#
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 \((1,6)\).