Σ TMUA Practice
Home›Topics›Straight lines

Straight lines

Equations of lines, gradients, parallel and perpendicular lines, distances and midpoints.

Spec MM3.1, M4.9, M4.10Section 1 Part 1: AS-level pure mathematics15 questions7 real past questions
📖 Revision notes🧩 Solution bank⚡ Quick-fire this topicFormula cardsSet 1 questions (PDF)Set 1 solutionsSet 2 questions (PDF)Set 2 solutions

Click an option to answer. A correct first attempt marks the question secure; a wrong one marks it needs work and opens the worked solution. Keys A-H answer the question nearest the top of the screen.

1Straight lines · spec MM3.1 · Paper 1 style · ★☆☆

Find the equation of the line through \((2, -3)\) that is perpendicular to \(3x - 2y + 5 = 0\).

  1. A\(2x - 3y - 13 = 0\)
  2. B\(3x + 2y = 0\)
  3. C\(2x + 3y + 5 = 0\)
  4. D\(2x + 3y - 13 = 0\)
  5. E\(2x + 3y - 5 = 0\)
  6. F\(3x - 2y - 12 = 0\)
Worked solution

The given line has gradient \(\frac32\), so the perpendicular has gradient \(-\frac23\): \(y + 3 = -\frac23(x - 2) \Rightarrow 3y + 9 = -2x + 4 \Rightarrow 2x + 3y + 5 = 0\).

Full method — how and why
Perpendicular line through a point: find the gradient of the given line, take the negative reciprocal, then use the point-gradient form.
  1. Rearrange \(3x - 2y + 5 = 0\) into \(y = \frac32 x + \frac52\), so its gradient is \(\frac32\).
    Alternatively use the shortcut: \(ax + by + c = 0\) has gradient \(-\frac ab = -\frac{3}{-2} = \frac32\).
  2. Perpendicular gradient: \(-\dfrac{1}{3/2} = -\dfrac23\).
    Perpendicular gradients multiply to \(-1\); invert and change sign.
  3. Use \(y - y_1 = m(x - x_1)\) at \((2, -3)\): \(y + 3 = -\frac23(x - 2)\).
    \(y - (-3)\) is \(y + 3\) - the commonest sign slip in this whole topic.
  4. Clear fractions and tidy: \(3y + 9 = -2x + 4 \Rightarrow 2x + 3y + 5 = 0\).
    Multiply through by \(3\) first, then collect everything on one side.
Where it goes wrong
  • Using \(y - 3\) instead of \(y + 3\), which gives \(2x + 3y - 5 = 0\) - an offered option.
  • Taking the perpendicular gradient as \(-\frac32\) or \(\frac23\).
Takeaway Negative reciprocal, and substitute the point with its own sign: \(y - (-3) = y + 3\).
Answer C · Straight lines (MM3.1) · question LIN-01
2Straight lines · spec MM3.1 · Paper 1 style · ★☆☆

The points \(A(1, 2)\), \(B(5, 4)\) and \(C(k, 8)\) are collinear. Find \(k\).

  1. A\(12\)
  2. B\(11\)
  3. C\(10\)
  4. D\(9\)
  5. E\(7\)
  6. F\(13\)
Worked solution

Gradient \(AB = \frac{2}{4} = \frac12\). Gradient \(AC = \frac{6}{k-1} = \frac12 \Rightarrow k - 1 = 12 \Rightarrow k = 13\).

