Question

Triangle PQR is transformed to triangle P'Q'R'. Triangle PQR has vertices P(8, 0), Q(6, 2), and R(−2, −4). Triangle P'Q'R' has vertices P'(4, 0), Q'(3, 1), and R'(−1, −2).

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. (4 points)

Part B: Write the coordinates of triangle P"Q"R" obtained after P'Q'R' is reflected about the y-axis. (4 points)

Part C: Are the two triangles PQR and P''Q''R'' congruent? Explain your answer. (2 points)

Answer 1

@Michele_Laino

Answer 2

first step
we have to compute the distances PQ and P'Q'

Answer 3

ok..

Answer 4

we have this:
\[PQ = \sqrt {{{\left( {8 - 6} \right)}^2} + {{\left( {0 - 2} \right)}^2}} = ...\]

Answer 5

ok..

Answer 6

please complete the computation

Answer 7

ok ill try :)

Answer 8

ok! :)

Answer 9

PQ= \[\sqrt{8}\]

Answer 10

that's right!
we get this:
\[PQ = \sqrt {{{\left( {8 - 6} \right)}^2} + {{\left( {0 - 2} \right)}^2}} = \sqrt {4 + 4} = \sqrt 8 = 2\sqrt 2 \]

Answer 11

ok!

Answer 12

next, we have:
\[P'Q' = \sqrt {{{\left( {4 - 3} \right)}^2} + {{\left( {0 - 1} \right)}^2}} = ...\]
again, please complete the computation

Answer 13

ok!

Answer 14

P'Q'=2

Answer 15

are you sure?

Answer 16

yes the square root of 2

Answer 17

that's right!
\[P'Q' = \sqrt {{{\left( {4 - 3} \right)}^2} + {{\left( {0 - 1} \right)}^2}} = \sqrt 2 \]

Answer 18

now, the dilation factor, is defined as the ratio between the lengths of corresponding sides, so we can write:
\[scale\;factor = \frac{{PQ}}{{P'Q'}} = \frac{{2\sqrt 2 }}{{\sqrt 2 }} = ...\]

Answer 19

please simplify

Answer 20

ok let me try..:)

Answer 21

2.8 / 1.4?

Answer 22

it is simple: