User interface language: English | Español

Date May 2014 Marks available 3 Reference code 14M.1.sl.TZ1.9
Level SL only Paper 1 Time zone TZ1
Command term Deduce Question number 9 Adapted from N/A

Question

The diagram shows the points M(a, 18) and B(24, 10) . The straight line BM intersects the y-axis at A(0, 26). M is the midpoint of the line segment AB.


Write down the value of \(a\).

[1]
a.

Find the gradient of the line AB.

[2]
b.

Decide whether triangle OAM is a right-angled triangle. Justify your answer.

[3]
c.

Markscheme

12     (A1)     (C1)

 

Note: Award (A1) for \(\left( {12,18} \right)\).

 

[1 mark]

a.

\(\frac{{26 - 10}}{{0 - 24}}\)     (M1)

 

Note: Accept \(\frac{{26 - 18}}{{0 - 12}}\)   or   \(\frac{{18 - 10}}{{12 - 24}}\)   (or equivalent).

 

\( =  - \frac{2}{3}{\text{ }}\left( { - \frac{{16}}{{24}},{\text{ }} - 0.666666 \ldots } \right)\)     (A1)     (C2)

 

Note: If either of the alternative fractions is used, follow through from their answer to part (a).

     The answer is now (A1)(ft).

 

[2 marks]

b.

gradient of \({\text{OM}} = \frac{3}{2}\)     (A1)(ft)

 

Note: Follow through from their answer to part (b).

 

\( - \frac{2}{3} \times \frac{3}{2}\)     (M1)

 

Note: Award (M1) for multiplying their gradients.

 

Since the product is \(-1\), OAM is a right-angled triangle     (R1)(ft)

 

Notes: Award the final (R1) only if their conclusion is consistent with their answer for the product of the gradients.

     The statement that OAM is a right-angled triangle without justification is awarded no marks.

 

OR

\({(26 - 18)^2} + {12^2}\) and \({12^2} + {18^2}\)     (A1)(ft)

\(\left( {{{(26 - 18)}^2} + {{12}^2}} \right) + ({12^2} + {18^2}) = {26^2}\)     (M1)

 

Note: This method can also be applied to triangle OMB.

     Follow through from (a).

 

Hence a right angled triangle     (R1)(ft)

 

Note: Award the final (R1) only if their conclusion is consistent with their (M1) mark.

 

OR

\(OA = OB = 26\) (cm) an isosceles triangle     (A1)

 

Note:     Award (A1) for \(OA = 26\) (cm) and \(OB = 26\) (cm).

 

Line drawn from vertex to midpoint of base is perpendicular to the base     (M1)

Conclusion     (R1)     (C3)

 

Note: Award, at most (A1)(M0)(R0) for stating that OAB is an isosceles triangle without any calculations.

 

[3 marks]

c.

Examiners report

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

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.1 » Perpendicular lines, \({m_1} \times {m_2} = - 1\) .
Show 24 related questions

View options