Full method — how and why
Collinear points have equal gradients between any pair. Set two gradients equal and solve.
  1. Gradient of \(AB\): \(\dfrac{4 - 2}{5 - 1} = \dfrac{2}{4} = \dfrac12\).
    Any two of the three points determine the line, so compute the gradient from the pair with known coordinates.
  2. Gradient of \(AC\): \(\dfrac{8 - 2}{k - 1} = \dfrac{6}{k-1}\).
    Use the same starting point \(A\) for both gradients to avoid sign confusion.
  3. Set them equal: \(\dfrac{6}{k-1} = \dfrac12\).
    Collinear means the gradient between any pair is the same.
  4. Cross-multiply: \(k - 1 = 12\), so \(k = 13\).
    Check: from \(B(5,4)\) to \(C(13,8)\) the gradient is \(\frac48 = \frac12\) ✓.
Where it goes wrong
  • Forgetting the \(-1\) and answering \(12\).
  • Inverting the gradient and solving \(\frac{k-1}{6} = \frac12\), which gives \(k = 4\).
Takeaway Collinearity is an equal-gradient condition; always verify with the third pair of points.
Answer F · Straight lines (MM3.1) · question LIN-02
3Straight lines · spec MM3.1 · Paper 1 style · ★☆☆

Which of the following lines is parallel to \(4x + 6y = 9\)?

  1. A\(2x + 3y = 1\)
  2. B\(3x - 2y = 4\)
  3. C\(y = \frac{3}{2}x + 1\)
  4. D\(6x + 4y = 9\)
  5. E\(4x - 6y = 9\)
Worked solution

\(4x + 6y = 9\) has gradient \(-\frac23\); so does \(2x + 3y = 1\).

Full method — how and why
Parallel means equal gradient. Convert each option to gradient form - or compare the coefficient ratios directly.
  1. Gradient of \(4x + 6y = 9\) is \(-\frac{4}{6} = -\frac23\).
    For \(ax + by = c\) the gradient is \(-\frac ab\).
  2. Test the options: \(2x + 3y = 1\) has gradient \(-\frac23\) ✓.
    Equivalently, the coefficients \((2,3)\) are proportional to \((4,6)\) - the constant term is irrelevant to direction.
  3. Reject the rest: \(3x - 2y = 4\) gives \(\frac32\) (perpendicular), \(4x - 6y = 9\) gives \(\frac23\), \(6x + 4y = 9\) gives \(-\frac32\), and \(y = \frac32x+1\) gives \(\frac32\).
    Note how many distractors are the perpendicular gradient - a deliberate trap.
Where it goes wrong
  • Choosing \(6x + 4y = 9\) because the numbers \(4\) and \(6\) reappear - but swapped, so the gradient changes.
  • Forgetting the minus sign in \(-\frac ab\).
Takeaway Parallel \(\Leftrightarrow\) the coefficients of \(x\) and \(y\) are in the same ratio; the constant plays no part.
Answer A · Straight lines (MM3.1) · question LIN-05
4Straight lines · spec MM3.1 · Paper 1 style · ★☆☆

The perpendicular bisector of the line segment joining \((1, 5)\) and \((7, 1)\) has equation

  1. A\(3x - 2y = 6\)
  2. B\(2x - 3y = -1\)
  3. C\(3x + 2y = 18\)
  4. D\(2x + 3y = 17\)
  5. E\(y = \frac32 x - 3\)
  6. F\(3x - 2y = 18\)
Worked solution

Midpoint \((4, 3)\); gradient of segment \(\frac{-4}{6} = -\frac23\), so the bisector has gradient \(\frac32\): \(y - 3 = \frac32(x - 4) \Rightarrow 2y - 6 = 3x - 12 \Rightarrow 3x - 2y = 6\).

Full method — how and why
Perpendicular bisector: through the midpoint, perpendicular to the segment. Two small calculations, then the point-gradient form.
  1. Midpoint: \(\left(\frac{1+7}{2}, \frac{5+1}{2}\right) = (4, 3)\).
    The bisector must pass through the middle of the segment.
  2. Gradient of the segment: \(\dfrac{1 - 5}{7 - 1} = -\dfrac{4}{6} = -\dfrac23\).
    Keep the order of subtraction consistent in numerator and denominator.
  3. Perpendicular gradient: \(\dfrac32\).
    Negative reciprocal of \(-\frac23\).
  4. Line through \((4,3)\): \(y - 3 = \frac32(x - 4) \Rightarrow 2y - 6 = 3x - 12 \Rightarrow 3x - 2y = 6\).
    Multiply by \(2\) to clear the fraction, then collect.
Where it goes wrong
  • Rearranging to \(3x - 2y = 18\) by a sign slip on the constant.
  • Using the segment's gradient instead of the perpendicular one.
Takeaway Perpendicular bisector \(=\) midpoint \(+\) negative reciprocal gradient; it is the locus of points equidistant from the two ends.
Answer A · Straight lines (MM3.1) · question LIN-06
5Straight lines · spec MM3.1 · Paper 1 style · ★☆☆

