C1 June 2012 Q9
9. The line \(L_1\) has equation \(4y + 3 = 2x\)
The point \(A\,(p, 4)\) lies on \(L_1\)
The line \(L_2\) passes through the point \(C\,(2, 4)\) and is perpendicular to \(L_1\)
The line \(L_1\) and the line \(L_2\) intersect at the point \(D\).
A point \(B\) lies on \(L_1\) and the length of \(AB = \sqrt{(80)}\)
The point \(E\) lies on \(L_2\) such that the length of the line \(CDE = 3\) times the length of \(CD\).
| Scheme | Marks |
|---|---|
| \(L_1: 4y + 3 = 2x \Rightarrow y = \dfrac{1}{2}x - \dfrac{3}{4};\ A(p, 4)\) lies on \(L_1\). | |
| \(\{p =\}\ 9\dfrac{1}{2}\) or \(\dfrac{19}{2}\) or 9.5 | B1 |
| (1) |
Notes
B1: 9.5 oe.
| Scheme | Marks |
|---|---|
| \(\{4y + 3 = 2x\} \Rightarrow y = \dfrac{2x - 3}{4} \Rightarrow m(L_1) = \dfrac{1}{2}\) or \(\dfrac{2}{4}\) | M1 A1 |
| So \(m(L_2) = -2\) | B1ft |
| \(L_2\): \(y - 4 = -2(x - 2)\) | M1 |
| \(L_2\): \(2x + y - 8 = 0\) or \(L_2\): \(2x + 1y - 8 = 0\) | A1 |
| (5) |
Notes
1st M1: for an attempt to rearrange \(4y + 3 = 2x\) into \(y = mx + c\).
This mark can be implied by the correct gradient of \(L_1\) or \(L_2\).
1st A1: for gradient of \(L_1 = \tfrac{1}{2}\) or \(\tfrac{2}{4}\). Stating \(m(L_1) = \tfrac{1}{2}\) without working is M1A1.
B1ft: for applying \(m(L_2) = \dfrac{-1}{\text{their } m(L_1)}\). Need not be simplified.
Note: Writing down \(m(L_2) = -2\) with no earlier incorrect working gains M1A1B1
2nd M1: for applying \(y - 4 = \pm\lambda(x - 2)\) where \(\lambda\) is a numerical value, \(\lambda \neq 0\).
or full method of \(y = mx + c\), with \(x = 2\), \(y = 4\) and (their \(\pm\lambda\)) to find \(c\).
2nd A1: \(2x + y - 8 = 0\) or \(-2x - y + 8 = 0\) or \(y + 2x - 8 = 0\) or \(4x + 2y - 16 = 0\) or \(2x + 1y - 8 = 0\) etc. Must be “\(= 0\)”. So do not allow \(2x + y = 8\) etc.
Note: Condone the error of incorrectly rearranging \(L_1\) to give \(y = \dfrac{1}{2}x - 3 \Rightarrow m(L_1) = \dfrac{1}{2}\).| Scheme | Marks |
|---|---|
| \(\{L_1 = L_2 \Rightarrow\}\ 4(8 - 2x) + 3 = 2x\) or \(-2x + 8 = \dfrac{1}{2}x - \dfrac{3}{4}\) | M1 |
| \(x = 3.5,\ y = 1\) | A1, A1 cso |
| (3) |
Notes
M1: for an attempt to solve. Must form a linear equation in one variable.
1st A1: for \(x = 3.5\) (correct solution only).
2nd A1: for \(y = 1\) (correct solution only).
Note: If \(x = 3.5,\ y = 1\) is found from no working, then send to review.Note: Use of trial and error to find one of \(x\) or \(y\) and then substitution into one of \(L_1\) or \(L_2\) can achieve M1A1A1.
| Scheme | Marks |
|---|---|
| \(CD^2 = (\text{"}3.5\text{"} - 2)^2 + (\text{"}1\text{"} - 4)^2\) | “M1” |
| \(CD = \sqrt{(\text{"}3.5\text{"} - 2)^2 + (\text{"}1\text{"} - 4)^2}\) | A1 ft |
| \(= \sqrt{1.5^2 + 3^2} = 1.5\sqrt{1^2 + 2^2} = 1.5\sqrt{5}\) or \(\dfrac{3}{2}\sqrt{5} \quad (*)\) | A1 cso |
| (3) |
Notes
M1: for an attempt at \(CD^2\) - ft their point \(D\). Eg: \((\text{"}3.5\text{"} - 2)^2 + (\text{"}1\text{"} - 4)^2\) or simplified.
This mark can be implied by finding \(CD\).
1st A1ft: for finding their \(CD\) - ft their point \(D\). Eg: \(\sqrt{(\text{"}3.5\text{"} - 2)^2 + (\text{"}1\text{"} - 4)^2}\) or correctly simplified.
