Date | May 2013 | Marks available | 2 | Reference code | 13M.1.sl.TZ2.8 |
Level | SL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 8 | Adapted from | N/A |
Question
The equation of a line L1 is \(2x + 5y = −4\).
Write down the gradient of the line L1.
A second line L2 is perpendicular to L1.
Write down the gradient of L2.
The point (5, 3) is on L2.
Determine the equation of L2.
Lines L1 and L2 intersect at point P.
Using your graphic display calculator or otherwise, find the coordinates of P.
Markscheme
\(\frac{-2}{{5}}\) (A1) (C1)
\(\frac{5}{{2}}\) (A1)(ft) (C1)
Note: Follow through from their answer to part (a).
\(3 = \frac{5}{2} \times 5 + c\) (M1)
Notes: Award (M1) for correct substitution of their gradient into equation of line. Follow through from their answer to part (b).
\(y = \frac{5}{2}x - \frac{19}{2}\) (A1)(ft)
OR
\(y - 3 = \frac{5}{2}(x - 5)\) (M1)(A1)(ft) (C2)
Notes: Award (M1) for correct substitution of their gradient into equation of line. Follow through from their answer to part (b).
(3, −2) (A1)(ft)(A1)(ft) (C2)
Notes: If parentheses not seen award at most (A0)(A1)(ft). Accept x = 3, y = −2. Follow through from their answer to part (c), even if no working is seen. Award (M1)(A1)(ft) for a sensible attempt to solve \(2x + 5y = −4\) and their \(y = \frac{5}{2}x - \frac{19}{2}\) or equivalent, simultaneously.