C1 June 2010 Q8
8.
The point \(C\) has coordinates \((2, t)\), where \(t > 0\), and \(AC = AB\).
| Scheme | Marks |
|---|---|
| \(m_{AB} = \dfrac{4 - 0}{7 - 2} \ \left(= \dfrac{4}{5}\right)\) | M1 |
| Equation of \(AB\) is: \(\quad y - 0 = \dfrac{4}{5}(x - 2)\) or \(y - 4 = \dfrac{4}{5}(x - 7)\) (o.e.) | M1 |
| \(\underline{4x - 5y - 8 = 0}\) (o.e.) | A1 |
| (3) |
Notes
Apply the usual rules for quoting formulae here.
For a correctly quoted formula with some correct substitution award M1
If no formula is quoted then a fully correct expression is needed for the M mark
1st M1: for attempt at gradient of \(AB\). Some correct substitution in correct formula.
2nd M1: for an attempt at equation of \(AB\). Follow through their gradient, not e.g. \(-\dfrac{1}{m}\)
Using \(y = mx + c\) scores this mark when \(c\) is found.
Use of \(\dfrac{y - y_1}{y_2 - y_1} = \dfrac{x - x_1}{x_2 - x_1}\) scores 1st M1 for denominator, 2nd M1 for use of a correct point
A1: requires integer form but allow \(5y + 8 = 4x\) etc. Must have an “=” or A0
| Scheme | Marks |
|---|---|
| \(\left(AB = \right)\sqrt{(7 - 2)^2 + (4 - 0)^2}\) | M1 |
| \(= \sqrt{41}\) | A1 |
| (2) |
Notes
M1: for an expression for \(AB\) or \(AB^2\). Ignore what is “left” of the equals sign
| Scheme | Marks |
|---|---|
| Using isos triangle with \(AB = AC\) then \(t = 2\times y_A = 2\times 4 = 8\) | B1 |
| (1) |
Notes
B1: for \(t = 8\). May be implied by correct coordinates \((2, 8)\) or the value appearing in (d)
| Scheme | Marks |
|---|---|
| Area of triangle \(= \tfrac{1}{2}t\times(7 - 2)\) | M1 |
| \(= \underline{20}\) | A1 |
| (2) | |
| (8 marks) |
Notes
M1: for an expression for the area of the triangle, follow through their \(t\ (\ne 0)\) but must have the \((7 - 2)\) or 5 and the \(\tfrac{1}{2}\).
DET e.g. \(\begin{matrix} 2 & 7 & 2 & 2 \\ 0 & 4 & t & 0 \end{matrix}\) Area \(= \tfrac{1}{2}\left[8 + 7t + 0 - (0 + 8 + 2t)\right]\) Must have the \(\tfrac{1}{2}\) for M1