C1 June 2011 Q3
3. The points \(P\) and \(Q\) have coordinates \((-1, 6)\) and \((9, 0)\) respectively.
The line \(l\) is perpendicular to \(PQ\) and passes through the mid-point of \(PQ\).
Find an equation for \(l\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. (5)
| Scheme | Marks |
|---|---|
| Mid-point of \(PQ\) is \((4, 3)\) | B1 |
| \(PQ\): \(m = \dfrac{0 - 6}{9 - (-1)},\ \left(= -\dfrac{3}{5}\right)\) | B1 |
| Gradient perpendicular to \(PQ\) \(= -\dfrac{1}{m} \quad \left(= \dfrac{5}{3}\right)\) | M1 |
| \(y - 3 = \dfrac{5}{3}(x - 4)\) | M1 |
| \(5x - 3y - 11 = 0\) or \(3y - 5x + 11 = 0\) or multiples e.g. \(10x - 6y - 22 = 0\) | A1 |
| (5) | |
| (5 marks) |
Notes
B1: correct midpoint.
B1: correct numerical expression for gradient – need not be simplified
1st M: Negative reciprocal of their numerical value for \(m\)
2nd M: Equation of a line through their \((4, 3)\) with any gradient except 0 or \(\infty\).
If the 4 and 3 are the wrong way round the 2nd M mark can still be given if a correct formula (e.g. \(y - y_1 = m(x - x_1)\)) is seen, otherwise M0.
If \((4, 3)\) is substituted into \(y = mx + c\) to find \(c\), the 2nd M mark is for attempting this.
A1: Requires integer form with an = zero (see examples above)