2nd A1:cso: for no incorrect working seen.
Note: A candidate initially writing down \(\sqrt{1.5^2 + 3^2}\) can be awarded M1A1.Alternatives part (d): Final accuracy
1. \(\left\{\sqrt{1.5^2 + 3^2} =\right\}\ \sqrt{\dfrac{9}{4} + 9} = \sqrt{\dfrac{9}{4} + \dfrac{36}{4}} = \sqrt{\dfrac{45}{4}} = \dfrac{3\sqrt{5}}{2}\)
2. \(\left\{\sqrt{1.5^2 + 3^2} =\right\}\ \sqrt{11.25} = \sqrt{2.25}\sqrt{5} = 1.5\sqrt{5}\)
| Scheme | Marks |
|---|---|
| Area = triangle \(ABC\) + triangle \(ABE\) | |
| \(= \dfrac{1}{2}\times\dfrac{3}{2}\sqrt{5}\times\sqrt{80} + \dfrac{1}{2}\times 3\sqrt{5}\times\sqrt{80}\) Finding the area of any triangle. | M1 |
| \(= \dfrac{3}{4}\sqrt{5}\times 4\sqrt{5} + \dfrac{3}{2}\sqrt{5}\times 4\sqrt{5}\) \(= \dfrac{3}{4}(20) + \dfrac{3}{2}(20)\) | B1 |
| \(= 45\) | A1 |
| (3) | |
| (15 marks) |
Notes
M1: for an attempt at finding the area of either triangle \(ABC\) or triangle \(ABE\).
B1: Correct method for removing a square root. Eg: \(\sqrt{80}\sqrt{5} = \sqrt{400} = 20\) or \(\sqrt{5}\times 4\sqrt{5} = 20\)
Note: This mark can be implied.
A1: for 45 only.
Alternative 1 to part (e)
Area \(= \dfrac{1}{2}\left(\dfrac{3}{2}\sqrt{5} + 3\sqrt{5}\right)\left(\sqrt{80}\right) = \dfrac{1}{2}(30 + 60) = 45\)
M1: \(\dfrac{1}{2}(AB)(CE)\). B1: Evidence of correct surd removal. A1: for 45.
Note: Multiplying the diagonals (usually to find 90) is M0, B1 if surds are removed correctly, A0.
Alternative 2 to part (e)
Area = triangle \(DAC\) + triangle \(DCB\) + triangle \(DEA\) + triangle \(DBE\)\[\begin{aligned}&= \left(\tfrac{1}{2}\times\tfrac{3}{2}\sqrt{5}\times\sqrt{45}\right) + \left(\tfrac{1}{2}\times\tfrac{3}{2}\sqrt{5}\times\left(\sqrt{80} - \sqrt{45}\right)\right) + \left(\tfrac{1}{2}\times 3\sqrt{5}\times\sqrt{45}\right) + \left(\tfrac{1}{2}\times 3\sqrt{5}\times\left(\sqrt{80} - \sqrt{45}\right)\right)\\ &= \left(\tfrac{1}{2}\times\tfrac{3}{2}(15)\right) + \left(\tfrac{1}{2}\times\tfrac{3}{2}(5)\right) + \left(\tfrac{1}{2}\times 3(15)\right) + \left(\tfrac{1}{2}\times 3(5)\right)\\ &= \left(\tfrac{45}{4}\right) + \left(\tfrac{15}{4}\right) + \left(\tfrac{45}{2}\right) + \left(\tfrac{15}{2}\right)\\ &= 45\end{aligned}\]M1: For finding the area of one of the four triangles. B1: Evidence of correct surd removal. A1: for 45.
Alternative 3 to part (e)
\(\left\{CE = CD + DE = \dfrac{3}{2}\sqrt{5} + 3\sqrt{5} = \dfrac{9}{2}\sqrt{5}\right\},\ \left\{BD = DA + \underline{AB} = 3\sqrt{5} + \underline{4\sqrt{5}} = 7\sqrt{5}\right\}\)
Area = triangle \(BCE\) – triangle \(ACE\) \(= \dfrac{1}{2}(CE)(BD) - \dfrac{1}{2}(CE)(BD)\)
\(= \dfrac{1}{2}\times\dfrac{9}{2}\sqrt{5}\times 7\sqrt{5} - \dfrac{1}{2}\times\dfrac{9}{2}\sqrt{5}\times 3\sqrt{5}\) M1: for an attempt at the area of triangle \(BCE\) or triangle \(ACE\).
\(= \dfrac{63(5)}{4} - \dfrac{27(5)}{4} = \dfrac{36(5)}{4} = 9(5)\) B1: Evidence of correct surd removal.
\(= 45\) A1: for 45 only.