The line \(ax + by = 12\) has \(x\)-intercept \(3\) and \(y\)-intercept \(-4\). Find \(a - b\).

  1. A\(12\)
  2. B\(7\)
  3. C\(-1\)
  4. D\(0\)
  5. E\(-7\)
  6. F\(1\)
Worked solution

\((3, 0)\): \(3a = 12 \Rightarrow a = 4\). \((0, -4)\): \(-4b = 12 \Rightarrow b = -3\). \(a - b = 7\).

Full method — how and why
Intercepts are points on the line. Substitute each one to get an equation for \(a\) and for \(b\) separately.
  1. \(x\)-intercept \(3\) means the point \((3, 0)\) lies on the line: \(3a + 0 = 12\), so \(a = 4\).
    At an \(x\)-intercept, \(y = 0\).
  2. \(y\)-intercept \(-4\) means \((0, -4)\) lies on it: \(0 - 4b = 12\), so \(b = -3\).
    At a \(y\)-intercept, \(x = 0\). The negative intercept makes \(b\) negative.
  3. Compute \(a - b = 4 - (-3) = 7\).
    Subtracting a negative adds.
Where it goes wrong
  • Writing \(b = 3\) and answering \(1\).
  • Mixing up which intercept is which and substituting \((0,3)\) and \((-4,0)\).
Takeaway An intercept is just a point with one coordinate zero - substitute it into the equation.
Answer B · Straight lines (MM3.1) · question LIN-07
6Straight lines · spec MM3.1 · Paper 1 style · ★☆☆

A line passes through \((1, 2)\) with gradient \(m\). It crosses the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\), and the midpoint of \(AB\) is \((1, 2)\). Find \(m\).

  1. A\(-2\)
  2. B\(\dfrac12\)
  3. C\(2\)
  4. D\(-4\)
  5. E\(-\dfrac12\)
  6. F\(-1\)
Worked solution

Midpoint \((1, 2)\) means \(A = (2, 0)\) and \(B = (0, 4)\). Gradient \(= \frac{4 - 0}{0 - 2} = -2\).

Full method — how and why
The midpoint condition determines both intercepts, and two points give the gradient.
  1. Let \(A = (p, 0)\) and \(B = (0, q)\). Their midpoint is \(\left(\frac p2, \frac q2\right)\).
    One intercept on each axis, so each has a zero coordinate.
  2. Set the midpoint equal to \((1,2)\): \(\frac p2 = 1\) and \(\frac q2 = 2\), so \(p = 2\), \(q = 4\).
    Equate coordinates separately.
  3. Gradient through \((2,0)\) and \((0,4)\): \(\dfrac{4 - 0}{0 - 2} = -2\).
    Consistent order of subtraction in numerator and denominator.
Where it goes wrong
  • Halving instead of doubling, giving \(A = (\frac12, 0)\) and the wrong gradient.
  • Answering \(2\) by ignoring that the line falls from left to right.
Takeaway If \((1,2)\) is the midpoint of the intercepts, the intercepts are \((2,0)\) and \((0,4)\) - double each coordinate.
Answer A · Straight lines (MM3.1) · question LIN-14
7Straight lines · spec MM3.1 · Paper 2 style · ★☆☆

Two distinct lines have equations \(y = m_1 x + c_1\) and \(y = m_2 x + c_2\). Which of the following statements is/are true?

