\(\triangle ABC\) 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 \(C\).
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 \((1,6)\).