Question

name the line of reflection used to map each preimage to its image.
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1. b(-3, -4) is the image of b after a reflection in the y - axis. what are the coordinates of b?
2. r(-1, 9) is the image of r after a reflection in the x - axis. what are the coordinates of r?
3. k(6, 5) is the image of k after a reflection in the line y = 2. what are the coordinates of k?
4. z(5, -7) is the image of z after a reflection in the line y = x. what are the coordinates of z?

© gina wilson (all things algebra®, llc). 2015

Explanation:

Step1: Recall reflection rules

For reflection over y - axis, the rule is $(x,y)\to(-x,y)$. For reflection over x - axis, the rule is $(x,y)\to(x, - y)$. For reflection over $y = a$, the rule is $(x,y)\to(x,2a - y)$. For reflection over $y=x$, the rule is $(x,y)\to(y,x)$.

Step2: Solve for B

Given $B'(-3,-4)$ is the image of $B$ after reflection in the $y$ - axis. Using the rule $(x,y)\to(-x,y)$, if $-x=-3$ and $y = - 4$, then $x = 3$ and $y=-4$. So $B(3,-4)$.

Step3: Solve for R

Given $R'(-1,9)$ is the image of $R$ after reflection in the $x$ - axis. Using the rule $(x,y)\to(x,-y)$, if $x=-1$ and $-y = 9$, then $y=-9$. So $R(-1,-9)$.

Step4: Solve for K

Given $K'(6,5)$ is the image of $K$ after reflection in the line $y = 2$. Using the rule $(x,y)\to(x,2a - y)$ with $a = 2$, we have $x = 6$ and $2a-y=5$. Substituting $a = 2$ into $2a - y=5$ gives $4 - y=5$, so $y=-1$. So $K(6,-1)$.

Step5: Solve for Z

Given $Z'(5,-7)$ is the image of $Z$ after reflection in the line $y=x$. Using the rule $(x,y)\to(y,x)$, if $y = 5$ and $x=-7$. So $Z(-7,5)$.

Answer:

1. $B(3,-4)$
2. $R(-1,-9)$
3. $K(6,-1)$
4. $Z(-7,5)$