Equation of medians and parallel lines
Working with equations of lines
Remember that to find the equation of a straight line you need to know its gradient and a point on the line.
In some instances you are told the value of the gradient, however there are times where you may be required to calculate the gradient using the relevant formulae or your knowledge of parallel and perpendicular lines.
Example 1
Find the equation of the line through \((3,1)\) and parallel to the line with equation \(2x - y = 4\).
Solution
We need to use \(y - b = m(x - a)\)
First find the gradient of the parallel line.
\(2x - y = 4\)
\(- y = - 2x + 4\)
\(y = 2x - 4\)
\(m = 2\)
Parallel lines have equal gradients.
Now we can use \(y - b = m(x - a)\) with \(m = 2\) and \((a,b) = (3,1)\)
\(y - b = m(x - a)\)
\(y - 1 = 2(x - 3)\)
\(y - 1 = 2x - 6\)
\(y = 2x - 5\)
The equation of the parallel line is \(y = 2x - 5\) or \(2x-y-5=0\)
Example 2
Find the equation of the line perpendicular to \(AB\) where \(A = (1,5)\) and \(B = (3, - 8)\) and passing through the midpoint of AB.
Solution
Firstly find the gradient of AB.
\({m_{AB}} = \frac{{ - 8 - 5}}{{3 - 1}} = - \frac{{13}}{2}\)
Perpendicular lines have gradients which multiply to give -1.
\({m_{perp}} = \frac{2}{{13}}\,\,\,\,\,\,\,\,\,\,\,\, (since - \frac{{13}}{2} \times \frac{2}{{13}} = - 1)\)
Now that we have the gradient, we next need to find the midpoint of AB so that we have a point on the line.
Midpoint\((AB) = \left( {\frac{{1 + 3}}{2},\frac{{5 + ( - 8)}}{2}} \right)\)
\(= \left( {2,\frac{{ - 3}}{2}} \right)\)
Now we can find the equation of the line with gradient \(\frac{2}{{13}}\), passing through the point \(\left( {2,\frac{{ - 3}}{2}} \right)\)
\(y - b = m(x - a)\)
\(y - \left( { - \frac{3}{2}} \right) = \frac{2}{{13}}(x - 2)\)
\(y + \frac{3}{2} = \frac{2}{{13}}(x - 2)\)
If it is required for the next part of a question then multiply both sides by 26 to clear the fractions. The lowest common multiple of 2 and 13 is 26.
\(26 \times y + 26 \times \frac{3}{2} = 26 \times \frac{2}{{13}}(x - 2)\)
\(26y + 39 = 4x - 8\)
\(- 4x + 26y + 47 = 0\)
Therefore the equation of the line is \(4x - 26y - 47 = 0\)