  1. IIf \(m_1 m_2 = -1\) then the lines meet at exactly one point.
  2. IIIf \(m_1 = m_2\) then the lines do not meet.
  3. IIIIf the lines meet on the \(y\)-axis then \(c_1 = c_2\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I: perpendicular lines are not parallel, so they meet once. II: equal gradients and distinct lines means parallel and different, so no intersection. III: on the \(y\)-axis \(x = 0\), so \(y = c_1 = c_2\).

Full method — how and why
Three claims about two distinct straight lines. Each follows from what gradients and intercepts mean.
  1. I: \(m_1m_2 = -1\) means the lines are perpendicular, so \(m_1 \neq m_2\) and the lines are not parallel - they meet exactly once. True.
    Two non-parallel straight lines always meet at exactly one point.
  2. II: \(m_1 = m_2\) with the lines distinct means \(c_1 \neq c_2\): parallel and separate, so they never meet. True.
    The word 'distinct' in the question is doing real work - without it the lines could coincide.
  3. III: on the \(y\)-axis \(x = 0\), so the two \(y\)-values are \(c_1\) and \(c_2\). Meeting there forces \(c_1 = c_2\). True.
    Substituting \(x = 0\) into both equations.
  4. So all three.
    Match to the list.
Where it goes wrong
  • Rejecting II by imagining the lines could coincide - 'distinct' rules that out.
  • Rejecting III by thinking they might meet elsewhere; the claim conditions on them meeting on the \(y\)-axis.
Takeaway Same gradient and distinct \(\Rightarrow\) parallel, never meet. Different gradients \(\Rightarrow\) exactly one intersection. Read 'distinct' carefully.
Answer H · Straight lines (MM3.1) · question X2-05
8Straight lines · spec MM3.1 · Paper 1 style · ★★☆

The line \(L\) passes through \((4, 0)\) and \((0, 6)\). Find the perpendicular distance from the origin to \(L\).

  1. A\(5\)
  2. B\(\dfrac{24}{13}\)
  3. C\(\dfrac{12}{13}\)
  4. D\(\dfrac{12}{\sqrt{13}}\)
  5. E\(2\sqrt{13}\)
  6. F\(\sqrt{13}\)
Worked solution

Area of triangle \(= \frac12 \cdot 4 \cdot 6 = 12\). Hypotenuse \(= \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}\). Distance \(d\) satisfies \(\frac12 \cdot 2\sqrt{13} \cdot d = 12\), so \(d = \frac{12}{\sqrt{13}}\).

Full method — how and why
Perpendicular distance from a point to a line. The 'area two ways' method avoids memorising the distance formula.
  1. The line cuts the axes at \((4,0)\) and \((0,6)\), so with the origin it forms a right-angled triangle of area \(\frac12 \times 4 \times 6 = 12\).
    The two intercepts and the origin make a triangle whose legs lie along the axes - its area is easy.
  2. The hypotenuse (the segment of \(L\)) has length \(\sqrt{4^2 + 6^2} = \sqrt{52} = 2\sqrt{13}\).
    Pythagoras on the same triangle.
  3. Compute the area again using the hypotenuse as base and the required distance \(d\) as height: \(\frac12 \times 2\sqrt{13} \times d = 12\).
    The perpendicular distance from the origin to \(L\) is the height of that triangle on that base.
  4. Solve: \(d = \dfrac{12}{\sqrt{13}}\).
    Equivalently, the formula \(\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}\) with \(3x + 2y = 12\) gives \(\frac{12}{\sqrt{13}}\).
Where it goes wrong
  • Using the distance to one of the intercepts (\(4\), \(6\) or \(2\sqrt{13}\)) instead of the perpendicular distance.
  • Forgetting the square root and answering \(\frac{12}{13}\).
Takeaway Area computed two ways turns a distance problem into arithmetic; or use \(\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}\).
Answer D · Straight lines (MM3.1) · question LIN-03
9Straight lines · spec MM3.1 · Paper 1 style · ★★☆

The lines \(y = 2x + 1\), \(y = -x + 7\) and the \(x\)-axis enclose a triangle. Find its area.

