C1 January 2009 Q10
10. The line \(l_1\) passes through the point \(A\,(2, 5)\) and has gradient \(-\dfrac{1}{2}\).
The point \(B\) has coordinates \((-2, 7)\).
The point \(C\) lies on \(l_1\) and has \(x\)-coordinate equal to \(p\).
The length of \(AC\) is 5 units.
| Scheme | Marks |
|---|---|
| \(y - 5 = -\tfrac{1}{2}(x - 2)\) or equivalent, e.g. \(\dfrac{y - 5}{x - 2} = -\dfrac{1}{2}\), | M1A1, |
| \(y = -\tfrac{1}{2}x + 6\) | A1cao |
| (3) |
Notes
M1 A1: The version in the scheme above can be written down directly (for 2 marks), and M1 A0 can be allowed if there is just one slip (sign or number).
If the 5 and 2 are the wrong way round the M mark can still be given if a correct formula (e.g. \(y - y_1 = m(x - x_1)\)) is seen, otherwise M0.
If \((2, 5)\) is substituted into \(y = mx + c\) to find \(c\), the M mark is for attempting this and the 1st A mark is for \(c = 6\).
Correct answer without working or from a sketch scores full marks.
| Scheme | Marks |
|---|---|
| \(x = -2 \Rightarrow \quad y = -\tfrac{1}{2}(-2) + 6 \; = \; 7\) (therefore \(B\) lies on the line) (or equivalent verification methods) | B1 |
| (1) |
Notes
A conclusion/comment is not required, except when the method used is to establish that the line through \((-2, 7)\) with gradient \(-\tfrac{1}{2}\) has the same eqn. as found in part (a), or to establish that the line through \((-2, 7)\) and \((2, 5)\) has gradient \(-\tfrac{1}{2}\). In these cases a comment ‘same equation’ or ‘same gradient’ or ‘therefore on same line’ is sufficient.
| Scheme | Marks |
|---|---|
| \(\left(AB^2 =\right)\ (2 - -2)^2 + (7 - 5)^2, \quad = 16 + 4 = 20, \quad AB = \sqrt{20} = 2\sqrt{5}\) | M1, A1, A1 |
| (3) |
Notes
M1: for attempting \(AB^2\) or \(AB\). Allow one slip (sign or number) inside a bracket, i.e. do not allow \((2 - -2)^2 - (7 - 5)^2\).
1st A1: for 20 (condone bracketing slips such as \(-2^2 = 4\))
2nd A1: for \(2\sqrt{5}\) or \(k = 2\) (Ignore \(\pm\) here).
| Scheme | Marks |
|---|---|
| \(C\) is \(\left(p,\ -\tfrac{1}{2}p + 6\right)\), so \(\quad AC^2 = (p - 2)^2 + \left(-\dfrac{1}{2}p + 6 - 5\right)^2\) | M1 |
| Therefore \(\quad 25 = p^2 - 4p + 4 + \tfrac{1}{4}p^2 - p + 1\) | M1 |
| \(25 = 1.25p^2 - 5p + 5\) or \(100 = 5p^2 - 20p + 20\) (or better, RHS simplified to 3 terms) | A1 |
| Leading to: \(\quad 0 = p^2 - 4p - 16\) (*) | A1cso |
| (4) | |
| (11 marks) |
Notes
1st M1: for \((p - 2)^2 + (\text{linear function of } p)^2\). The linear function may be unsimplified but must be equivalent to \(ap + b,\ a \ne 0,\ b \ne 0\).
2nd M1: (dependent on 1st M) for forming an equation in \(p\) (using 25 or 5) and attempting (perhaps not very well) to multiply out both brackets.
1st A1: for collecting like \(p\) terms and having a correct expression.
2nd A1: for correct work leading to printed answer.
Alternative, using the result:
Solve the quadratic \(\left(p = 2 \pm 2\sqrt{5}\right)\) and use one or both of the two solutions to find the length of \(AC^2\) or \({C_1C_2}^2\): e.g. \(AC^2 = \left(2 + 2\sqrt{5} - 2\right)^2 + \left(5 - \sqrt{5} - 5\right)^2\) scores 1st M1, and 1st A1 if fully correct.
Finding the length of \(AC\) or \(AC^2\) for both values of \(p\), or finding \(C_1C_2\) with some evidence of halving (or intending to halve) scores the 2nd M1.
Getting \(AC = 5\) for both values of \(p\), or showing \(\dfrac{1}{2}C_1C_2 = 5\) scores the 2nd A1 (cso).