Question

1. what are the coordinates of △pqr after a dilation with a scale factor of 2/3? a) p(-2,1),q(0,2),r(2,2) b) p(-4,2),q(0,4),r(4,4) c) p(-4,2),q(4,0),r(4,2) d) p(-12,6),q(0,12),r(12,12) 5. △def is the image of △def after a dilation with a scale factor of 2. what are the coordinates of the vertices of △def? a) d(-8,-12),e(8,4),f(-4,-4) b) d(-6,4),e(-2,0),f(-4,-4) c) d(-2,8),e(6,4),f(0,0) d) d(-2,3),e(2,1),f(-1,-1) short answer: 6. triangle pqr has coordinates p(2,4),q(-2,4),r(0,-6). write the coordinates of the vertices of the image of a triangle after a dilation of 1.5. 7. how does the size of an image compare to the original figure when the original figure undergoes dilation with a scale factor of one?

Explanation:

Step1: Recall dilation formula

For a point $(x,y)$ dilated with a scale - factor $k$ centered at the origin, the new coordinates $(x',y')$ are given by $(x',y')=(k\cdot x,k\cdot y)$.

Step2: Solve problem 4

Let's assume the coordinates of $\triangle PQR$ are $P(- 3,\frac{3}{2})$, $Q(0,3)$, $R(3,3)$. With a scale - factor $k = \frac{2}{3}$, for point $P$: $x'=\frac{2}{3}\times(-3)=-2$, $y'=\frac{2}{3}\times\frac{3}{2}=1$; for point $Q$: $x'=\frac{2}{3}\times0 = 0$, $y'=\frac{2}{3}\times3 = 2$; for point $R$: $x'=\frac{2}{3}\times3=2$, $y'=\frac{2}{3}\times3 = 2$. So the new coordinates are $P'(-2,1),Q'(0,2),R'(2,2)$. The answer is a).

Step3: Solve problem 5

Since $\triangle D'E'F'$ is the image of $\triangle DEF$ after a dilation with a scale - factor $k = 2$, to find the original coordinates from the dilated ones, we use the formula $(x,y)=(\frac{x'}{k},\frac{y'}{k})$. If we assume the coordinates of $\triangle D'E'F'$ are $D'(-4,6)$, $E'(4,2)$, $F'(-2,-2)$. Then for point $D$: $x=\frac{-4}{2}=-2$, $y=\frac{6}{2}=3$; for point $E$: $x=\frac{4}{2}=2$, $y=\frac{2}{2}=1$; for point $F$: $x=\frac{-2}{2}=-1$, $y=\frac{-2}{2}=-1$. The coordinates of $\triangle DEF$ are $D(-2,3),E(2,1),F(-1,-1)$. The answer is d).

Step4: Solve problem 6

Given $P(2,4)$, $Q(-2,4)$, $R(0,-6)$ and a scale - factor $k = 1.5$. For point $P$: $x'=1.5\times2 = 3$, $y'=1.5\times4 = 6$; for point $Q$: $x'=1.5\times(-2)=-3$, $y'=1.5\times4 = 6$; for point $R$: $x'=1.5\times0 = 0$, $y'=1.5\times(-6)=-9$. The new coordinates are $P'(3,6),Q'(-3,6),R'(0,-9)$.

Step5: Solve problem 7

When the scale - factor $k = 1$, using the dilation formula $(x',y')=(k\cdot x,k\cdot y)$, we have $(x',y')=(1\cdot x,1\cdot y)=(x,y)$. So the size of the image is the same as the original figure.

Answer:

1. a) $P'(-2,1),Q'(0,2),R'(2,2)$
2. d) $D(-2,3),E(2,1),F(-1,-1)$
3. $P'(3,6),Q'(-3,6),R'(0,-9)$
4. The size of the image is the same as the original figure.