In the \(x\)–\(y\) plane, the point \(A\) has coordinates \((a,0)\) and the point \(B\) has coordinates \((0,b)\), where \(a\) and \(b\) are positive. The point \(P\), which is distinct from \(A\) and \(B\), has coordinates \((s,t)\). \(X\) and \(Y\) are the feet of the perpendiculars from \(P\) to the \(x\)-axis and \(y\)-axis respectively, and \(N\) is the foot of the perpendicular from \(P\) to the line \(AB\). Show that the coordinates \((x,y)\) of \(N\) are given by \[x = \frac{ab^2 - a(bt - as)}{a^2 + b^2}, \quad y = \frac{a^2b + b(bt - as)}{a^2 + b^2}.\]
Show that, if \[\left(\frac{t - b}{s}\right)\left(\frac{t}{s - a}\right) = -1,\] then \(N\) lies on the line \(XY\).
Give a geometrical interpretation of this result.
