| Date | May 2017 | Marks available | 2 | Reference code | 17M.1.sl.TZ1.11 |
| Level | SL only | Paper | 1 | Time zone | TZ1 |
| Command term | Find | Question number | 11 | Adapted from | N/A |
Question
The equation of line \({L_1}\) is \(y = - \frac{2}{3}x - 2\).
Point P lies on \({L_1}\) and has \(x\)-coordinate \( - 6\).
The line \({L_2}\) is perpendicular to \({L_1}\) and intersects \({L_1}\) when \(x = - 6\).
Write down the gradient of \({L_1}\).
Find the \(y\)-coordinate of P.
Determine the equation of \({L_2}\). Give your answer in the form \(ax + by + d = 0\), where \(a\), \(b\) and \(d\) are integers.
Markscheme
\( - \frac{2}{3}\) (A1) (C1)
[1 mark]
\(y = - \frac{2}{3}( - 6) - 2\) (M1)
Note: Award (M1) for correctly substituting \( - 6\) into the formula for \({L_1}\).
\((y = ){\text{ }}2\) (A1) (C2)
Note: Award (A0)(A1) for \(( - 6,{\text{ }}2)\) with or without working.
[2 marks]
gradient of \({L_2}\) is \(\frac{3}{2}\) (A1)(ft)
Note: Follow through from part (a).
\(2 = \frac{3}{2}( - 6) + c\)\(\,\,\,\)OR\(\,\,\,\)\(y - 2 = \frac{3}{2}\left( {x - ( - 6)} \right)\) (M1)
Note: Award (M1) for substituting their part (b), their gradient and \( - 6\) into equation of a straight line.
\(3x - 2y + 22 = 0\) (A1)(ft) (C3)
Note: Follow through from parts (a) and (b). Accept any integer multiple.
Award (A1)(M1)(A0) for \(y = \frac{3}{2}x + 11\).
[3 marks]
