Practise Straight-line graphs. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
The equation of a straight line is \(y = 3x + 2\).
(a) Write down the gradient of the line.[1]
(b) Write down the coordinates of the point where the line crosses the \(y\)-axis.[1]
(c) Does the point \((6, 19)\) lie on the line? You must show how you get your answer.[2]
Show the answer and mark scheme
(a)Answer: \(3\)
B1 for 3
Worked solution: The gradient is the coefficient of \(x\): \(3\).
(b)Answer: \((0, 2)\)
B1 for \((0, 2)\)
Worked solution: When \(x = 0\), \(y = 2\).
(c)Answer: No: when \(x = 6\), \(y = 20\)
M1 for substituting \(x = 6\) to get \(y = 20\)
C1 for no with a correct reason
Worked solution: \(y = 3 \times 6 + 2 = 20\), not 19, so the point is not on the line.
Question 2Medium3 marks
Line L has equation \(2x + 3y = 6\).
(a) Find the gradient of line L.[2]
(b) Write down the coordinates of the point where line L crosses the \(y\)-axis.[1]
Show the answer and mark scheme
(a)Answer: \(-\frac{2}{3}\)
M1 for rearranging to \(y = -\frac{2}{3}x + 2\) or \(3y = -2x + 6\)
A1 for \(-\frac{2}{3}\) oe
Worked solution: \(3y = -2x + 6\) so \(y = -\frac{2}{3}x + 2\). The gradient is \(-\frac{2}{3}\).
(b)Answer: \((0, 2)\)
B1 for \((0, 2)\)
Worked solution: When \(x = 0\): \(3y = 6\) so \(y = 2\).
Question 3Hard5 marks
\(A\), \(B\) and \(C\) are the points \(A(1, 2)\), \(B(5, 4)\) and \(C(3, 8)\).
(a) Prove that triangle \(ABC\) is a right-angled triangle.[3]
(b) Work out the area of triangle \(ABC\).[2]
Show the answer and mark scheme
(a)Answer: Gradient of \(AB = \frac{1}{2}\), gradient of \(BC = -2\); \(\frac{1}{2} \times (-2) = -1\), so \(AB\) is perpendicular to \(BC\): right angle at \(B\).
M1 for gradient of \(AB\) \(= \frac{4 - 2}{5 - 1} = \frac{1}{2}\)
M1 for gradient of \(BC\) \(= \frac{8 - 4}{3 - 5} = -2\)
C1 for \(\frac{1}{2} \times (-2) = -1\) (or −2 is the negative reciprocal of \(\frac{1}{2}\)), so the angle at \(B\) is 90°
Worked solution: Gradient \(AB = \frac{2}{4} = \frac{1}{2}\), gradient \(BC = \frac{4}{-2} = -2\). \(\frac{1}{2} \times (-2) = -1\), so \(AB\) and \(BC\) are perpendicular and angle \(ABC = 90^\circ\).
(b)Answer: 10 square units
M1 for \(AB = BC = \sqrt{20}\) and \(\frac{1}{2} \times \sqrt{20} \times \sqrt{20}\)
A1 for 10
Worked solution: \(AB^2 = 4^2 + 2^2 = 20\) and \(BC^2 = 2^2 + 4^2 = 20\). The right angle is at \(B\), so area \(= \frac{1}{2} \times \sqrt{20} \times \sqrt{20} = 10\).