4MA0 Higher June 2014 Paper 4 Q13

EdexcelHigherOld spec5 marksLinear Graphs & Gradients

13 The line L passes through the points (0, −2) and (6, 1)

(a) Find an equation of the line L. (3)

(b) Find an equation of the line that is parallel to L and which passes through the point (4, −2) (2)

Mark scheme (a)

Scheme	Marks
eg \(\dfrac{1 - -2}{6 - 0}\) oe \(\left(= \dfrac{1}{2}\text{ oe}\right)\)	M1
\(y = \text{“}\tfrac{1}{2}\text{”}x - 2\) or \(y = mx - 2\) or \(y = \text{“}\tfrac{1}{2}\text{”}x + c\)	M1
\(y = \dfrac{1}{2}x - 2\) oe	A1
	(3)

Notes

M1: for any correct method to find gradient

M1: for “\(\frac{1}{2}\)” substituted for \(m\) or −2 substituted for \(c\) in \(y = mx + c\) or
\(y - 1 = \text{“}\tfrac{1}{2}\text{”}(x - 6)\) oe or
\(y - -2 = \text{“}\tfrac{1}{2}\text{”}(x - 0)\) oe

A1: NB Award M2A0 for a final answer of \(0.5x - 2\) or L = \(0.5x - 2\)

Alternative

Scheme	Marks
\(-2 = 0 + c\) ; \(1 = 6m + c\)	M1
\(1 = 6m + -2\)	M1
\(y = \dfrac{1}{2}x - 2\) oe	A1

Notes

M1: form two simultaneous equations

M1: substitute for \(c\)

A1: NB Award M2A0 for a final answer of \(0.5x - 2\) or L = \(0.5x - 2\)

Mark scheme (b)

Scheme	Marks
	M1
\(y = \dfrac{1}{2}x - 4\) oe	A1
	(2)
	(5 marks)

Notes

M1: for correct substitution of (4, −2) into \(y = \text{“}\tfrac{1}{2}\text{”}x + c\) oe using their gradient found in (a)

A1: for \(y = \frac{1}{2}x - 4\) oe follow through with their gradient found in (a)
NB Award M1A0 for a final answer of \(0.5x - 4\)