Explain why point P is not on the line L.

Find the gradient of line L.

(i) Write down the gradient of a line perpendicular to line L.

(ii) Find the equation of the line perpendicular to L and passing through point P.

3 × (–5) + 2 × 6 ≠ 9     (A1)     (C1)

Note: Also accept 3 × (–5) + 2x = 9 gives x =12 ≠ 6 or 3y + 2 × (6) = 9 gives y = –1 ≠ –5.

[1 mark]

3y = –2x + 9     (M1)

Note: Award (M1) for 3y = –2x + 9 or \(y = \frac{{ - 2}}{3}x + 3\) or \(y = \frac{{( - 2x + 9)}}{3}\).

\({\text{gradient}} =  - \frac{2}{3}( - 0.667)( - 0.666666...)\)     (A1)     (C2)

[2 marks]

(i) gradient of perpendicular line \( = \frac{3}{2}(1.5)\)     (A1)(ft)

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

(ii) \(y = \frac{3}{2}x + c\)

\( - 5 = \frac{3}{2} \times 6 + c\)     (M1)

Note: Award (M1) for substitution of their perpendicular gradient and the point (6, –5) into the equation of their line.

\(y = \frac{3}{2}x - 14\)     (A1)(ft)

Note: Follow through from their perpendicular gradient. Accept equivalent forms.

OR

\(y + 5 = \frac{3}{2}(x - 6)\)     (M1)(A1)(ft)     (C3)

Notes: Award (M1) for substitution of their perpendicular gradient and the point (6, –5) into the equation of their line. Follow through from their perpendicular gradient.

[3 marks]

In part (a), the word 'explain' required more than simply stating that 'I put the coordinates into my GDC and it did not work'. A written statement, showing the substitution of one or both of the coordinates leading to an inequality was required for this first mark. It was pleasing to see that many scripts showed correct methodology for calculating the gradient of L and the gradient of a line perpendicular to L. The correct equation of the line perpendicular to L passing through P proved to be more elusive as poor arithmetic spoilt what could have been excellent work. In particular \( - 5 = \frac{3}{2} \times 6 + c\) leading to \(c = ( \pm )4\) proved to be a popular but erroneous calculation.

In part (a), the word 'explain' required more than simply stating that 'I put the coordinates into my GDC and it did not work'. A written statement, showing the substitution of one or both of the coordinates leading to an inequality was required for this first mark. It was pleasing to see that many scripts showed correct methodology for calculating the gradient of L and the gradient of a line perpendicular to L. The correct equation of the line perpendicular to L passing through P proved to be more elusive as poor arithmetic spoilt what could have been excellent work. In particular \( - 5 = \frac{3}{2} \times 6 + c\) leading to \(c = ( \pm )4\) proved to be a popular but erroneous calculation.

In part (a), the word 'explain' required more than simply stating that 'I put the coordinates into my GDC and it did not work'. A written statement, showing the substitution of one or both of the coordinates leading to an inequality was required for this first mark. It was pleasing to see that many scripts showed correct methodology for calculating the gradient of L and the gradient of a line perpendicular to L. The correct equation of the line perpendicular to L passing through P proved to be more elusive as poor arithmetic spoilt what could have been excellent work. In particular \( - 5 = \frac{3}{2} \times 6 + c\) leading to \(c = ( \pm )4\) proved to be a popular but erroneous calculation.