Calculate the gradient of line CD.

Show that line AD is perpendicular to line CD.

Find the equation of line CD. Give your answer in the form \(ax + by = c\) where \(a,{\text{ }}b,{\text{ }}c \in \mathbb{Z}\).

Lines AB and CD intersect at point E. The equation of line AB is \(x + 3y = 6\).

Find the coordinates of E.

Lines AB and CD intersect at point E. The equation of line AB is \(x + 3y = 6\).

Find the distance between A and D.

The distance between D and E is \(\sqrt{20}\).

Find the area of triangle ADE.

\({\text{Gradient of CD}} = \frac{{1 - ( - 1)}}{{ - 2 - ( - 1)}}\)     (M1)

\( =  - 2\)     (A1)(G2)

Note: Award (M1) for correct substitution in gradient formula.

[2 marks]

\({\text{Gradient of AD}} = \frac{1}{2}\)     (A1)

\( - 2 \times \frac{1}{2} =  - 1\) or \(\frac{1}{2}\) is negative reciprocal of –2     (M1)

Hence AD is perpendicular for CD.     (AG)

Note: Last line must be seen for the (M1) to be awarded.

[2 marks]

\(y = -2x - 3\)     (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for their (a), (A1)(ft) for –3.

If part (a) incorrect award (A1)(ft) for their y-intercept only if working is seen.

OR

\(y - 1 = -2(x + 2)\)     (A1)(ft)(A1)

OR

\(y + 1 = -2(x + 1)\)     (A1)(ft)(A1)

Note: Award (A1)(ft) for their (a), (A1) for correct substitution of point.

\(2x + y = -3\)     (A1)(ft)

Note: The final (A1)(ft) is for their equation in the stated form.

[3 marks]

E (−3, 3) (Accept x = −3, y = 3)     (G2)

OR

Award (M1) for solving the pair of simultaneous equations by hand. (A1)(ft) for correct answer, (ft) from their (c).     (M1)(A1)(ft)

OR

Award (M1) for having extended the lines in their own graph seen drawn on answer paper. (A1) for correct answer.     (M1)(A1)

Note: Missing coordinate brackets receive (G1)(G0) or (M1)(A0).

[2 marks]

Distance between A and D = \(\sqrt {4^2 + 2^2}\)     (M1)

\( = \sqrt{20}\) OR \(2 \sqrt {5}\) OR 4.47 (3 s.f.)     (A1)(G2)

Note: Award (M1) for correct substitution into the distance formula, (A1) for correct answer.

[2 marks]

Area of ADE = \(\frac{1}{2}\sqrt {20}  \times \sqrt {20} \)     (M1)

= 10     (A1)(ft)(G2)

Follow through from (e).

[2 marks]

This was well done overall. Almost all students could calculate the gradient of the straight line. Gradient of perpendicular line was found, but some candidates failed to communicate the requirement, in terms of gradients for two lines to be perpendicular (Example: They are perpendicular because their gradients are opposite and reciprocal or they are perpendicular because the product of their gradients is −1)

Distance between points and area of triangle was answered well by most candidates. Both formulae for the area of the triangle were correctly used.

This was well done overall. Almost all students could calculate the gradient of the straight line. Gradient of perpendicular line was found, but some candidates failed to communicate the requirement, in terms of gradients for two lines to be perpendicular (Example: They are perpendicular because their gradients are opposite and reciprocal or they are perpendicular because the product of their gradients is −1)

Distance between points and area of triangle was answered well by most candidates. Both formulae for the area of the triangle were correctly used.

This was well done overall. Almost all students could calculate the gradient of the straight line. Gradient of perpendicular line was found, but some candidates failed to communicate the requirement, in terms of gradients for two lines to be perpendicular (Example: They are perpendicular because their gradients are opposite and reciprocal or they are perpendicular because the product of their gradients is −1)

Distance between points and area of triangle was answered well by most candidates. Both formulae for the area of the triangle were correctly used.

This was well done overall. Almost all students could calculate the gradient of the straight line. Gradient of perpendicular line was found, but some candidates failed to communicate the requirement, in terms of gradients for two lines to be perpendicular (Example: They are perpendicular because their gradients are opposite and reciprocal or they are perpendicular because the product of their gradients is −1)

Distance between points and area of triangle was answered well by most candidates. Both formulae for the area of the triangle were correctly used.

This was well done overall. Almost all students could calculate the gradient of the straight line. Gradient of perpendicular line was found, but some candidates failed to communicate the requirement, in terms of gradients for two lines to be perpendicular (Example: They are perpendicular because their gradients are opposite and reciprocal or they are perpendicular because the product of their gradients is −1)

Distance between points and area of triangle was answered well by most candidates. Both formulae for the area of the triangle were correctly used.

This was well done overall. Almost all students could calculate the gradient of the straight line. Gradient of perpendicular line was found, but some candidates failed to communicate the requirement, in terms of gradients for two lines to be perpendicular (Example: They are perpendicular because their gradients are opposite and reciprocal or they are perpendicular because the product of their gradients is −1)

Distance between points and area of triangle was answered well by most candidates. Both formulae for the area of the triangle were correctly used.