Higher June 2017 Paper 1 Q17

AQACurrent spec4 marksLinear Graphs & Gradients

17 A is the point \((2, -5)\)

B is the point \((4, -9)\)

(a) Show that the gradient of the straight line passing through A and B is \(-2\) [2 marks]

(b) C is the point \((-301, 601)\)

Does C lie on the straight line passing through A and B?

You must show your working. [2 marks]

Mark scheme (a)

Answer	Mark	Comments
Alternative method 1
\(\dfrac{-9 - -5}{4 - 2}\) or \(\dfrac{-5 - -9}{2 - 4}\) or \((2, -5) - (4, -9) = (-2, 4)\) or \((4, -9) - (2, -5) = (2, -4)\) or \(\dfrac{\text{change in } y}{\text{change in } x}\) or \(\dfrac{\Delta y}{\Delta x}\) or triangle drawn with points A and B and side lengths of 4 and (–)2 identified or correct explanation of pattern of graph and \(\dfrac{-4}{2}\) \(= -2\) or \(\dfrac{4}{-2}\) \(= -2\)	B2	oe fraction eg \(\dfrac{-9 + 5}{4 - 2}\) or \(\dfrac{-5 + 9}{2 - 4}\) B1 for \(\dfrac{-9 - -5}{4 - 2}\) or \(\dfrac{-5 - -9}{2 - 4}\) or \((2, -5) - (4, -9) = (-2, 4)\) or \((4, -9) - (2, -5) = (2, -4)\) or \(\dfrac{\text{change in } y}{\text{change in } x}\) or \(\dfrac{\Delta y}{\Delta x}\) or triangle drawn with points A and B and side lengths of 4 and (–)2 identified or correct explanation of pattern of graph or \(\dfrac{-4}{2}\) \(= -2\) or \(\dfrac{4}{-2}\) \(= -2\)
Alternative method 2
Gives \(y = -2x + c\) and substitutes \((2, -5)\) or \((4, -9)\) to find \(c = -1\) or \(y - -5 = -2(x - 2)\) or \(y + 5 = -2(x - 2)\) or \(y - -9 = -2(x - 4)\) or \(y + 9 = -2(x - 4)\) and gives \(y = -2x - 1\) and correctly substitutes and evaluates with the other pair of coordinates to check	B2	B1 for \((2, -5)\) or \((4, -9)\) to find \(c = -1\) or \(y - -5 = -2(x - 2)\) or \(y + 5 = -2(x - 2)\) or \(y - -9 = -2(x - 4)\) or \(y + 9 = -2(x - 4)\) or gives \(y = -2x - 1\) and correctly substitutes and evaluates with one or both pair(s) of coordinates
Alternative method 3
\(-5 = 2m + c\) and \(-9 = 4m + c\) and works out \(m = -2\) using a correct algebraic method	B2	oe equations B1 for \(-5 = 2m + c\) and \(-9 = 4m + c\)
Alternative method 4
\(-5 = -2(2) + c\) and \(-9 = -2(4) + c\) and works out \(c = -1\) for both	B2	oe equations B1 for \(-5 = -2(2) + c\) and \(-9 = -2(4) + c\)

Additional guidance

In alt 1, examples of correct explanation are: 2 left and 4 up 2 right and 4 down
In alt 1, points A and B can be identified on a diagram by their coordinates
In alt 2, accept rearrangements of \(y = -2x - 1\) eg \(\;2x + y = -1\)
\(\dfrac{-5 - 9}{2 - 4}\) or \(\dfrac{-9 - 5}{4 - 2}\) \(\;(= -2\) or \(= 2)\)	B0

Mark scheme (b)

Answer	Mark	Comments
Alternative method 1 – uses given point with one from (a) to show gradient \(= -2\)
\(\dfrac{601 - -9}{-301 - 4}\) or \(\dfrac{601 - -5}{-301 - 2}\)	M1	oe eg \(\dfrac{610}{-305}\) or \(\dfrac{606}{-303}\)
\(-2\) and Yes	A1	Must see working for M1
Alternative method 2 – correct or no equation shown in (a)
Correct method to find \(y = -2x - 1\)	M1	May be seen in part (a)
\(y = -2x - 1\) and shows that \(601 = -2(-301) - 1\) and Yes	A1
Alternative method 3 – incorrect equation shown in (a)
Substitutes \(-301\) and 601 into their equation from (a)	M1	equation must involve \(x\) and \(y\)
Correct evaluation and No	A1ft
Alternative method 4 – have gained two marks in (a) by any method
uses \((2, -5)\) or \((4, -9)\) to work out \(c = -1\)	M1
\(601 = -2(-301) + c\) and \(c = -1\) and Yes	A1
Alternative method 5 – have shown that \(c = -1\) for both points in (a)
\(601 = -2(-301) + c\)	M1
\(601 = -2(-301) + c\) and \(c = -1\) and Yes	A1

Additional guidance

\(y = -2x - 1\) given in (a) but not used in (b)	M0 for equation
Correct method in (a) to show that the gradient is \(-2\), but followed by incorrect equation. Incorrect equation then used correctly in (b)	B2 in (a) M1A0 in (b)