\[-1\]

, then the triangle is right angled, We can also use this condition to find the coordinates of a vertex. The triangle below has a right angled at \[C(4,k)\]

and we want to find the value of \[k\]

.The gradients of AC and CB multiply to give

\[-1\]

The gradient of AC is

\[\frac{k-5}{4-2}=\frac{k-5}{2}\]

The gradient of CB is

\[\frac{4-k}{7-4}=\frac{4-k}{3}\]

Hence

\[\frac{k-5}{2} \frac{4-k}{3} =-1\]

Rearranging gives

\[(k-5)(4-k) =-6 \rightarrow -k^2+9k-20=-6 \rightarrow k^2-9k+14=0 \]

This factorises to give

\[(k-7)(k-2)=0\]

.Hence

\[k=7, \: 2\]

.Obviously, from the diagram above

\[k=2\]

and the point C has coordinates \[(4,2)\]