Find the gradient of ${L_1}$ .

The equation of line ${L_2}$ is $y = mx$ . ${L_2}$ is perpendicular to ${L_1}$ .

Find the value of $m$.

The equation of line ${L_2}$ is $y = mx$ . ${L_2}$ is perpendicular to ${L_1}$ .

Find the coordinates of the point of intersection of the lines ${L_1}$ and ${L_2}$ .

$4y = - x - 34$ or similar rearrangement     (M1)

${\text{Gradient}} = - \frac{1}{4}$     (A1)     (C2)

$m = 4$     (A1)(ft)(A1)(ft)     (C2)

Note: (A1) Change of sign
(A1) Use of reciprocal

[2 marks]

Reasonable attempt to solve equations simultaneously     (M1)

$( - 2{\text{, }} - 8)$     (A1)(ft)     (C2)

Note: Accept $x = - 2$  $y = - 8$. Award (A0) if brackets not included.

[2 marks]

The omission of the negative sign was a common fault.

Most candidates managed to answer this correctly from their (a).

This part proved challenging for the majority. Once again, the use of the GDC was expected.