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21N.1.SL.TZ0.4

pestleMathematicsAISLPaper 121N· sl-2-1-equations-of-straight-lines-parallel-and-perpendicularsource ↗

Dilara is designing a kite $ABCD$ on a set of coordinate axes in which one unit represents $10 cm$ .

The coordinates of $A$ , $B$ and $C$ are $(2, 0), (0, 4)$ and $(4, 6)$ respectively. Point $D$ lies on the $x$ -axis. $[AC]$ is perpendicular to $[BD]$ . This information is shown in the following diagram.

Find the gradient of the line through $A$ and $C$ .

[2]

a.

Write down the gradient of the line through $B$ and $D$ .

[1]

b.

Find the equation of the line through $B$ and $D$ . Give your answer in the form $a x + b y + d = 0$ , where $a, b$ and $d$ are integers.

[2]

c.

Write down the $x$ -coordinate of point $D$ .

[1]

d.

Markscheme / solution

$m = \frac{6 - 0}{4 - 2} = 3$ (M1)A1

[2 marks]

a.

$(m =) - \frac{1}{3} (- 0.333, - 0.333333 \dots)$ A1

[1 mark]

b.

an equation of line with a correct intercept and either of their gradients from (a) or (b) (M1)

e.g. $y = - \frac{1}{3} x + 4$ OR $y - 4 = - \frac{1}{3} (x - 0)$

Note: Award (M1) for substituting either of their gradients from parts (a) or (b) and point $B$ or $(3, 3)$ into equation of a line.

$x + 3 y - 12 = 0$ or any integer multiple A1

[2 marks]

c.

$(x =) 12$ A1

[1 mark]

d.

Examiners’ report

This question was overall well done by most candidates. In part (a) calculating the gradient was correctly done with few errors noted where candidates swapped the $y$ and $x$ coordinates in the gradient formula. Some candidates left their answer as $\frac{6}{3}$ which resulted in a loss of the final mark. There was some confusion with the gradient of a line and the gradient of the perpendicular line in part (a). Some candidates found the perpendicular gradient in part (a). Although many candidates were able to write an appropriate equation of the line through $BD$ , several did not express their answer in the required form $a x + b y + d = 0$ , where $a$ , $b$ and $d$ are integers. Many final answers were given as $y = \frac{1}{3} x + 4$ or $y - \frac{1}{3} x - 4 = 0$ . In part (d), writing the $x$ -coordinate of point $D$ was well done by most candidates. Some candidates wrote a coordinate pair rather than just the $x$ -coordinate as required.

a.

This question was overall well done by most candidates. In part (a) calculating the gradient was correctly done with few errors noted where candidates swapped the $y$ and $x$ coordinates in the gradient formula. Some candidates left their answer as $\frac{6}{3}$ which resulted in a loss of the final mark. There was some confusion with the gradient of a line and the gradient of the perpendicular line in part (a). Some candidates found the perpendicular gradient in part (a). Although many candidates were able to write an appropriate equation of the line through $BD$ , several did not express their answer in the required form $a x + b y + d = 0$ , where $a$ , $b$ and $d$ are integers. Many final answers were given as $y = \frac{1}{3} x + 4$ or $y - \frac{1}{3} x - 4 = 0$ . In part (d), writing the $x$ -coordinate of point $D$ was well done by most candidates. Some candidates wrote a coordinate pair rather than just the $x$ -coordinate as required.

b.

This question was overall well done by most candidates. In part (a) calculating the gradient was correctly done with few errors noted where candidates swapped the $y$ and $x$ coordinates in the gradient formula. Some candidates left their answer as $\frac{6}{3}$ which resulted in a loss of the final mark. There was some confusion with the gradient of a line and the gradient of the perpendicular line in part (a). Some candidates found the perpendicular gradient in part (a). Although many candidates were able to write an appropriate equation of the line through $BD$ , several did not express their answer in the required form $a x + b y + d = 0$ , where $a$ , $b$ and $d$ are integers. Many final answers were given as $y = \frac{1}{3} x + 4$ or $y - \frac{1}{3} x - 4 = 0$ . In part (d), writing the $x$ -coordinate of point $D$ was well done by most candidates. Some candidates wrote a coordinate pair rather than just the $x$ -coordinate as required.

c.

This question was overall well done by most candidates. In part (a) calculating the gradient was correctly done with few errors noted where candidates swapped the $y$ and $x$ coordinates in the gradient formula. Some candidates left their answer as $\frac{6}{3}$ which resulted in a loss of the final mark. There was some confusion with the gradient of a line and the gradient of the perpendicular line in part (a). Some candidates found the perpendicular gradient in part (a). Although many candidates were able to write an appropriate equation of the line through $BD$ , several did not express their answer in the required form $a x + b y + d = 0$ , where $a$ , $b$ and $d$ are integers. Many final answers were given as $y = \frac{1}{3} x + 4$ or $y - \frac{1}{3} x - 4 = 0$ . In part (d), writing the $x$ -coordinate of point $D$ was well done by most candidates. Some candidates wrote a coordinate pair rather than just the $x$ -coordinate as required.

d.