| Date | May 2015 | Marks available | 2 | Reference code | 15M.1.sl.TZ2.3 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | Find | Question number | 3 | Adapted from | N/A |
Question
The equation of the line \({L_1}\) is \(2x + y = 10\).
Write down
(i) the gradient of \({L_1}\);
(ii) the \(y\)-intercept of \({L_1}\).
The line \({L_2}\) is parallel to \({L_1}\) and passes through the point \({\text{P}}(0,{\text{ }}3)\).
Write down the equation of \({L_2}\).
The line \({L_2}\) is parallel to \({L_1}\) and passes through the point \({\text{P}}(0,{\text{ }}3)\).
Find the \(x\)-coordinate of the point where \({L_2}\) crosses the \(x\)-axis.
Markscheme
(i) \( - 2\) (A1) (C1)
(ii) \(10\) (A1) (C1)
\(2x + y - 3 = 0\) (A1)(ft)(A1) (C2)
Notes: Award (A1)(ft) for gradient, (A1) for correct \(y\)-intercept.
The answer must be an equation.
\( - 2x + 3 = 0\;\;\;\)or equivalent (M1)
\((x = ){\text{ }}1.5\) (A1)(ft) (C2)
Notes: Follow through from their equation in part (b). If answer given as coordinates \((1.5,{\text{ }}0)\) award at most (M1)(A0) if working seen or (A1)(A0) if no working seen.
