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Date May 2011 Marks available 2 Reference code 11M.1.sl.TZ1.10
Level SL only Paper 1 Time zone TZ1
Command term Find Question number 10 Adapted from N/A

Question

The diagram shows the straight lines \({L_1}\) and \({L_2}\) . The equation of \({L_2}\) is \(y = x\) .

Find
(i)     the gradient of \({L_1}\) ;
(ii)    the equation of \({L_1}\) .

[3]
a.

Find the area of the shaded triangle.

[2]
b.

Markscheme

(i)     \(\frac{{0 - 2}}{{6 - 0}}\)     (M1)
\( =  - \frac{1}{3}{\text{ }}\left( { - \frac{2}{6}{\text{, }} - 0.333} \right)\)     (A1)     (C2)


(ii)    \(y =  - \frac{1}{3}x + 2\)     (A1)(ft)     (C1)

Notes: Follow through from their gradient in part (a)(i). Accept equivalent forms for the equation of a line.

 

[3 marks]

a.

\({\text{area}} = \frac{{6 \times 1.5}}{2}\)     (A1)(M1)

Note: Award (A1) for \(1.5\) seen, (M1) for use of triangle formula with \(6\) seen.

 

\( = 4.5\)     (A1)     (C3)

[2 marks]

b.

Examiners report

In this question, many candidates did not use the \(x\) and \(y\) intercepts to find the slope and attempted to read ordered pairs from the graph.

a.

Part b proved difficult for many candidates, often using trigonometry rather than the more straight forward area of the triangle.

b.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.0 » Simple two-dimensional shapes and their properties, including perimeters and areas of circles, triangles, quadrilaterals and compound shapes.
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