C1 June 2006 Q11
11. The line \(l_1\) passes through the points \(P(-1, 2)\) and \(Q(11, 8)\).
The line \(l_2\) passes through the point \(R(10, 0)\) and is perpendicular to \(l_1\). The lines \(l_1\) and \(l_2\) intersect at the point \(S\).
| Scheme | Marks |
|---|---|
| \(m = \dfrac{8 - 2}{11 + 1}\ \left(= \dfrac{1}{2}\right)\) | M1 A1 |
| \(y - 2 = \dfrac{1}{2}(x - -1)\) or \(y - 8 = \tfrac{1}{2}(x - 11)\) o.e. | M1 |
| \(y = \dfrac{1}{2}x + \dfrac{5}{2}\) accept exact equivalents e.g. \(\dfrac{6}{12}\) | A1c.a.o. |
| (4) |
Notes
1st M1 for attempting \(\dfrac{y_1 - y_2}{x_1 - x_2}\), must be \(y\) over \(x\). No formula condone one sign slip, but if formula is quoted then there must be some correct substitution.
1st A1 for a fully correct expression, needn’t be simplified.
2nd M1 for attempting to find equation of \(l_1\).
| Scheme | Marks |
|---|---|
| Gradient of \(l_2 = -2\) | M1 |
| Equation of \(l_2\): \(y - 0 = -2(x - 10)\) \([y = -2x + 20]\) | M1 |
| \(\dfrac{1}{2}x + \dfrac{5}{2} = -2x + 20\) | M1 |
| \(\underline{x = 7 \text{ and } y = 6}\) depend on all 3 Ms | A1, A1 |
| (5) |
Notes
1st M1 for using the perpendicular gradient rule
2nd M1 for attempting to find equation of \(l_2\). Follow their gradient provided different.
3rd M1 for forming a suitable equation to find \(S\).
| Scheme | Marks |
|---|---|
| \(RS^2 = (10 - 7)^2 + (0 - 6)^2\ (= 3^2 + 6^2)\) | M1 |
| \(RS = \sqrt{45} = 3\sqrt{5}\) (*) | A1c.s.o. |
| (2) |
Notes
M1 for expression for \(RS\) or \(RS^2\). Ft their \(S\) coordinates
| Scheme | Marks |
|---|---|
| \(PQ = \sqrt{12^2 + 6^2},\ = 6\sqrt{5}\) or \(\sqrt{180}\) or \(PS = 4\sqrt{5}\) and \(SQ = 2\sqrt{5}\) | M1,A1 |
| Area \(= \dfrac{1}{2}PQ\times RS = \dfrac{1}{2}6\sqrt{5}\times 3\sqrt{5}\) | dM1 |
| \(= \underline{45}\) | A1 c.a.o. |
| (4) | |
| (15 marks) |
Notes
1st M1 for expression for \(PQ\) or \(PQ^2\). \(PQ^2 = 12^2 + 6^2\) is M1 but \(PQ = 12^2 + 6^2\) is M0
Allow one numerical slip.
2nd dM1 for a full, correct attempt at area of triangle. Dependent on previous M1.