Equation of altitudes and perpendicular bisectors
Now use the information you already have about finding the equation of a line and apply it to the dashed lines shown in these triangles.
Remember, to find the equation of a line you need a point \((a,b)\) and the gradient \(m\).
Equation of Median
Question
Find the equation of the median through A in triangle ABC.
Median passes through the midpoint of BC.
\(= \left( {\frac{{6 + 4}}{2},\frac{{3 + 11}}{2}} \right)\)
\(= (5,7)\)
Now we can use the midpoint and the coordinates of point A to find the gradient \(m\).
\(m = \frac{{7 - 2}}{{5 - 1}} = \frac{5}{4}\)
Use either \((1,2)\) or \((5,7)\) and \(m\) to find the equation.
\(y - b = m(x - a)\)
\(y - 2 = \frac{5}{4}(x - 1)\)
\(4y - 8 = 5x - 5\)
\(4y = 5x + 3\)
\(5x-4y+3=0\)
Watch this video to learn about equations of altitudes.
Equation of Altitude
Question
Find the equation of the altitude through B in Triangle ABC.
Altitude is perpendicular to AC:
\({m_{AC}} = \frac{{11 - 2}}{{4 - 1}} = \frac{9}{3} = 3\)
Perpendicular gradients are connected by \({m_1}\times {m_2} = - 1\)
\({m_{perp}} = - \frac{1}{3}\,\,\,\,\,\,\,\,\,\,\,\,\, (since- \frac{1}{3} \times 3 = - 1)\)
This determines the gradient of the required line.
Use B\((6,3)\) and \(m\).
\(y - b = m(x - a)\)
\(y - 3 = - \frac{1}{3}(x - 6)\)
\(3y - 9 = - x + 6\)
\(3y + x = 15\)
\(x+3y-15=0\)
Watch this video to learn about the equation of a perpendicular bisector.
Equations of Perpendicular Bisector
Question
Find the equation of the perpendicular bisector of AC in triangle ABC.
The perpendicular bisector passes through the midpoint of AC. This determines the point on the required line.
The midpoint of AC \(= \left( {\frac{{1 + 4}}{2},\frac{{2 + 11}}{2}} \right) = \left( {\frac{5}{2},\frac{{13}}{2}} \right)\)
Perpendicular bisector is perpendicular to AC.
\({m_{AC}} = \frac{{11 - 2}}{{4 - 1}} = \frac{9}{3} = 3\)
Perpendicular gradients are connected by \({m_1}\times{m_2} = - 1\). This determines the gradient of the required line.
\({m_{perp}} = - \frac{1}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,(since- \frac{1}{3} \times 3 = - 1)\)
Use \(\left( {\frac{5}{2},\frac{{13}}{2}} \right)\) and \(m\).
\(y - b = m(x - a)\)
\(y - \frac{{13}}{2} = - \frac{1}{3}\left( {x - \frac{5}{2}} \right)\)
\(6y - 6 \times \frac{{13}}{2} = 6 \times \left( { - \frac{1}{3}\left( {x - \frac{5}{2}} \right)} \right)\)
\(6y - 39 = - 2\left( {x - \frac{5}{2}} \right)\)
\(6y - 39 = - 2x + 5\)
\(6y = - 2x + 44\)
\(3y + x = 22\)
\(x + 3y - 22 = 0\)
