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Date May 2018 Marks available 2 Reference code 18M.1.SL.TZ2.T_6
Level Standard Level Paper Paper 1 Time zone Time zone 2
Command term Find Question number T_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.

[2]
a.

Find the coordinates of R.

[2]
b.

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]
c.

Markscheme

* 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.

OR

0 5 10 0      (M1)

Note: Award (M1) for correctly substituting any two points on L1 into the gradient formula.

5 10 ( 1 2 , 0.5 )      (A1) (C2)

[2 marks]

a.

( 1.33 , 5.67 ) ( ( 4 3 , 17 13 ) , ( 1 1 3 , 5 2 3 ) , ( 1.33333 , 5.66666 ) )      (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]

b.

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]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 2—Functions » SL 2.5—Modelling functions
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