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Date May 2013 Marks available 1 Reference code 13M.1.sl.TZ2.8
Level SL only Paper 1 Time zone TZ2
Command term Write down Question number 8 Adapted from N/A

Question

The equation of a line L1 is \(2x + 5y = −4\).

Write down the gradient of the line L1.

[1]
a.

A second line L2 is perpendicular to L1.

Write down the gradient of L2.

[1]
b.

The point (5, 3) is on L2.

Determine the equation of L2.

[2]
c.

Lines L1 and L2 intersect at point P.

Using your graphic display calculator or otherwise, find the coordinates of P.

[2]
d.

Markscheme

\(\frac{-2}{{5}}\)     (A1)     (C1)

a.

\(\frac{5}{{2}}\)     (A1)(ft)     (C1)


Note: Follow through from their answer to part (a).

b.

\(3 = \frac{5}{2} \times 5 + c\)     (M1)

Notes: Award (M1) for correct substitution of their gradient into equation of line. Follow through from their answer to part (b).


\(y = \frac{5}{2}x - \frac{19}{2}\)     (A1)(ft)

OR

\(y - 3 = \frac{5}{2}(x - 5)\)     (M1)(A1)(ft)     (C2)


Notes: Award (M1) for correct substitution of their gradient into equation of line. Follow through from their answer to part (b).

c.

(3, −2)     (A1)(ft)(A1)(ft)     (C2)


Notes: If parentheses not seen award at most (A0)(A1)(ft). Accept x = 3, y = −2. Follow through from their answer to part (c), even if no working is seen. Award (M1)(A1)(ft) for a sensible attempt to solve \(2x + 5y = −4\) and their \(y = \frac{5}{2}x - \frac{19}{2}\) or equivalent, simultaneously.

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

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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