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Date May 2016 Marks available 2 Reference code 16M.1.sl.TZ2.12
Level SL only Paper 1 Time zone TZ2
Command term Find Question number 12 Adapted from N/A

Question

The equation of the straight line \({L_1}\) is \(y = 2x - 3.\)

Write down the \(y\)-intercept of \({L_1}\) .

[1]
a.

Write down the gradient of \({L_1}\) .

[1]
b.

The line \({L_2}\) is parallel to \({L_1}\) and passes through the point \((0,\,\,3)\) .

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

[1]
c.

The line \({L_3}\) is perpendicular to \({L_1}\) and passes through the point \(( - 2,\,\,6).\)

Write down the gradient of \({L_3}.\)

[1]
d.

Find the equation of \({L_3}\) . Give your answer in the form \(ax + by + d = 0\) , where \(a\) , \(b\) and \(d\) are integers.

[2]
e.

Markscheme

\((0,\,\, - 3)\)       (A1)     (C1)

Note: Accept \( - 3\) or \(y =  - 3.\)

a.

\(2\)       (A1)     (C1)

b.

\(y = 2x + 3\)        (A1)(ft)     (C1)

Note: Award (A1)(ft) for correct equation. Follow through from part (b)
Award (A0) for \({L_2} = 2x + 3\).

c.

\( - \frac{1}{2}\)             (A1)(ft)     (C1)

Note: Follow through from part (b).

d.

\(6 =  - \frac{1}{2}( - 2) + c\)        (M1)

\(c = 5\)  (may be implied)

OR

\(y - 6 =  - \frac{1}{2}(x + 2)\)        (M1)

Note: Award (M1) for correct substitution of their gradient in part (d) and the point \(( - 2,\,\,6)\). Follow through from part (d).

\(x + 2y - 10 = 0\)  (or any integer multiple)        (A1)(ft)     (C2)

Note: Follow through from (d). The answer must be in the form \(ax + by + d = 0\) for the (A1)(ft) to be awarded. Accept any integer multiple.

e.

Examiners report

Question 12: Linear function.

Many candidates demonstrated a good understanding of linear functions so successfully found the \(y\)-intercepts, gradient and equation in the form \(y = mx + c\). However only the very best were able to rewrite this in the form \(ax + by + d = 0\) where \(a\), \(b\) and \(d\) are integers.

a.

Question 12: Linear function.

Many candidates demonstrated a good understanding of linear functions so successfully found the \(y\)-intercepts, gradient and equation in the form \(y = mx + c\). However only the very best were able to rewrite this in the form \(ax + by + d = 0\) where \(a\), \(b\) and \(d\) are integers.

b.

Question 12: Linear function.

Many candidates demonstrated a good understanding of linear functions so successfully found the \(y\)-intercepts, gradient and equation in the form \(y = mx + c\). However only the very best were able to rewrite this in the form \(ax + by + d = 0\) where \(a\), \(b\) and \(d\) are integers.

c.

Question 12: Linear function.

Many candidates demonstrated a good understanding of linear functions so successfully found the \(y\)-intercepts, gradient and equation in the form \(y = mx + c\). However only the very best were able to rewrite this in the form \(ax + by + d = 0\) where \(a\), \(b\) and \(d\) are integers.

d.

Question 12: Linear function.

Many candidates demonstrated a good understanding of linear functions so successfully found the \(y\)-intercepts, gradient and equation in the form \(y = mx + c\). However only the very best were able to rewrite this in the form \(ax + by + d = 0\) where \(a\), \(b\) and \(d\) are integers.

e.

Syllabus sections

Topic 7 - Introduction to differential calculus » 7.3 » Equation of the tangent at a given point.
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