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Date	May 2017	Marks available	2	Reference code	17M.1.SL.TZ2.T_4
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 2
Command term	Find	Question number	T_4	Adapted from	N/A

Question

Line $L$ intersects the $x$ -axis at point A and the $y$ -axis at point B, as shown on the diagram.

M17/5/MATSD/SP1/ENG/TZ2/04

The length of line segment OB is three times the length of line segment OA, where O is the origin.

Point ${\text{(2, 6)}}$ lies on $L$ .

Find the gradient of $L$ .

[2]

Find the equation of $L$ in the form $y = mx + c$ .

[2]

Find the $x$ -coordinate of point A.

[2]

Markscheme

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

$- 3$ (A1)(A1) (C2)

Notes: Award (A1) for 3 and (A1) for a negative value.

Award (A1)(A0) for either $3x$ or $- 3x$ .

[2 marks]

$6 = - 3(2) + c$ $\,\,\,$ OR $\,\,\,$ $(y - 6) = - 3(x - 2)$ (M1)

Note: Award (M1) for substitution of their gradient from part (a) into a correct equation with the coordinates $(2,{\text{ }}6)$ correctly substituted.

$y = - 3x + 12$ (A1)(ft) (C2)

Notes: Award (A1)(ft) for their correct equation. Follow through from part (a).

If no method seen, award (A1)(A0) for $y = - 3x$ .

Award (A1)(A0) for $- 3x + 12$ .

[2 marks]

$0 = - 3x + 12$ (M1)

Note: Award (M1) for substitution of $y = 0$ in their equation from part (b).

$(x = ){\text{ }}4$ (A1)(ft) (C2)

Notes: Follow through from their equation from part (b). Do not follow through if no method seen. Do not award the final (A1) if the value of $x$ is negative or zero.

[2 marks]

Examiners report

[N/A]

Syllabus sections

Topic 2—Functions » SL 2.1—Equations of straight lines, parallel and perpendicular

Show 26 related questions

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22M.2.AHL.TZ2.4c:
(i) the gradient of this line in terms of $c$ ;

(ii) the $y$ -intercept of this line in terms of $A$ .
22M.2.AHL.TZ2.4d:
Find the equation of the regression line for $\ln k$ on $\frac{1}{T}$ .
18N.2.SL.TZ0.T_4b.iii:
Use your graphic display calculator to find the equation of the tangent to the graph of y = f (x) at the point (–2, 38.75).

Give your answer in the form y = mx + c.
21N.1.SL.TZ0.4c:
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22M.1.SL.TZ2.11c:
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21N.1.SL.TZ0.4a:
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22M.1.SL.TZ1.4a:
Find the equation of the line that the road follows.
22M.1.SL.TZ1.4b:
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21N.1.SL.TZ0.4b:
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18N.2.SL.TZ0.T_4b.i:
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18M.1.SL.TZ2.T_5b:
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19N.2.SL.TZ0.T_2a:
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19N.2.SL.TZ0.T_2d:
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17M.1.SL.TZ2.T_4a:
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18N.2.SL.TZ0.T_4b.ii:
Use your graphic display calculator to find the coordinates of the local minimum point.
17M.1.SL.TZ2.T_4c:
Find the $x$ -coordinate of point A.
19N.2.SL.TZ0.T_2e:
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17N.1.SL.TZ0.T_2c.ii:
Write down, in the form $y = mx + c$ , the equation of ${L_2}$ .
19N.2.SL.TZ0.T_2c:
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18M.1.SL.TZ2.T_5a.ii:
Write down the amount of interest he earned after 5 years.
21M.2.AHL.TZ1.4a:
Find the equation of the line passing through these two points.
18M.1.SL.TZ2.T_5a.i:
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19N.2.SL.TZ0.T_2f:
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21N.1.SL.TZ0.4d:
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