Question

Triangle PQR is transformed to triangle P′Q′R′. Triangle PQR has vertices P(4, 0), Q(0, −4), and R(−8, −4). Triangle P′Q′R′ has vertices P′(1, 0), Q′(0, −1), and R′(−2, −1). Plot triangles PQR and P′Q′R′ on your own coordinate grid. Part A: What is the scale factor of the dilation that transforms triangle PQR to triangle P′Q′R′? Explain your answer

Answer 1

Answer:

Scale factor is $\frac{1}{4}$.

Step-by-step explanation:

Given the coordinates of $\triangle PQR$ as:

$P(4, 0)\\Q(0, -4)\\R(-8, -4)$

And the coordinates of $\triangle P' Q' R'$ as:

$P'(1, 0)\\Q'(0, -1)\\R'(-2, -1)$

To find:

The scaling factor of the dilation to transform the $\triangle PQR$ to $\triangle P' Q' R'$.

Solution:

First of all, let us find the distance between the vertices i.e. the sides of the triangle.

Distance formula:

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

Where $(x_1,y_1), (x_2,y_2)$ are the coordinates of two points between which the distance is to be calculated.

$PQ = \sqrt{4^2+4^2} = 4\sqrt2$

$QR = \sqrt{8^2+0^2} = 8$

$PR = \sqrt{4^2+12^2} = 4\sqrt{10}$

Now, let us find the sides of the $\triangle P' Q' R'$:

$P' Q' = \sqrt{1^2+1^2} = \sqrt{2}$

$Q' R' = \sqrt{2^2+0^2} = 2$

$P' R' = \sqrt{3^2+1^2} = \sqrt{10}$

We can clearly see that, the sides of $\triangle PQR$ are four times the corresponding sides of $\triangle P' Q' R'$.

Therefore, the scaling factor is $\frac{1}{4}$.

Please refer to the attache image in the answer area.