Straight lines - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 15 exam-style questions on Straight lines for Edexcel A-Level Maths (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

15 questions
2–9 marks each
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15 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

The point P has coordinates \((k, 2k - 1)\), where k is a constant.
Given that P lies on the line with equation \(5x - 3y = 11\), find the value of k.

Model answer

\(5k - 3(2k - 1) = 11\)
\(5k - 6k + 3 = 11\)
\(-k = 8\)
\(k = -8\)

Explanation

A point lies on a line when its coordinates satisfy the line's equation, so substitute \(x = k\) and \(y = 2k - 1\). The method mark is for the substitution and the accuracy mark for \(k = -8\). A common error is a sign slip when expanding \(-3(2k - 1)\).

Mark scheme

1 mark
M1 Substitutes the coordinates of P into the equation: \(5k - 3(2k - 1) = 11\); or a correct final answer of \(k = -8\) with no working
An answer of \(k = -8\) with no working scores 2 marks.
1 mark
A1 \(k = -8\)
A common error is \(-3(2k - 1) = -6k - 3\), giving \(k = -14\): M1 A0.

Question 2

5 marks

The points \(A(-1, 2)\), \(B(5, 4)\) and \(C(9, 8)\) are three of the vertices of the trapezium ABCD, in which the side AB is parallel to the side DC. The vertex D lies on the y-axis.

Find an equation of the line through C and D, giving your answer in the form \(ax + by + c = 0\), where a, b and c are integers.

Find the coordinates of D, and show that ABCD is not a parallelogram.

Model answer

Gradient of AB \(= \dfrac{4 - 2}{5 - (-1)} = \dfrac{2}{6} = \dfrac{1}{3}\)
DC is parallel to AB, so the line through C has gradient \(\dfrac{1}{3}\):
\(y - 8 = \dfrac{1}{3}(x - 9)\)
\(3y - 24 = x - 9\)
\(x - 3y + 15 = 0\)
D is on the y-axis, so \(x = 0\): \(-3y + 15 = 0\), \(y = 5\). D is \((0, 5)\).
Gradient of AD \(= \dfrac{5 - 2}{0 - (-1)} = 3\) and gradient of BC \(= \dfrac{8 - 4}{9 - 5} = 1\).
These are not equal, so AD is not parallel to BC, and ABCD is not a parallelogram.

Explanation

Since DC is parallel to AB, the line through C has the same gradient as AB, \(\dfrac{1}{3}\), and D is where this line meets the y-axis. A parallelogram needs both pairs of opposite sides to be parallel, but AD and BC have gradients 3 and 1. Common errors are using the gradient of BC for the line through C and leaving the equation without '= 0' or with fractions.

Mark scheme

1 mark
M1 Finds the gradient of AB, \(\dfrac{2}{6} = \dfrac{1}{3}\), and uses it as the gradient of DC; correct final answers of \(x - 3y + 15 = 0\) and \(D(0, 5)\) with no working score 4 marks
An answer of \(x - 3y + 15 = 0\) and \(D(0, 5)\) with no working scores 4 marks; showing that ABCD is not a parallelogram needs the gradients (or lengths) to be seen.
1 mark
M1 (dM1) Forms the equation of the line through \(C(9, 8)\) with their gradient: \(y - 8 = \dfrac{1}{3}(x - 9)\)
Depends on the previous M mark.
1 mark
A1 \(x - 3y + 15 = 0\)
Accept any integer multiple, such as \(-x + 3y - 15 = 0\). The '= 0' is needed. \(y = \dfrac{1}{3}x + 5\) is A0 (not in the form asked).
1 mark
B1 \(D(0, 5)\)
From \(x = 0\) in their equation, or correct.
1 mark
B1 Shows that AD and BC are not parallel (gradients 3 and 1), or that \(AB = \sqrt{40}\) and \(DC = \sqrt{90}\) are not equal, and concludes that ABCD is not a parallelogram
Correct values and a conclusion are needed. A sketch or 'it does not look like one' alone is B0.

Question 3

6 marks

The points A and B have coordinates (2, 8) and (6, 4) respectively.

