A straight line passes through the point P(-1, 3). Another line which passes through Q(-4, 4) intersects the first line at the point R(k, 5), where k is a constant. If \(<PRQ = 90°\), find the values of k.
The point of intersection is \(R(k,5),\) with \(P(-1,3)\) and \(Q(-4,4).\) The condition \(\angle PRQ=90^\circ\) means the segments \(RP\) and \(RQ\) are perpendicular at R, so the vectors \(\overrightarrow{RP}\) and \(\overrightarrow{RQ}\) have zero dot product.
\[\overrightarrow{RP}=P-R=(-1-k,\,3-5)=(-1-k,\,-2),\]
\[\overrightarrow{RQ}=Q-R=(-4-k,\,4-5)=(-4-k,\,-1).\]
Set the dot product to zero:
\[(-1-k)(-4-k)+(-2)(-1)=0.\]
Expand \((-1-k)(-4-k)=(1+k)(4+k)=k^2+5k+4,\) so
\[k^2+5k+4+2=0\Rightarrow k^2+5k+6=0.\]
Factorise:
\[(k+2)(k+3)=0\Rightarrow k=-2\ \text{or}\ k=-3.\]
Both values give a right angle at R, so the required values are \(k=-2\) and \(k=-3.\)