  1. A\(\dfrac{25}{2}\)
  2. B\(\dfrac{75}{4}\)
  3. C\(\dfrac{15}{2}\)
  4. D\(\dfrac{35}{2}\)
  5. E\(15\)
  6. F\(\dfrac{45}{4}\)
Worked solution

Intersection: \(2x + 1 = 7 - x \Rightarrow x = 2\), \(y = 5\). \(x\)-intercepts: \(-\frac12\) and \(7\), base \(= \frac{15}{2}\). Area \(= \frac12 \cdot \frac{15}{2} \cdot 5 = \frac{75}{4}\).

Full method — how and why
A triangle bounded by two lines and an axis. Find the three vertices, then use the axis segment as the base.
  1. Find where the two lines meet: \(2x + 1 = -x + 7 \Rightarrow 3x = 6 \Rightarrow x = 2\), \(y = 5\). Apex \((2,5)\).
    Equate the two expressions for \(y\).
  2. Find the \(x\)-intercepts: \(2x + 1 = 0\) gives \(x = -\frac12\); \(-x + 7 = 0\) gives \(x = 7\).
    The third side is along the \(x\)-axis, so the other two vertices are the \(x\)-intercepts.
  3. Base length: from \(-\frac12\) to \(7\) is \(7 - (-\frac12) = \frac{15}{2}\).
    Subtract, do not add; the negative intercept makes the base longer, not shorter.
  4. Height is the \(y\)-coordinate of the apex, \(5\): area \(= \frac12 \times \frac{15}{2} \times 5 = \frac{75}{4}\).
    With a horizontal base, the height is just the apex's distance from the axis.
Where it goes wrong
  • Using base \(7 - \frac12 = \frac{13}{2}\) by mishandling the negative intercept.
  • Using the apex \(x\)-coordinate \(2\) as the height.
Takeaway Put the base on an axis whenever you can: the height is then simply a coordinate.
Answer B · Straight lines (MM3.1) · question LIN-04
10Straight lines · spec MM3.1 · Paper 1 style · ★★☆

For which value of \(k\) are the lines \(kx + 3y = 2\) and \(2x - (k-1)y = 5\) perpendicular?

  1. A\(3\)
  2. B\(-1\)
  3. C\(\dfrac{3}{2}\)
  4. D\(6\)
  5. E\(-3\)
  6. F\(1\)
Worked solution

Gradients \(-\frac k3\) and \(\frac{2}{k-1}\). Perpendicular when the product is \(-1\): \(\frac{-2k}{3(k-1)} = -1 \Rightarrow 2k = 3k - 3 \Rightarrow k = 3\).

Full method — how and why
A perpendicularity condition with the parameter in both equations. Write both gradients, multiply, set to \(-1\).
  1. First line \(kx + 3y = 2\) has gradient \(-\dfrac k3\).
    Using \(-\frac ab\) avoids rearranging.
  2. Second line \(2x - (k-1)y = 5\) has gradient \(-\dfrac{2}{-(k-1)} = \dfrac{2}{k-1}\).
    Here \(b = -(k-1)\), so the two minus signs cancel.
  3. Set the product to \(-1\): \(-\dfrac{k}{3} \times \dfrac{2}{k-1} = -1\), i.e. \(\dfrac{-2k}{3(k-1)} = -1\).
    Perpendicular lines have gradients multiplying to \(-1\).
  4. Solve: \(-2k = -3(k-1) \Rightarrow -2k = -3k + 3 \Rightarrow k = 3\).
    Check: gradients \(-1\) and \(1\), product \(-1\) ✓.
Where it goes wrong
  • Mishandling the sign of \(b\) in the second line and getting \(k = -3\) - which is offered.
  • Setting the product to \(+1\) or the gradients equal (that would be the parallel condition).
Takeaway Read gradients as \(-\frac ab\) straight from \(ax + by = c\), watching the sign of \(b\) carefully.
Answer A · Straight lines (MM3.1) · question LIN-08
11Straight lines · spec MM3.1 · Paper 1 style · ★★☆

The point \(P\) lies on the line \(y = 2x - 1\) and is equidistant from \((0, 0)\) and \((4, 2)\). Find the coordinates of \(P\).

