Two distinct lines are drawn such that the first line passes through point \(P\) on the positive \(y\)-axis and the second line passes through point \(Q\) on the positive \(x\)-axis. Line segment \(PQ\) is perpendicular to both lines.
If the line through \(P\) has equation \(y=mx+k\), then determine the \(y\)-intercept of the line through \(Q\) in terms of \(m\) and \(k\).
Suggestion: If you are finding the general problem difficult to start, consider first solving a problem with a specific example for the line through \(P\), like \(y=4x+3\), and then attempt the more general problem.
We let \(l_1\) represent the first line, which passes through \(P\), and let \(l_2\) represent the second line, which passes through \(Q\).
Since \(l_1\) has equation \(y=mx+k\), we know that the slope of \(l_1\) is \(m\) and the \(y\)-intercept is \(k\). Therefore, \(P\) has coordinates \((0,k)\).
Since \(PQ\) is perpendicular to \(l_1\), the slope of \(PQ\) is the negative reciprocal of the slope of \(l_1\). Therefore, the slope of \(PQ\) is \(-\frac{1}{m}\). Since \(k\) is the \(y\)-intercept of segment \(PQ\) and the slope of \(PQ\) is \(-\frac{1}{m}\), the equation of the line through \(PQ\) is \(y=-\frac{1}{m}x+k\).
The \(x\)-coordinate of \(Q\) is equal to the \(x\)-intercept of the line with equation \(y=-\frac{1}{m}x+k\). To determine the \(x\)-intercept, we set \(y=0\) and solve for \(x\). If \(y=0\), then \(0=-\frac{1}{m}x+k\) and \(\frac{1}{m}x=k\). The result \(x=mk\) follows. Therefore, the \(x\)-intercept of the line through \(PQ\) is \(mk\) and \(Q\) has coordinates \((mk,0)\).
Since \(PQ\) is perpendicular to \(l_2\) and perpendicular to \(l_1\), it follows that \(l_2\) is parallel to \(l_1\). Thus, the slope of \(l_2\) is \(m\). Let \(b\) represent the \(y\)-intercept of \(l_2\).
Thus, \(l_2\) has equation \(y=mx+b\). Since \(Q(mk,0)\) lies on \(l_2\), we have \(0=m(mk)+b\) which simplifies to \(b=-m^2k\).
Therefore, the \(y\)-intercept of \(l_2\), the line through \(Q\), is \(-m^2k\).
For the student who solved the problem using \(y=4x+3\) as the equation of \(l_1\), you should have obtained the answer \(-48\) for the \(y\)-intercept of \(l_2\). A solution follows.
Let \(l_1\) represent the line with equation \(y=4x+3\). Let \(l_2\) represent the second line, which passes through \(Q\).
From the equation of \(l_1\) we know that the slope is \(4\) and the \(y\)-intercept is \(3\). Therefore \(P\) has coordinates \((0,3)\).
Since \(PQ\) is perpendicular to \(l_1\), the slope of \(PQ\) is the negative reciprocal of the slope of \(l_1\). Therefore, the slope of \(PQ\) is \(-\frac{1}{4}\). Since \(3\) is the \(y\)-intercept of segment \(PQ\) and the slope of \(PQ\) is \(-\frac{1}{4}\), the equation of the line through \(PQ\) is \(y=-\frac{1}{4}x+3\).
The \(x\)-coordinate of \(Q\) is equal to the \(x\)-intercept of the line with equation \(y=-\frac{1}{4}x+3\). To determine the \(x\)-intercept, we set \(y=0\) and solve for \(x\). If \(y=0\), then \(0=-\frac{1}{4}x+3\) and \(\frac{1}{4}x=3\). The result \(x=12\) follows. Therefore, the \(x\)-intercept of the line through \(PQ\) is \(12\) and \(Q\) has coordinates \((12,0)\).
Since \(PQ\) is perpendicular to \(l_2\) and perpendicular to \(l_1\), it follows that \(l_2\) is parallel to \(l_1\). Thus, the slope of \(l_2\) is \(4\). Let \(b\) represent the \(y\)-intercept of \(l_2\).
Thus, \(l_2\) has equation \(y=4x+b\). Since \(Q(12,0)\) lies on \(l_2\), we have \(0=4(12)+b\) which simplifies to \(b=-48\).
Therefore, the equation of \(l_2\) is \(y=4x-48\) and the \(y\)-intercept of \(l_2\) is \(-48\).