Straight lines - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 16 exam-style questions on Straight lines for Edexcel A-Level Maths (Paper 1). 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.

16 questions
2–10 marks each
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16 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

3 marks

The line \(l_1\) has equation \(4x - 3y + 12 = 0\).

The line \(l_2\) is perpendicular to \(l_1\) and passes through the point \(P(8, -1)\).

Find an equation for \(l_2\), giving your answer in the form \(y = mx + c\).

Model answer

\(3y = 4x + 12 \Rightarrow y = \dfrac{4}{3}x + 4\), so the gradient of \(l_1\) is \(\dfrac{4}{3}\).
The gradient of \(l_2\) is \(-\dfrac{3}{4}\), since \(\dfrac{4}{3} \times \left(-\dfrac{3}{4}\right) = -1\).
\(y + 1 = -\dfrac{3}{4}(x - 8)\)
\(y = -\dfrac{3}{4}x + 6 - 1\)
\(y = -\dfrac{3}{4}x + 5\)

Explanation

Perpendicular gradients multiply to give -1, so the gradient of \(l_2\) is the negative reciprocal of \(\tfrac{4}{3}\), which is \(-\tfrac{3}{4}\). Substituting this and P into \(y - y_1 = m(x - x_1)\) and rearranging gives the equation in the form asked for. Common errors are taking the gradient of \(l_1\) as 4 and using only the reciprocal without changing the sign.

Mark scheme

1 mark
B1 Gradient of \(l_1\) is \(\dfrac{4}{3}\); a correct final answer of \(y = -\dfrac{3}{4}x + 5\) with no working scores 3 marks
An answer of \(y = -\tfrac{3}{4}x + 5\) with no working scores 3 marks. May be implied by a correct perpendicular gradient.
1 mark
M1 Uses the perpendicular gradient \(-\dfrac{1}{m}\) (with their m) and P in \(y - y_1 = m(x - x_1)\) or y = mx + c: \(y + 1 = -\dfrac{3}{4}(x - 8)\)
Using the gradient \(\tfrac{4}{3}\) itself (a parallel line) or \(\tfrac{3}{4}\) is M0.
1 mark
A1 \(y = -\dfrac{3}{4}x + 5\)
Accept y = -0.75x + 5. Must be in the form y = mx + c: \(3x + 4y - 20 = 0\) alone is A0. If y = mx + c is used, c must be found.

Question 2

5 marks

The line l is parallel to the line with equation \(3x + 4y = 5\), and \((4, 3)\) lies on l.

Write an equation for l in the form \(px + qy = r\), where p, q and r are integers.

The line l meets the x-axis at the point A and the y-axis at the point B. Find the area of the triangle OAB, where O is the origin, and the length AB.

Model answer

Parallel lines have equations of the form \(3x + 4y = k\).
Through (4, 3): \(k = 3(4) + 4(3) = 24\), so l is \(3x + 4y = 24\).
\(y = 0\): \(x = 8\), so A(8, 0); \(x = 0\): \(y = 6\), so B(0, 6).
Area of OAB \(= \tfrac{1}{2} \times 8 \times 6 = 24\)
\(AB = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\)

Explanation

A line parallel to \(3x + 4y = 5\) has the same x and y coefficients, so l is \(3x + 4y = k\), and the point (4, 3) gives k = 24. The intercepts give a right-angled triangle with the axes, so its area is half the product of the intercepts and AB comes from Pythagoras. A common error is to use the perpendicular gradient \(\tfrac{4}{3}\) instead of the parallel one.

Mark scheme

1 mark
M1 Uses the gradient \(-\dfrac{3}{4}\) with the point (4, 3), or writes \(3x + 4y = k\) and substitutes (4, 3); or a correct final answer of \(3x + 4y = 24\), area 24 and AB = 10 with no working
An answer of \(3x + 4y = 24\), an area of 24 and a length of 10 with no working scores 5 marks.
1 mark
A1 \(3x + 4y = 24\)
Accept any integer multiple, e.g. \(6x + 8y = 48\). \(y = -\tfrac{3}{4}x + 6\) is A0 unless rewritten in the form asked for.
1 mark
B1ft A(8, 0) and B(0, 6)
ft their equation. Both points needed.
1 mark
B1ft Area 24
\(\tfrac{1}{2} \times 8 \times 6\); ft their A and B. Units are not needed.
1 mark
B1ft AB = 10
By Pythagoras, \(\sqrt{8^2 + 6^2}\); ft their A and B. Accept \(\sqrt{100}\).

