Can we reflect this point in this line?

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Let \(L\) be the line with equation \(3x + 4y = 50\).

The point \(P(3,4)\) and its reflection \(P'\) in the line \(L\) are related in two ways: the line \(PP'\) is perpendicular to the line \(L\), and \(P\) and \(P'\) are equidistant (equal distances) from \(L\). (So in terms of vectors, \(\overrightarrow{PM}=\overrightarrow{MP'}\), where \(M\) is the point of intersection of \(PP'\) and \(L\).)

Now \(L\) has gradient \(-\frac{3}{4}\), so any line perpendicular to \(L\) will have gradient \(\frac{4}{3}\).

The perpendicular line \(PP'\) passing through \(P(3,4)\) therefore has equation \[y - 4 = \frac{4}{3}(x - 3),\] that is, \[y = \frac{4}{3}x.\]

From substituting this into the equation for \(L\), we see that \(L\) and \(PP'\) intersect at \(M\), where \[3x + 4 \times \frac{4}{3}x = 50,\] and rearranging this gives \[\frac{25}{3} x = 50,\] that is, \[x = 6.\] Then substituting this into either equation gives \(y = 8\), so \(M\) lies at \((6,8)\).

We must now use this information to find the reflected point. We can see that the difference in \(x\)-coordinates between \((3,4)\) and \((6,8)\) is \(3\), and the difference in \(y\)-coordinates is \(4\). So to get the reflected point \(P'\), we must add \(3\) to the \(x\)-coordinate and \(4\) to the \(y\)-coordinate of \(M(6,8)\), as illustrated below:

This tells us that the coordinate of the reflected point is \((9,12)\).

The answer is (a).