  1. A\((2, 1)\)
  2. B\(\left(\dfrac32, 2\right)\)
  3. C\((2, 3)\)
  4. D\(\left(\dfrac{5}{4}, \dfrac{3}{2}\right)\)
  5. E\((3, 5)\)
  6. F\((1, 1)\)
Worked solution

Equidistant means \(P\) is on the perpendicular bisector of \((0,0)\)-\((4,2)\): midpoint \((2, 1)\), gradient \(-2\): \(y = -2x + 5\). With \(y = 2x - 1\): \(4x = 6\), \(x = \frac32\), \(y = 2\). So \(P = (\frac32, 2)\).

Full method — how and why
'Equidistant from two points' means 'on their perpendicular bisector'. Intersect that with the given line.
  1. Find the perpendicular bisector of \((0,0)\) and \((4,2)\): midpoint \((2,1)\), segment gradient \(\frac12\), so bisector gradient \(-2\).
    Equidistant points form the perpendicular bisector - far quicker than setting two distance formulas equal and squaring.
  2. Its equation: \(y - 1 = -2(x - 2)\), i.e. \(y = -2x + 5\).
    Point-gradient form, then tidy.
  3. Solve simultaneously with \(y = 2x - 1\): \(2x - 1 = -2x + 5 \Rightarrow 4x = 6 \Rightarrow x = \frac32\).
    Equate the two expressions for \(y\).
  4. Then \(y = 2(\frac32) - 1 = 2\), so \(P = \left(\frac32, 2\right)\).
    Check: distance to \((0,0)\) is \(\sqrt{\frac94 + 4}\) and to \((4,2)\) is \(\sqrt{\frac{25}{4} + 0}\) - both \(\frac52\) ✓.
Where it goes wrong
  • Using the midpoint \((2,1)\) as the answer without checking it lies on \(y = 2x - 1\) (it does not).
  • Algebra slips when squaring distances, if you take that longer route.
Takeaway 'Equidistant from two points' \(\Rightarrow\) perpendicular bisector; then intersect with whatever else is given.
Answer B · Straight lines (MM3.1) · question LIN-09
12Straight lines · spec MM3.1 · Paper 1 style · ★★☆

The vertices of a triangle are \(A(0, 0)\), \(B(6, 0)\) and \(C(2, 4)\). Find the equation of the line through \(C\) that bisects the area of the triangle.

  1. A\(y = x + 2\)
  2. B\(y = -2x + 8\)
  3. C\(x = 2\)
  4. D\(y = 2x\)
  5. E\(y = 4x - 4\)
  6. F\(y = -4x + 12\)
Worked solution

A line through a vertex bisects the area iff it passes through the midpoint of the opposite side, \((3, 0)\). Through \((2, 4)\) and \((3, 0)\): gradient \(-4\), \(y = -4x + 12\).

Full method — how and why
A line through a vertex bisects a triangle's area exactly when it hits the midpoint of the opposite side - the two halves then have equal bases and the same height.
  1. Identify the side opposite \(C\): it is \(AB\), from \((0,0)\) to \((6,0)\).
    The line must pass through \(C\), so it cuts the opposite side.
  2. Its midpoint is \((3, 0)\).
    A median divides the triangle into two triangles with equal bases (\(3\) each) and the same apex height, hence equal areas.
  3. Find the line through \(C(2,4)\) and \((3,0)\): gradient \(= \dfrac{0-4}{3-2} = -4\).
    Two points give the gradient directly.
  4. Equation: \(y - 0 = -4(x - 3)\), i.e. \(y = -4x + 12\).
    Using the simpler point \((3,0)\) keeps the arithmetic clean.
Where it goes wrong
  • Using the midpoint of the wrong side, or the centroid.
  • Sign slip giving \(y = 4x - 4\), which passes through \((1,0)\) - an offered option.
Takeaway A median (vertex to opposite midpoint) always bisects the area, whatever the triangle's shape.
Answer F · Straight lines (MM3.1) · question LIN-11
13Straight lines · spec MM3.1 · Paper 1 style · ★★☆

The lines \(y = mx + 4\) and \(y = 3x - 2\) intersect at a point with positive \(x\)- and \(y\)-coordinates. Find the complete set of values of \(m\).

