C1 June 2013 (R) Q4
4. The line \(L_1\) has equation \(4x + 2y - 3 = 0\)
(a) Find the gradient of \(L_1\). (2)
The line \(L_2\) is perpendicular to \(L_1\) and passes through the point \((2, 5)\).
(b) Find the equation of \(L_2\) in the form \(y = mx + c\), where \(m\) and \(c\) are constants. (3)
| Scheme | Marks |
|---|---|
| \(4x + 2y - 3 = 0 \Rightarrow y = -2x + \dfrac{3}{2}\) | M1 |
| \(\Rightarrow\) gradient \(= -2\) | A1 |
| Answer only scores M1A1 | |
| (2) |
Notes
M1: Attempt to write in the form \(y =\)
A1: Accept any un-simplified form and allow even with an incorrect value of “c”
(a) Way 2
| Scheme | Marks |
|---|---|
| Alternative: \(4 + 2\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | M1 |
| \(\Rightarrow\) gradient \(= -2\) | A1 |
M1: Attempt to differentiate
Allow \(p \pm q\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0,\ p, q \neq 0\)
A1: Accept any un-simplified form
| Scheme | Marks |
|---|---|
| Using \(m_N = -\dfrac{1}{m_T}\) | M1 |
| \(y - 5 = \text{‘}\tfrac{1}{2}\text{’}(x - 2)\) or Uses \(y = mx + c\) in an attempt to find c | M1 |
| \(y = \dfrac{1}{2}x + 4\) | A1 |
| (3) | |
| (5 marks) |
Notes
M1: Attempt to use \(m_N = -\dfrac{1}{\textit{gradient from (a)}}\)
M1: Correct straight line method using a ‘changed’ gradient and the point \((2, 5)\)
A1: Cao (Isw)