2)=RQ
RQ=± 2sqrt(2)
Since RQ is a length and lengths must be , we know that the value of RQ needs to be positive. The length of RQ is 2sqrt(2) units.
| Side Lengths of â–³ PQR (units)
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| RP=2
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| PQ=2
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| RQ=2sqrt(2)
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Next, we can find the lengths of \overline{P_\text{d}Q_\text{d}} and \overline{P_\text{d}R_\text{d}} by counting the grids in the graph. We can see that the legs are 4 units long.
Next, we can find the length of \overline{R_\text{d}Q_\text{d}} by using the Pythagorean Theorem again.
\overline{R_\text{d}P_\text{d}}^2+\overline{P_\text{d}Q_\text{d}}^2=\overline{R_\text{d}Q_\text{d}}^2
4^2+ 4^2=RQ^2
16+16=RQ^2
32=RQ^2
sqrt(32)=sqrt(RQ^2)
± sqrt(32)=RQ
± sqrt(2*2*2*2*2)=RQ
± sqrt(2^2*2^2*2)=RQ
± sqrt(2^2)*sqrt(2^2)*sqrt(2)=RQ
± 2*2*sqrt(2)=RQ
± 4sqrt(2)=RQ
RQ=± 4sqrt(2)
A length cannot be negative, so we can find \overline{R_\text{d}Q_\text{d}} as 4sqrt(2) units after following similar steps with finding the length of RQ. As a result, we found the lengths of each side of the preimage and the image.
| Side Lengths of â–³ PQR (units)
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Side Lengths of \triangle P_\text{d}Q_\text{d}R_\text{d} (units)
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| RP=2
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\overline{R_\text{d}P_\text{d}}=4
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| PQ=2
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\overline{P_\text{d}Q_\text{d}}=4
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| RQ=2sqrt(2)
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\quad \overline{R_\text{d}Q_\text{d}}=4\sqrt{2}
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Decide Whether the Figures are Congruent
Now we want to decide whether the figures are . Recall that congruent figures have the same shape and size. Although the shapes of the preimage and the image triangles are the same, their sizes are different.
| Side Lengths of â–³ PQR (units)
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Side Lengths of \triangle P_\text{d}Q_\text{d}R_\text{d} (units)
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Are Lengths of the Side Equal?
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| RP=2
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\overline{R_\text{d}P_\text{d}}=4
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2≠4 *
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| PQ=2
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\overline{P_\text{d}Q_\text{d}}=4
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2≠4 *
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| RQ=2sqrt(2)
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\quad \overline{R_\text{d}Q_\text{d}}=4\sqrt{2}
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2sqrt(2)≠4sqrt(2) *
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As a result, they are not congruent.