  1. A\(m < 3\)
  2. B\(m < -6\) or \(m > 3\)
  3. C\(-6 < m < 3\)
  4. D\(m > -6\)
  5. E\(0 < m < 3\)
  6. F\(-6 < m < 0\)
Worked solution

\(mx + 4 = 3x - 2 \Rightarrow x = \frac{6}{3 - m}\), which is positive iff \(m < 3\). Then \(y = 3x - 2 > 0\) iff \(x > \frac23\), i.e. \(\frac{6}{3-m} > \frac23\); with \(3 - m > 0\) this is \(18 > 6 - 2m\), i.e. \(m > -6\). So \(-6 < m < 3\).

Full method — how and why
Two conditions on the intersection point. Find the intersection in terms of \(m\), then impose positivity on each coordinate.
  1. Find the intersection: \(mx + 4 = 3x - 2 \Rightarrow x(m - 3) = -6 \Rightarrow x = \dfrac{6}{3 - m}\).
    Keep the denominator as \(3 - m\) so that positivity is easy to read.
  2. \(x > 0\) requires \(3 - m > 0\), i.e. \(m < 3\).
    The numerator \(6\) is positive, so the sign of \(x\) is the sign of \(3 - m\).
  3. \(y > 0\): since \(y = 3x - 2\), this needs \(x > \frac23\), i.e. \(\dfrac{6}{3-m} > \dfrac23\).
    Use whichever line is simpler to express \(y\) in terms of \(x\).
  4. With \(3 - m > 0\) we may multiply safely: \(18 > 2(3 - m) \Rightarrow 18 > 6 - 2m \Rightarrow m > -6\).
    Multiplying by a positive quantity is safe - and we already know \(3 - m > 0\) from the first condition.
  5. Combine: \(-6 < m < 3\).
    Both conditions must hold at once.
Where it goes wrong
  • Checking only \(x > 0\) and answering \(m < 3\).
  • Multiplying by \(3 - m\) without noting its sign - legitimate here only because the first condition established it.
Takeaway Impose one condition at a time, and use the first to justify the algebra needed for the second.
Answer C · Straight lines (MM3.1) · question LIN-12
14Straight lines · spec MM3.1 · Paper 1 style · ★★☆

The distance between the parallel lines \(3x + 4y = 5\) and \(3x + 4y = 20\) is

  1. A\(15\)
  2. B\(3\)
  3. C\(4\)
  4. D\(\dfrac{15}{7}\)
  5. E\(1\)
  6. F\(5\)
Worked solution

The lines are \(3x + 4y = c\); the perpendicular direction is \((3, 4)\) of length \(5\). Distance \(= \frac{|20 - 5|}{5} = 3\). (Alternatively, the point \((0, \frac54)\) on the first line is \(3\) from the second.)

Full method — how and why
Distance between parallel lines. Either use the coefficient formula or take any point on one line and measure to the other.
  1. Note both lines have the same left-hand side \(3x + 4y\), so they are parallel with normal direction \((3, 4)\), of length \(5\).
    The vector of coefficients is perpendicular to the line; its length normalises the distance formula.
  2. Use \(\text{distance} = \dfrac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \dfrac{|20 - 5|}{5}\).
    This is the point-to-line formula applied to any point on the first line.
  3. Evaluate: \(\dfrac{15}{5} = 3\).
    Check with a concrete point: \((0, \frac54)\) is on the first line, and its distance to \(3x + 4y = 20\) is \(\frac{|5 - 20|}{5} = 3\) ✓.
Where it goes wrong
  • Answering \(15\) by forgetting to divide by \(\sqrt{a^2+b^2}\).
  • Using \(\sqrt{3^2+4^2} = 7\) (adding instead of squaring), giving \(\frac{15}{7}\) - an offered option.
Takeaway For \(ax+by=c_1\) and \(ax+by=c_2\): distance \(= \frac{|c_1-c_2|}{\sqrt{a^2+b^2}}\); the coefficients must match exactly first.
Answer B · Straight lines (MM3.1) · question LIN-13
15Straight lines · spec MM3.1 · Paper 2 style · ★★★

The line through \((a, 0)\) and \((0, b)\), with \(a, b > 0\), passes through \((2, 3)\). Which of the following statements is/are true?

