Find the gradient of AB.

L₁ passes through C and is parallel to AB.

Write down the y-intercept of L₁.

L₂ passes through A and is perpendicular to AB.

Write down the equation of L₂. Give your answer in the form ax + by + d = 0 where a, b and d \( \in \mathbb{Z}\).

Write down the coordinates of the point D, the intersection of L₁ and L₂.

There is a point R on L₁ such that ABRD is a rectangle.

Write down the coordinates of R.

The distance between A and D is \(\sqrt {45} \).

(i) Find the distance between D and R .

(ii) Find the area of the triangle BDR .

\(\frac{{9 - 1}}{{0 - ( - 4)}}\)     (M1)

= 2    (A1)(G2)

Notes: Award (M1) for correct substitution into the gradient formula.

[2 marks]

–6     (A1)

Note: Accept (0, –6) .

[1 mark]

\(y =  - \frac{1}{2}x - 1\)  (or equivalent)     (A1)(ft)(A1)

Notes: Award (A1)(ft) for gradient, (A1) for correct y-intercept. Follow through from their gradient in (a).

x + 2y + 2 = 0     (A1)(ft)

Notes: Award (A1)(ft) from their gradient and their y-intercept. Accept any multiple of this equation with integer coefficients.

OR

\(y - 1 =  - \frac{1}{2}(x + 4)\) (or equivalent)     (A1)(ft)(A1)

Note: Award (A1)(ft) for gradient, (A1) for any point on the line correctly substituted in equation.

x + 2y + 2 = 0     (A1)(ft)

Notes: Award (A1)(ft) from their equation. Accept any multiple of this equation with integer coefficients.

[3 marks]

D(2, –2) or x = 2, y = –2     (A1)

Note: Award (A0) if brackets not present.

[1 mark]

R(6, 6) or x = 6, y = 6     (A1)(A1)

Note: Award at most (A0)(A1)(ft) if brackets not present and absence of brackets has not already been penalised in part (d).

[2 marks]

(i) \({\text{DR}} = \sqrt {{8^2} + {4^2}} \)     (M1)

\({\text{DR}} = \sqrt {80} \)  (8.94)     (A1)(ft)(G2)

Note: Award (M1) for correct substitution into the distance formula. Follow through from their D and R.

(ii) \({\text{Area}} = \frac{{\sqrt {80}  \times \sqrt {45} }}{2}\)     (M1)

= 30 (30.0)     (A1)(ft)(G2)

Note: Award (M1) for correct substitution in the area of triangle formula. Follow through from their answer to part (f) (i).

[4 marks]

This question was in general well answered. In part (a) the gradient of the line AB was correctly found although some candidates did not substitute well in the gradient formula and found answers as \(\frac{1}{2}\) or –2. Also some students read B as (0, 8) instead of (0, 9). In part (b) many students again did not make good use of time as they found the equation of the line instead of just extending it to find the y - intercept. The equation of L₂ in (c) was correctly found in the form y = mx + c but very few students were able to rearrange the equation in the form ax + by + d = 0 where a, b, d \( \in \mathbb{Z}\). In (d) many candidates found the coordinates of point D by solving simultaneous equations which led again to a waste of time. The last two parts of this question were well done by those students that attempted them.

This question was in general well answered. In part (a) the gradient of the line AB was correctly found although some candidates did not substitute well in the gradient formula and found answers as \(\frac{1}{2}\) or –2. Also some students read B as (0, 8) instead of (0, 9). In part (b) many students again did not make good use of time as they found the equation of the line instead of just extending it to find the y - intercept. The equation of L₂ in (c) was correctly found in the form y = mx + c but very few students were able to rearrange the equation in the form ax + by + d = 0 where a, b, d \( \in \mathbb{Z}\). In (d) many candidates found the coordinates of point D by solving simultaneous equations which led again to a waste of time. The last two parts of this question were well done by those students that attempted them.

This question was in general well answered. In part (a) the gradient of the line AB was correctly found although some candidates did not substitute well in the gradient formula and found answers as \(\frac{1}{2}\) or –2. Also some students read B as (0, 8) instead of (0, 9). In part (b) many students again did not make good use of time as they found the equation of the line instead of just extending it to find the y - intercept. The equation of L₂ in (c) was correctly found in the form y = mx + c but very few students were able to rearrange the equation in the form ax + by + d = 0 where a, b, d \( \in \mathbb{Z}\). In (d) many candidates found the coordinates of point D by solving simultaneous equations which led again to a waste of time. The last two parts of this question were well done by those students that attempted them.

This question was in general well answered. In part (a) the gradient of the line AB was correctly found although some candidates did not substitute well in the gradient formula and found answers as \(\frac{1}{2}\) or –2. Also some students read B as (0, 8) instead of (0, 9). In part (b) many students again did not make good use of time as they found the equation of the line instead of just extending it to find the y - intercept. The equation of L₂ in (c) was correctly found in the form y = mx + c but very few students were able to rearrange the equation in the form ax + by + d = 0 where a, b, d \( \in \mathbb{Z}\). In (d) many candidates found the coordinates of point D by solving simultaneous equations which led again to a waste of time. The last two parts of this question were well done by those students that attempted them.

This question was in general well answered. In part (a) the gradient of the line AB was correctly found although some candidates did not substitute well in the gradient formula and found answers as \(\frac{1}{2}\) or –2. Also some students read B as (0, 8) instead of (0, 9). In part (b) many students again did not make good use of time as they found the equation of the line instead of just extending it to find the y - intercept. The equation of L₂ in (c) was correctly found in the form y = mx + c but very few students were able to rearrange the equation in the form ax + by + d = 0 where a, b, d \( \in \mathbb{Z}\). In (d) many candidates found the coordinates of point D by solving simultaneous equations which led again to a waste of time. The last two parts of this question were well done by those students that attempted them.

This question was in general well answered. In part (a) the gradient of the line AB was correctly found although some candidates did not substitute well in the gradient formula and found answers as \(\frac{1}{2}\) or –2. Also some students read B as (0, 8) instead of (0, 9). In part (b) many students again did not make good use of time as they found the equation of the line instead of just extending it to find the y - intercept. The equation of L₂ in (c) was correctly found in the form y = mx + c but very few students were able to rearrange the equation in the form ax + by + d = 0 where a, b, d \( \in \mathbb{Z}\). In (d) many candidates found the coordinates of point D by solving simultaneous equations which led again to a waste of time. The last two parts of this question were well done by those students that attempted them.