Question

A coordinate plane is shown. Triangle PQR has vertices P at negative 6 comma 6, Q at negative 6 comma 3, and R at negative 3 comma 3. Triangle P prime Q prime R prime has vertices at P prime negative 2 comma 2, Q prime at negative 2 comma 1, and R prime at negative 1 comma 1. What is the scale factor of dilation?

Answer 1

Answer:

The scale factor of dilation from ΔPQR to ΔP'Q'R' is 1/3

Step-by-step explanation:

The given information are;

The vertices of ΔPQR are;

P(-6, 6), Q(-6, 3), and R(-3, 3) and the coordinates of ΔP'Q'R' are P'(-2, 2), Q'(-2, 1), and R'(-1, 1)

Given that the length, l, of each side of the triangles is given by the formula for calculating length given the vertices we have;

$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

For side PQ, we have;

(x₁, y₁) = (-6, 6), (x₂, y₂) = (-6, 3)

Given that the points have equal y-values, the length is the difference of their x-values as follows;

$l_{PQ}$ = 6 - 3 = 3

Similarly, length QR is the difference in the y-values, since the x-values are equal. therefore, we have;

$l_{QR}$ = -3 - (-6) = 3

Length PR is given as follows;

(x₁, y₁) = (-6, 6), (x₂, y₂) = (-3, 3)

$l_{PR} = \sqrt{\left (3-6 \right )^{2}+\left (-3-(-6) \right )^{2}} = \sqrt{18} = 3\cdot \sqrt{2}$

$l_{PR}$ = 3·√2

For P'Q', we have

P'(-2, 2), Q'(-2, 1)

$l_{P'Q'}$ = 2 - 1 = 1

$l_{P'Q'}$ = 1

For Q'R', we have

Q'(-2, 1), R'(-1, 1)

$l_{Q'R'}$ = -1 - (-2) = -1 + 2 = 1

$l_{Q'R'}$ = 1

For P'R', we have;

P'(-2, 2), R'(-1, 1)

$l_{P'R'} = \sqrt{\left (1-2 \right )^{2}+\left (-1-(-2) \right )^{2}} = \sqrt{2} = \sqrt{2}$

$l_{P'R'}$ = √2

Therefore, given that we have

PQ/(P'Q') = PR/(P'R') = 3/1 = 3 and QR/(Q'R') = 3·√2/(√2) = 3

The scale factor of dilation from ΔPQR to ΔP'Q'R' = 1/3.