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Date November 2017 Marks available 2 Reference code 17N.1.sl.TZ0.2
Level SL only Paper 1 Time zone TZ0
Command term Find Question number 2 Adapted from N/A

Question

The coordinates of point A are \((6,{\text{ }} - 7)\) and the coordinates of point B are \(( - 6,{\text{ }}2)\). Point M is the midpoint of AB.

\({L_1}\) is the line through A and B.

The line \({L_2}\) is perpendicular to \({L_1}\) and passes through M.

Find the coordinates of M.

[2]
a.

Find the gradient of \({L_1}\).

[2]
b.

Write down the gradient of \({L_2}\).

[1]
c.i.

Write down, in the form \(y = mx + c\), the equation of \({L_2}\).

[1]
c.ii.

Markscheme

\((0,{\text{ }}2.5)\)\(\,\,\,\)OR\(\,\,\,\)\(\left( {0,{\text{ }} - \frac{5}{2}} \right)\)     (A1)(A1)     (C2)

 

Note:     Award (A1) for 0 and (A1) for –2.5 written as a coordinate pair. Award at most (A1)(A0) if brackets are missing. Accept “\(x = 0\) and \(y =  - 2.5\)”.

 

[2 marks]

a.

\(\frac{{2 - ( - 7)}}{{ - 6 - 6}}\)     (M1)

 

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

 

\( =  - \frac{3}{4}{\text{ }}( - 0.75)\)     (A1)     (C2)

[2 marks]

b.

\(\frac{4}{3}{\text{ }}(1.33333 \ldots )\)     (A1)(ft)     (C1)

 

Note:     Award (A0) for \(\frac{1}{{0.75}}\). Follow through from part (b).

 

[1 mark]

c.i.

\(y = \frac{4}{3}x - \frac{5}{2}{\text{ }}(y = 1.33 \ldots x - 2.5)\)     (A1)(ft)     (C1)

 

Note:     Follow through from parts (c)(i) and (a). Award (A0) if final answer is not written in the form \(y = mx + c\).

[1 mark]

c.ii.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.1 » Equation of a line in two dimensions: the forms \(y = mx + c\) and \(ax + by + d = 0\) .
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