  1. I\(\dfrac{2}{a} + \dfrac{3}{b} = 1\)
  2. II\(a > 2\) and \(b > 3\)
  3. IIIThe smallest possible value of \(ab\) is \(24\)
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

The line is \(\frac xa + \frac yb = 1\), so I holds. Since \(\frac2a < 1\) and \(\frac3b < 1\), II holds. \(b = \frac{3a}{a-2}\), so \(ab = \frac{3a^2}{a-2}\); setting \(a = 2 + t\) gives \(\frac{3(2+t)^2}{t} = 3\left(\frac4t + 4 + t\right) \geq 3(4 + 4) = 24\) by AM-GM (equality at \(t = 2\), \(a = 4\), \(b = 6\)). III holds.

Full method — how and why
The intercept form of a line, plus an optimisation. Statements I and II are quick; III needs a minimisation argument.
  1. I: a line with intercepts \(a\) and \(b\) has equation \(\frac xa + \frac yb = 1\); substituting \((2,3)\) gives \(\frac2a + \frac3b = 1\). True.
    The intercept form is worth knowing precisely for questions like this.
  2. II: since \(\frac2a\) and \(\frac3b\) are both positive and sum to \(1\), each is less than \(1\), so \(a > 2\) and \(b > 3\). True.
    Two positive numbers summing to \(1\) are each below \(1\) - a clean argument with no algebra.
  3. III: from I, \(b = \dfrac{3a}{a - 2}\), so \(ab = \dfrac{3a^2}{a-2}\). Put \(a = 2 + t\) with \(t > 0\): \(ab = \dfrac{3(2+t)^2}{t} = 3\left(\dfrac4t + 4 + t\right)\).
    The substitution \(a = 2 + t\) clears the awkward denominator and exposes the structure \(\frac4t + t\).
  4. By AM-GM, \(\dfrac4t + t \geq 2\sqrt{4} = 4\), with equality when \(t = 2\). So \(ab \geq 3(4 + 4) = 24\), attained at \(a = 4\), \(b = 6\). True.
    \(u + \frac ku \geq 2\sqrt k\) for positive \(u\) - here it gives both the bound and the case of equality, which is what 'smallest possible value' demands.
  5. All three are true.
    Match to the options.
Where it goes wrong
  • Assuming II fails because \(a\) and \(b\) 'could be anything' - the positivity constraint forces both bounds.
  • Claiming III without checking equality is attainable; a lower bound is only the minimum if some case achieves it.
Takeaway \(\frac xa + \frac yb = 1\) is the intercept form; and \(t + \frac4t \geq 4\) (AM-GM) is the standard tool for such minimisations.
Answer H · Straight lines (MM3.1) · question LIN-10

How this topic has been examined

Real TMUA questions on Straight lines, newest first. Links open the official paper at the question and the online marking page.

2023 · Paper 1Question 16right-angled triangle from coordinatesquestion papersit online
2023 · Paper 2Question 14three lines ax + by + c = 0 ...question papersit online
2021 · Paper 2Question 2diagonal of a squarequestion papersit online
2021 · Paper 2Question 7line bisecting a circle and a squarequestion papersit online
November 2017 · Paper 1Question 3area enclosed by two perpendicular lines and the x-axisquestion papersit online
November 2017 · Paper 2Question 2rectangle PQRS from two verticesquestion papersit online
Specimen papers · Paper 1Question 3perpendicular bisector meets an axisquestion papersit online