| Date | May 2018 | Marks available | 2 | Reference code | 18M.1.sl.TZ2.6 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | Find | Question number | 6 | Adapted from | N/A |
Question
Consider the straight lines L1 and L2 . R is the point of intersection of these lines.
The equation of line L1 is y = ax + 5.
The equation of line L2 is y = −2x + 3.
Find the value of a.
Find the coordinates of R.
Line L3 is parallel to line L2 and passes through the point (2, 3).
Find the equation of line L3. Give your answer in the form y = mx + c.
Markscheme
0 = 10a + 5 (M1)
Note: Award (M1) for correctly substituting any point from L1 into the equation.
OR
\(\frac{{0 - 5}}{{10 - 0}}\) (M1)
Note: Award (M1) for correctly substituting any two points on L1 into the gradient formula.
\( - \frac{5}{{10}}\left( { - \frac{1}{2},\,\, - 0.5} \right)\) (A1) (C2)
[2 marks]
\(\left( { - 1.33,\,\,5.67} \right)\,\,\left( {\left( { - \frac{4}{3},\,\,\frac{{17}}{{13}}} \right),\,\,\left( { - 1\frac{1}{3},\,\,5\frac{2}{3}} \right),\,\,\left( { - 1.33333 \ldots ,\,\,5.66666 \ldots } \right)} \right)\) (A1)(ft)(A1)(ft) (C2)
Note: Award (A1) for x-coordinate and (A1) for y-coordinate. Follow through from their part (a). Award (A1)(A0) if brackets are missing. Accept x = −1.33, y = 5.67.
[2 marks]
3 = −2(2) + c (M1)
Note: Award (M1) for correctly substituting –2 and the given point into the equation of a line.
y = −2x + 7 (A1) (C2)
Note: Award (A0) if the equation is not written in the form y = mx + c.
[2 marks]
