Higher June 2019 Paper 1R Q15
15 The straight line L1 has equation \(2y = 6x - 5\)
The straight line L2 is perpendicular to L1 and passes through the point \((9, -1)\)
Find an equation for L2
Give your answer in the form \(ay + bx = c\)
(4)
| Scheme | Marks |
|---|---|
| (Gradient of L1 =) 6 ÷ 2 (=3) | M1 |
| \(m \times \text{``}{3}\text{''} = -1\) or \(m = -\dfrac{1}{\text{``}{3}\text{''}}\) | M1 |
| \(-1 = \text{``}{-\dfrac{1}{3}}\text{''} \times 9 + c\) or \(y - -1 = \text{``}{-\dfrac{1}{3}}\text{''}(x - 9)\) or \(c = 2\) | M1 |
| \(y + \dfrac{1}{3}x = 2\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: could be seen as part of an equation. Ignore constant term if candidate rearranges L1
M1: for use of \(m_1m_2 = -1\)
could be seen as part of an equation
A1: oe in required form eg \(3y + x = 6\), \(6y + 2x = 12\) etc