Question 3

7 marks

Three points are \(A(-2, 1)\), \(B(2, 3)\) and \(C(5, -3)\).

Show that angle ABC is a right angle.

Find the exact area of the triangle ABC.

A fourth point D is chosen so that ABCD is a rectangle. Find the coordinates of D.

Model answer

Gradient of AB \(= \dfrac{3 - 1}{2 - (-2)} = \dfrac{1}{2}\); gradient of BC \(= \dfrac{-3 - 3}{5 - 2} = -2\)
\(\dfrac{1}{2} \times (-2) = -1\), so AB is perpendicular to BC: angle ABC is a right angle.
\(AB = \sqrt{4^2 + 2^2} = \sqrt{20} = 2\sqrt{5}\), \(BC = \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}\)
Area \(= \dfrac{1}{2} \times 2\sqrt{5} \times 3\sqrt{5} = 15\)
From B to C is \(+3\) in x and \(-6\) in y, so D is A moved by the same amounts:
\(D = (-2 + 3, 1 - 6) = (1, -5)\)

Explanation

Gradients whose product is -1 show that AB and BC are perpendicular, so the angle at B is 90 degrees and the two shorter sides give the area as half their product. In a rectangle the side AD is parallel and equal to BC, so D is found by applying the step from B to C to A. A common error is to use the step from A to B, which gives a point that does not complete the rectangle in the right order.

Mark scheme

1 mark
M1 Finds the gradients of AB and BC, \(\dfrac{3 - 1}{2 - (-2)}\) and \(\dfrac{-3 - 3}{5 - 2}\). An answer with no working scores 0 marks for the show-that.
An answer of area 15 and D(1, -5) with no working scores 5 marks, as a special case: M1 A1 A1 for the area and M1 A1 for D; the show-that scores 0 marks without working. Alternative: Pythagoras, \(AB^2 + BC^2 = 20 + 45 = 65 = AC^2\).
1 mark
A1* Gradients \(\dfrac{1}{2}\) and \(-2\), product \(\dfrac{1}{2} \times (-2) = -1\), so AB is perpendicular to BC and angle ABC = 90 degrees (with a conclusion)
Needs both gradients, the product -1 (or 'negative reciprocals') and a conclusion.
1 mark
M1 Finds the lengths AB and BC using Pythagoras (or uses another complete method for the area: see comments)
Other correct methods for the area are fine, e.g. the shoelace (determinant) formula, \(\tfrac{1}{2}\left|(-2)(3 - (-3)) + 2(-3 - 1) + 5(1 - 3)\right|\), or the surrounding rectangle minus three right-angled triangles, \(42 - 4 - 9 - 14\): M1 for the complete method, the next A1 for a correct expression (in place of the two lengths) and the final A1 for 15.
1 mark
A1 \(AB = \sqrt{20} = 2\sqrt{5}\) and \(BC = \sqrt{45} = 3\sqrt{5}\)
Both lengths, exact.
1 mark
A1 Area = 15
\(\tfrac{1}{2} \times 2\sqrt{5} \times 3\sqrt{5}\). Exact: 15.0 from rounded lengths is A0 unless exact working is seen.
1 mark
M1 Uses a correct method for D, e.g. D = A + (C - B), moving from A by the same step as from B to C, (+3, -6)
Or intersects the line through A parallel to BC with the line through C parallel to AB.
1 mark
A1 \(D(1, -5)\)

Question 4

10 marks

A phone is plugged in to charge. The charge in its battery t minutes after it is plugged in is \(P\)% of a full charge, where P is modelled by the equation \(P = pt + q\) and p and q are constants. The charge is 34% after 10 minutes and 76% after 25 minutes.

