18M.1.SL.TZ2.T_6

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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.

[2]

Find the coordinates of R.

[2]

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.

[2]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

0 = 10a + 5 (M1)

Note: Award (M1) for correctly substituting any point from L₁ into the equation.

$\frac{{0 - 5}}{{10 - 0}}$ (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]

Examiners’ report

[N/A]