The equation of line \({L_1}\) is \(y = 2.5x + k\). Point \({\text{A}}\) \(\,(3,\, - 2)\) lies on \({L_1}\).

Find the value of \(k\).

The line \({L_2}\) is perpendicular to \({L_1}\) and intersects \({L_1}\) at point \({\text{A}}\).

Write down the gradient of \({L_2}\).

Find the equation of \({L_2}\). Give your answer in the form \(y = mx + c\) .

Write your answer to part (c) in the form \(ax + by + d = 0\)  where \(a\), \(b\) and \(d \in \mathbb{Z}\).

\( - 2 = 2.5\, \times 3 + k\)       (M1)

Note: Award (M1) for correct substitution of \((3,\, - 2)\) into equation of \({L_1}\).

\((k = ) - 9.5\)       (A1) (C2)

\( - 0.4\,\left( { - \frac{2}{5}} \right)\)       (A1)  (C1)

\(y - ( - 2) =  - 0.4\,(x - 3)\)       (M1)

OR

\( - 2 =  - 0.4\,(3) + c\)       (M1)

Note: Award (M1) for their gradient and given point substituted into equation of a straight line. Follow through from part (b).

\(y =  - 0.4x - 0.8\)       \(\left( {y =  - \frac{2}{5}x - \frac{4}{5}} \right)\)       (A1)(ft)    (C2)

\(2x + 5y + 4 = 0\) (or any integer multiple)      (A1)(ft) (C1)

Note: Follow through from part (c).

Question 7: Perpendicular Line
The response to this question was mixed.
Part (a) was well attempted by the majority.

In part (b), the gradient was not fully calculated (being left as a reciprocal) by a large number of candidates.

In part (c), the common error was the use of c from part (a) in the line.

In part (d), the notation for integer was not understood by a large number of candidates.