# Problem of the Week Problem D and Solution Reflect on This

## Problem

$$\triangle ABC$$ has vertices $$A(2,1)$$, $$B(5,8)$$, and $$C(1,c)$$, where $$c>0$$.

Vertices $$A$$ and $$B$$ are reflected in the $$y$$-axis, and vertex $$C$$ is reflected in the $$x$$-axis. The three image points are collinear. That is, a line passes through the three image points.

Determine the coordinates of $$C$$.

## Solution

When a point is reflected in the $$y$$-axis, the image point has the same $$y$$-coordinate and the $$x$$-coordinate is $$-1$$ multiplied by the pre-image $$x$$-coordinate. Thus, the image of $$A(2,1)$$ is $$A'(-2,1)$$, and the image of $$B(5,8)$$ is $$B'(-5,8)$$.

When a point is reflected about the $$x$$-axis, the image point has the same $$x$$-coordinate and the $$y$$-coordinate is $$-1$$ multiplied by the pre-image $$y$$-coordinate. Thus, the image of $$C(1,c)$$ is $$C'(1,-c)$$.

The three image points, $$A'$$, $$B'$$, and $$C'$$, are collinear.

Solution 1

In this solution, we find the equation of the line through the three image points. We begin by first determining the slope of the line, and then the $$y$$-intercept.

$\text{slope}(A'B')=\frac{8 - 1}{-5-(-2)} = -\frac{7}{3}$

Since $$A'(-2,1)$$ lies on the line, we can substitute $$x=-2$$, $$y=1$$, and $$m=-\frac{7}{3}$$ into $$y=mx+b$$. \begin{aligned} 1&=-\frac{7}{3}\left(-2\right)+b\\ 1&=\frac{14}{3}+b \\ b&=1 -\frac{14}{3}\\ &=-\frac{11}{3} \end{aligned}

Thus, the equation of the line through the three image points is $$y=-\frac{7}{3}x-\frac{11}{3}$$. Since the point $$C'(1,-c)$$ lies on this line, we can substitute $$x=1$$ and $$y=-c$$ into the equation to solve for $$c$$.

Thus, $$-c=-\frac{7}{3}\left(1\right)-\frac{11}{3}=-\frac{18}{3}=-6$$ and $$c=6$$ follows.

Therefore, the coordinates of $$C$$ are $$(1,6)$$.

Solution 2

Since $$A'(-2,1)$$, $$B'(-5,8)$$, and $$C'(1,-c)$$ are collinear, slope$$(A'B')=$$ slope$$(B'C')$$. \begin{aligned} \mbox{slope}(A'B')&=\mbox{slope}(B'C')\\ \frac{8 - 1}{-5-(-2)}&=\frac{-c-8}{1-(-5)}\\ \frac{7}{-3}&=\frac{-c-8}{6}\\ 42&=3c+24\\ 18&=3c\\ 6&=c \end{aligned}

Therefore, the coordinates of $$C$$ are $$(1,6)$$.