Find an equation of the line through A and B.

The line \(l_m\) has equation \(y = mx + 2\), where m is a constant and \(m ≠ -1\).

Show that \(l_m\) meets the line through A and B at the point with x-coordinate \(\dfrac{8}{m + 1}\).

Hence, or otherwise, find the set of values of m for which \(l_m\) meets the line segment AB. Give your answer in set notation.

Model answer

Gradient of AB \(= \dfrac{4 - 8}{6 - 2} = -1\)
\(y - 8 = -(x - 2)\), so \(y = 10 - x\)
Where \(l_m\) meets AB: \(mx + 2 = 10 - x\)
\(mx + x = 8\)
\(x(m + 1) = 8\), so \(x = \dfrac{8}{m + 1}\) (as \(m ≠ -1\))
The meeting point is on the segment AB when \(2 \leqslant x \leqslant 6\):
\(2 \leqslant \dfrac{8}{m + 1} \leqslant 6\)
\(\dfrac{8}{m + 1} > 0\), so \(m + 1 > 0\), and then \(2(m + 1) \leqslant 8\) and \(6(m + 1) \geqslant 8\)
\(m \leqslant 3\) and \(m \geqslant \dfrac{1}{3}\)
\(\left\{m : \dfrac{1}{3} \leqslant m \leqslant 3\right\}\)

Explanation

The line AB has gradient \(-1\), and equating \(mx + 2\) with \(10 - x\) gives the x-coordinate \(\dfrac{8}{m + 1}\) of the meeting point. The meeting point lies on the segment only when this x-coordinate is from 2 to 6 inclusive, which gives \(\dfrac{1}{3} \leqslant m \leqslant 3\); equivalently, every line \(l_m\) passes through (0, 2), so its gradient must lie between the gradients of the lines from (0, 2) to B and to A. Common errors are using strict inequalities, although the segment includes A and B, and giving an interval instead of set notation.

Mark scheme

1 mark
M1 Finds the gradient of AB, \(\dfrac{4 - 8}{6 - 2} = -1\), and uses it with A or B, e.g. \(y - 8 = -(x - 2)\); or a correct final answer of \(y = 10 - x\) with no working, which scores 2 marks; an answer with no working scores 0 marks for the show-that
A correct equation \(y = 10 - x\) with no working scores 2 marks; an answer with no working scores 0 marks for the show-that; a correct final answer of \(\left\{m : \dfrac{1}{3} \leqslant m \leqslant 3\right\}\) with no working scores 1 mark (the final A1, as a special case). Using \(\dfrac{x_2 - x_1}{y_2 - y_1}\) for the gradient is M0.
1 mark
A1 \(y = 10 - x\)
Accept any correct form, e.g. \(x + y = 10\) or \(x + y - 10 = 0\).
1 mark
M1 Sets \(mx + 2\) equal to their expression for y on AB (or substitutes \(y = mx + 2\) into their equation of AB) and collects the x terms, e.g. \(mx + x = 8\)
Follow through their line AB of the form \(y = px + q\).
1 mark
A1* \(x(m + 1) = 8\), so \(x = \dfrac{8}{m + 1}\) cso
Requires the correct line AB. A factorised step such as \(x(m + 1) = 8\) must be seen before the given answer. Dividing by \(m + 1\) is valid because \(m ≠ -1\).
1 mark
M1 Uses the condition that the meeting point lies between A and B, \(2 \leqslant \dfrac{8}{m + 1} \leqslant 6\) (or \(4 \leqslant 10 - \dfrac{8}{m + 1} \leqslant 8\) for the y-coordinate), and solves for m; or finds the gradients of the lines from (0, 2) to A and to B, 3 and \(\dfrac{1}{3}\)
Alternative method: every line \(l_m\) passes through (0, 2), so it meets the segment AB when its gradient lies between the gradients of the lines from (0, 2) to B and to A, \(\dfrac{4 - 2}{6 - 0} = \dfrac{1}{3}\) and \(\dfrac{8 - 2}{2 - 0} = 3\). Using the y-coordinates 4 and 8 as limits for the x-coordinate is M0.
1 mark
A1 \(\left\{m : \dfrac{1}{3} \leqslant m \leqslant 3\right\}\)
Both ends are included, because the segment includes A and B: when \(m = 3\) the line passes through A, and when \(m = \dfrac{1}{3}\) it passes through B. Strict inequalities score A0. Accept \(\left\{m : m \geqslant \dfrac{1}{3}\right\} \cap \{m : m \leqslant 3\}\). \(\dfrac{1}{3} \leqslant m \leqslant 3\) not in set notation, or the interval \(\left[\dfrac{1}{3}, 3\right]\), scores A0.

