Write down the value of $a$.

Find the gradient of the line AB.

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

12     (A1)     (C1)

 

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

 

[1 mark]

$\frac{{26 - 10}}{{0 - 24}}$     (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]

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]

[N/A]

[N/A]

[N/A]