Rectilinear shapesWorked example

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 \(y - 0 = \frac{3}{2}\left( {x - ( - 3)} \right)\)

\(2y = 3(x + 3)\)

\(2y=3x+9\)

\(2y - 3x = 9\)

\(3x-2y+9=0\)

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 \(y - 4 = - \frac{1}{2}(x - ( - 3))\)

\(2y - 8 = - (x + 3)\)

\(2y - 8 = - x - 3\)

\(2y + x = 5\)

\(2y + x = 5\)

\(2y - 3x = 9\)

\(4x = - 4\)

\(\Rightarrow x = - 1\)

Substitute \(x = - 1\) in \(2y + x = 5\)

\(2y-1=5\)

\(2y = 6\)

\(y = 3\)

The point of intersection is \(S( - 1,3)\)

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 \(PR\).