Question 4

8 marks

The triangle ABC has vertices \(A(-7, 3)\), \(B(1, -5)\) and \(C(5, 7)\).
An altitude of a triangle is the straight line through one vertex that is perpendicular to the opposite side.

Find an equation of the altitude through A and an equation of the altitude through B, each in the form \(ax + by + c = 0\), where a, b and c are integers.

These two altitudes meet at the point H. Find the coordinates of H.

Show that H lies on the altitude through C.

Model answer

Gradient of BC \(= \dfrac{7 - (-5)}{5 - 1} = \dfrac{12}{4} = 3\), so the altitude through A has gradient \(-\dfrac{1}{3}\):
\(y - 3 = -\dfrac{1}{3}(x + 7) \Rightarrow 3y - 9 = -x - 7 \Rightarrow x + 3y - 2 = 0\)
Gradient of AC \(= \dfrac{7 - 3}{5 - (-7)} = \dfrac{4}{12} = \dfrac{1}{3}\), so the altitude through B has gradient \(-3\):
\(y + 5 = -3(x - 1) \Rightarrow y = -3x - 2 \Rightarrow 3x + y + 2 = 0\)
From the second equation, \(y = -3x - 2\). Substituting into the first: \(x + 3(-3x - 2) - 2 = 0 \Rightarrow -8x - 8 = 0 \Rightarrow x = -1\), \(y = 1\)
H is \((-1, 1)\).
Gradient of AB \(= \dfrac{-5 - 3}{1 - (-7)} = -1\), so the altitude through C has gradient 1: \(y - 7 = x - 5\), that is \(y = x + 2\).
At \(x = -1\): \(y = -1 + 2 = 1\), so \(H(-1, 1)\) lies on the altitude through C.

Explanation

Each altitude is perpendicular to the opposite side, so its gradient is the negative reciprocal of that side's gradient, and it passes through the vertex opposite that side. Solving two altitudes simultaneously gives H, and substituting H into the third altitude (or checking that CH is perpendicular to AB) shows that all three altitudes meet at H. A common error is to make the altitude through a vertex perpendicular to a side that passes through that vertex.

Mark scheme

1 mark
M1 Finds the gradient of BC, 3, and uses \(-\dfrac{1}{3}\) for the altitude through A. An answer with no working scores 0 marks for the show-that.
An answer with no working scores 0 marks for the show-that; correct equations \(x + 3y - 2 = 0\) and \(3x + y + 2 = 0\) and \(H(-1, 1)\) with no working score 6 marks.
1 mark
A1 \(x + 3y - 2 = 0\)
Any integer multiple, with '= 0'.
1 mark
M1 Finds the gradient of AC, \(\dfrac{1}{3}\), and uses \(-3\) for the altitude through B
1 mark
A1 \(3x + y + 2 = 0\)
Any integer multiple, with '= 0'.
1 mark
M1 Solves their two altitude equations simultaneously
1 mark
A1 \(H(-1, 1)\)
1 mark
M1 Finds the gradient of AB, \(-1\), and either the gradient of CH or the equation of the altitude through C, \(y = x + 2\)
1 mark
A1* Gradient of CH \(= 1\) and \(1 \times (-1) = -1\), so CH is perpendicular to AB (or H satisfies \(y = x + 2\)); so H lies on the altitude through C, with a conclusion
cso: correct values and a conclusion are needed.

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