Find the values of p and q, and interpret each of them in the context of the model.

After 30 minutes the charge is 87%. Evaluate the model in the light of this information.

Use the model to find the time, after the phone is plugged in, at which the battery would be fully charged. Give this time to the nearest second, in minutes and seconds, and comment on the reliability of your answer.

Model answer

\(10p + q = 34\) and \(25p + q = 76\)
Subtracting: \(15p = 42 \Rightarrow p = 2.8\); then \(q = 34 - 28 = 6\)
\(P = 2.8t + 6\)
p = 2.8: the charge rises by 2.8 percentage points each minute.
q = 6: the battery had 6% of a full charge when the phone was plugged in.
\(t = 30\): \(P = 2.8(30) + 6 = 90\). This is within 3 percentage points of the measured 87%, so the model is reasonably accurate after 30 minutes.
\(2.8t + 6 = 100 \Rightarrow 2.8t = 94 \Rightarrow t = 33.571...\) minutes
\(0.571... \times 60 = 34.3\) seconds, so the model predicts a full charge after 33 minutes 34 seconds.
This is unlikely to be reliable: a phone's charging slows down as the battery nears full charge, and the model already overestimated the charge at 30 minutes, so the real time is likely to be longer.

Explanation

Two data points give the gradient p = 2.8, the charge gained each minute, and the intercept q = 6, the charge when the phone was plugged in. To evaluate the model, compare its prediction of 90% with the measured 87% and give a verdict: the values are close, so the model is good at that time. Solving \(2.8t + 6 = 100\) gives 33.571... minutes, and a common error is to write this as 33 minutes 57 seconds instead of converting 0.571... minutes to 34 seconds; the prediction is doubtful because charging slows as a battery fills.

Mark scheme

1 mark
M1 Forms two equations, \(10p + q = 34\) and \(25p + q = 76\), or finds the gradient \(\dfrac{76 - 34}{25 - 10}\); or correct final answers p = 2.8 and q = 6, with interpretations, a correct evaluation, and 33 minutes 34 seconds with a comment, with no working
An answer of p = 2.8 and q = 6, with interpretations, an evaluation using 90%, and 33 minutes 34 seconds with a comment, with no working scores 10 marks. A calculator may be used to solve the simultaneous equations.
1 mark
A1 p = 2.8
Accept \(\tfrac{14}{5}\).
1 mark
A1 q = 6
Depends on the first M mark.
1 mark
B1 p: the charge rises by 2.8 percentage points (2.8% of a full charge) each minute, the rate of charging
Must refer to the increase in the charge per minute, in context (2.8 is exact). 'The gradient' alone is B0.
1 mark
B1 q: the charge in the battery when the phone was plugged in, 6%
Must refer to the charge at the start (t = 0). 'The intercept' alone is B0.
1 mark
M1 Substitutes t = 30 into their model: \(P = 2.8(30) + 6\)
ft their p and q (2.8 and 6 are exact).
1 mark
A1 The model gives 90%, which is close to 87% (3 percentage points too high), so the model is a good fit at t = 30
Needs the correct value 90, a comparison with 87 and a conclusion such as 'good', 'reasonable' or 'fairly accurate'. Do not accept 'the model is wrong because 90 is not 87' or 'it overestimates' with no verdict on the model.
1 mark
M1 Solves \(2.8t + 6 = 100\) (their p and q)
ft their p and q (2.8 and 6 are exact). Using a full charge other than 100 (e.g. P = 1) is M0.
1 mark
A1 33 minutes 34 seconds
From t = 33.571... minutes and \(0.571... \times 60 = 34.3\) seconds. Accept 33 min 34 s. 33.6 minutes alone, or 33 minutes 57 seconds (from reading the decimal part as seconds), is A0.
1 mark
B1 A valid comment on reliability, e.g. a phone's charging slows down as the battery nears full charge, so the charge will not keep rising at the same rate and the real time is likely to be longer; or the model already overestimated the charge at 30 minutes
Must refer to the model or the context. Also accept that the answer relies on extrapolation beyond the measured times. 'It is only a model' or 'phones vary' alone is B0.

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