4MA0 Higher June 2012 Paper 3 Q14

EdexcelHigherOld spec5 marksLinear Graphs & Gradients

(a) The equation of a line L is \(2x - 3y = 6\)
Find the gradient of L. (3)

(b) Find the equation of the line which is parallel to L and passes through the point (6, 9). (2)

Mark scheme (a)

Scheme	Marks
\(3y = 2x - 6\) or \(-3y = 6 - 2x\)	M1
\(y = \frac{2}{3}x - 2\) oe or \(y = \dfrac{2x - 6}{3}\) oe	M1
\(\frac{2}{3}\) oe	A1
	(3)

Notes

M1: May be implied by second M1 or by \(y = \frac{2}{3}x + c\) even if value of \(c\) is incorrect
or finds coordinates of 2 points on the line eg (3, 0), (0, −2), table, sketch showing line cutting \(x\)-axis at 3 and \(y\)-axis at −2

M1: for correct rearrangement of \(3y = 2x - 6\) with \(y\) as subject

or for clear attempt to use \(\dfrac{\text{vert difference}}{\text{horiz difference}}\) for their two points on L

A1: for \(\frac{2}{3}\) oe inc decimal equivalent rounded or truncated to at least 2 dp
Do not award A1 for \(\frac{2}{3}x\)

Apart from Questions 9, 13, 18, 20 and 21 (where the mark scheme states otherwise) the correct answer, unless clearly obtained by an incorrect method, should be taken to imply a correct method.

Mark scheme (b)

Scheme	Marks
\(9 = \text{“}\tfrac{2}{3}\text{”} \times 6 + c\)	M1
\(y = \frac{2}{3}x + 5\)	A1
	(2)
	(Total 5 marks)

Notes

M1: for correct substitution into \(y = \text{“}\tfrac{2}{3}\text{”}x + c\) using their answer to (a) oe

A1: for \(y = \frac{2}{3}x + 5\) oe inc \(2x - 3y = -15\)
ft from their answer to (a)

SC If M0 A0, award B1 for answer with ‘\(y =\)’ omitted which would otherwise score M1 A1 eg \(\frac{2}{3}x + 5\), \(2x - 3\) if ans to (a) is 2

SC Award B2 if \(y - 9 = \text{“}\tfrac{2}{3}\text{”}(x - 6)\) seen; then isw

SC Award B1 for \(2x - 3y = k\) where \(k \neq -15\) and \(k \neq 6\) with no working

SC If M0 A0, award B1 for \(y = \text{“}\tfrac{2}{3}\text{”}x + c\) where \(c \neq 5\) or \(c \neq 0\) (ie do not award this mark for \(y = \text{“}\tfrac{2}{3}\text{”}x + 5\) or \(y = \text{“}\tfrac{2}{3}\text{”}x\)) or does not ft from (a)