C1 June 2007 Q11
11. The line \(l_1\) has equation \(y = 3x + 2\) and the line \(l_2\) has equation \(3x + 2y - 8 = 0\).
The point of intersection of \(l_1\) and \(l_2\) is \(P\).
The lines \(l_1\) and \(l_2\) cross the line \(y = 1\) at the points \(A\) and \(B\) respectively.
| Scheme | Marks |
|---|---|
| \(y = -\dfrac{3}{2}x (+4) \qquad\) Gradient \(= -\dfrac{3}{2}\) | M1 A1 |
| (2) |
Notes
M1: for an attempt to write \(3x + 2y - 8 = 0\) in the form \(y = mx + c\)
or a full method that leads to \(m =\), e.g find 2 points, and attempt gradient using \(\dfrac{y_2 - y_1}{x_2 - x_1}\)
e.g. finding \(y = -1.5x + 4\) alone can score M1 (even if they go on to say \(m = 4\))
A1: for \(m = -\dfrac{3}{2}\) (can ignore the \(+c\)) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{3}{2}\)
| Scheme | Marks |
|---|---|
| \(3x + 2 = -\dfrac{3}{2}x + 4 \qquad x = \ldots,\ \underline{\dfrac{4}{9}}\) | M1, A1 |
| \(y = 3\left(\dfrac{4}{9}\right) + 2 = \underline{\dfrac{10}{3}}\ \left(= 3\dfrac{1}{3}\right)\) | A1 |
| (3) |
Notes
M1: for forming a suitable equation in one variable and attempting to solve leading to \(x = \ldots\) or \(y =\)
1st A1: for any exact correct value for \(x\)
2nd A1: for any exact correct value for \(y\)
(These 3 marks can be scored anywhere, they may treat (a) and (b) as a single part)
| Scheme | Marks |
|---|---|
| Where \(y = 1\), \(\qquad l_1: x_A = -\dfrac{1}{3} \qquad l_2: x_B = 2\) M: Attempt one of these | M1 A1 |
| Area \(= \dfrac{1}{2}\left(x_B - x_A\right)\left(y_P - 1\right)\) | M1 |
| \(= \dfrac{1}{2}\times\dfrac{7}{3}\times\dfrac{7}{3} = \underline{\dfrac{49}{18}} = 2\dfrac{13}{18}\) o.e. | A1 |
| (4) | |
| (9 marks) |
Notes
1st M1: for attempting the \(x\) coordinate of \(A\) or \(B\). One correct value seen scores M1.
1st A1: for \(x_A = -\tfrac{1}{3}\) and \(x_B = 2\)
2nd M1: for a full method for the area of the triangle – follow through their \(x_A, x_B, y_P\).
e.g. determinant approach \(\tfrac{1}{2}\begin{vmatrix} 2 & -\frac{1}{3} & \frac{4}{9} & 2 \\ 1 & 1 & \frac{10}{3} & 1 \end{vmatrix} = \tfrac{1}{2}\left|2 - \ldots - \left(-\tfrac{1}{3}\ldots\right)\right|\)
2nd A1: for \(\dfrac{49}{18}\) or an exact equivalent.
All accuracy marks require answers as single fractions or mixed numbers not necessarily in lowest terms.