Question

quadrilateral abcd has vertices a(-3, 1), b(-5, 3), c(-2, 5), and d(-1, 3). what are the coordinates of the vertices of the image after a translation along the vector <5, -3>? a. a(2, -2), b(0, 0), c(3, 2), and d(4, 0) b. a(-8, -2), b(-10, 0), c(-7, 2), and d(-6, 0) c. a(2, 4), b(0, 6), c(3, 8), and d(4, 6) d. a(-8, 4), b(-10, 6), c(-7, 8), and d(-6, 6)

Explanation:

Step1: Recall translation rule

To translate a point $(x,y)$ along the vector $\langle a,b
angle$, the new - point $(x',y')$ is given by $(x',y')=(x + a,y + b)$.

Step2: Translate point A

For point $A(-3,1)$ and vector $\langle5,-3
angle$, $x=-3,y = 1,a = 5,b=-3$. Then $x'=-3 + 5=2$ and $y'=1+( - 3)=-2$. So $A'=(2,-2)$.

Step3: Translate point B

For point $B(-5,3)$ and vector $\langle5,-3
angle$, $x=-5,y = 3,a = 5,b=-3$. Then $x'=-5 + 5=0$ and $y'=3+( - 3)=0$. So $B'=(0,0)$.

Step4: Translate point C

For point $C(-2,5)$ and vector $\langle5,-3
angle$, $x=-2,y = 5,a = 5,b=-3$. Then $x'=-2 + 5=3$ and $y'=5+( - 3)=2$. So $C'=(3,2)$.

Step5: Translate point D

For point $D(-1,3)$ and vector $\langle5,-3
angle$, $x=-1,y = 3,a = 5,b=-3$. Then $x'=-1 + 5=4$ and $y'=3+( - 3)=0$. So $D'=(4,0)$.

Answer:

A. $A'(2,-2),B'(0,0),C'(3,2)$, and $D'(4,0)$