Worked example
Triangle PQR has coordinates \(P( - 3, - 4)\), \(Q( - 3,4)\) and \(R(5,12)\)
Question
a) Find the equation of the median \(MR\)
a) \(M\left( {\frac{{ - 3 + ( - 3)}}{2},\frac{{4 + ( - 4)}}{2}} \right) = M( - 3,0)\)
For \(M( - 3,0)\) and \(R(5,12)\)
\({m_{MR}} = \frac{{12 - 0}}{{5 - ( - 3)}} = \frac{{12}}{8} = \frac{3}{2}\)
Gradient of median is \(\frac{3}{2}\)
Point on median is \(( - 3,0)\)
So equation is \(y - 0 = \frac{3}{2}\left( {x - ( - 3)} \right)\)
\(2y = 3(x + 3)\)
\(2y=3x+9\)
\(2y - 3x = 9\)
\(3x-2y+9=0\)
Question
b) Find the equation of the altitude \(NQ\)
b) For \(P( - 3, - 4)\) and \(R(5,12)\):
\({m_{PR}} = \frac{{12 - ( - 4)}}{{5 - ( - 3)}} = \frac{{16}}{8} = 2\)
\(\Rightarrow {m_ \bot } = - \frac{1}{2}\,\,\,\,\,\,\,\,\,\,(since\,\, m_{1}\times m_{2} = -1)\)
Gradient of altitude is \(- \frac{1}{2}\)
Point on altitude is \(Q( - 3,4)\)
So equation is \(y - 4 = - \frac{1}{2}(x - ( - 3))\)
\(2y - 8 = - (x + 3)\)
\(2y - 8 = - x - 3\)
\(2y + x = 5\)
Question
c) Median \(MR\) and altitude \(NQ\) intersect at point \(S\). Find the coordinates of \(S\).
c) To find the point of intersection solve:
\(2y + x = 5\)
\(2y - 3x = 9\)
Subtract:
\(4x = - 4\)
\(\Rightarrow x = - 1\)
Substitute \(x = - 1\) in \(2y + x = 5\)
\(2y-1=5\)
\(2y = 6\)
\(y = 3\)
The point of intersection is \(S( - 1,3)\)
Question
d) The point \(T( 2, 9)\) lies on \(QR\). Show that \(ST\) is parallel to \(PR\).
d) From b) \({m_{PR}} = 2\)
For \(S( - 1,3)\) and \(T(2,9)\)
\({m_{ST}} = \frac{{9 - 3}}{{2 - ( - 1)}} = \frac{6}{3} = 2\)
So \({m_{ST}} = {m_{PR}} = 2\)
And so \(ST\) is parallel to \(PR\).