18M.1.SL.TZ1.T_5

pestleMathematicsAASLPaper 118M· sl-5-1-introduction-of-differential-calculussource ↗

The point A has coordinates (4 , −8) and the point B has coordinates (−2 , 4).

The point D has coordinates (−3 , 1).

Write down the coordinates of C, the midpoint of line segment AB.

[2]

Find the gradient of the line DC.

[2]

Find the equation of the line DC. Write your answer in the form ax + by + d = 0 where a , b and d are integers.

[2]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(1, −2) (A1)(A1) (C2)
Note: Award (A1) for 1 and (A1) for −2, seen as a coordinate pair.

Accept x = 1, y = −2. Award (A1)(A0) if x and y coordinates are reversed.

[2 marks]

$\frac{{1 - \left( { - 2} \right)}}{{ - 3 - 1}}$ (M1)

Note: Award (M1) for correct substitution, of their part (a), into gradient formula.

$= - \frac{3}{4}\,\,\,\left( { - 0.75} \right)$ (A1)(ft) (C2)

Note: Follow through from part (a).

[2 marks]

$y - 1 = - \frac{3}{4}\left( {x + 3} \right)$ OR $y + 2 = - \frac{3}{4}\left( {x - 1} \right)$ OR $y = - \frac{3}{4}x - \frac{5}{4}$ (M1)

Note: Award (M1) for correct substitution of their part (b) and a given point.

$1 = - \frac{3}{4} \times - 3 + c$ OR $- 2 = - \frac{3}{4} \times 1 + c$ (M1)

Note: Award (M1) for correct substitution of their part (b) and a given point.

$3x + 4y + 5 = 0$ (accept any integer multiple, including negative multiples) (A1)(ft) (C2)

Note: Follow through from parts (a) and (b). Where the gradient in part (b) is found to be $\frac{5}{0}$ , award at most (M1)(A0) for either $x = - 3$ or $x + 3 = 0$ .

[2 marks]

Examiners’ report

[N/A]