TMUA: Straight lines
Straight lines in coordinate geometry.
- Section 1 Part 1 · MM3 Coordinate geometry in the (x, y)-plane · spec MM3.1
- Papers 1 and 2
- 8 practice questions
- No calculator
What the specification covers
- y - y1 = m(x - x1) and ax + by + c = 0
- Parallel and perpendicular gradients
- Intersections, distances and areas from coordinates
Key ideas
- \(y-y_1=m(x-x_1)\); \(ax+by+c=0\) has gradient \(-\tfrac ab\).
- Parallel lines have equal gradients; perpendicular gradients multiply to \(-1\).
- Intersections: solve the two equations simultaneously; axis intercepts: set \(x=0\) or \(y=0\).
- Area of a triangle from coordinates: often easiest as \(\tfrac12\times\) base on an axis \(\times\) height.
Common mistakes
- The gradient of \(3x-4y=7\) is \(\tfrac34\), not \(3\) or \(-\tfrac43\).
- A vertical line has no gradient; write it as \(x=k\).
Exam tip
Rearrange to \(y=mx+c\) before reading the gradient.
Worked example
Worked example
The line \(L\) passes through \((1,2)\) and is perpendicular to the line \(3x-4y=7\). At what value of \(x\) does \(L\) cross the \(x\)-axis?
- \(\frac{11}{4}\)
- \(-\frac{1}{2}\)
- \(\frac{5}{2}\)
- \(\frac{7}{3}\)
- \(\frac{5}{3}\)
Answer: C: \(\frac{5}{2}\)
\(3x-4y=7\) has gradient \(\tfrac34\), so \(L\) has gradient \(-\tfrac43\): \(y-2=-\tfrac43(x-1)\). Setting \(y=0\): \(x-1=\tfrac32\), \(x=\tfrac52\).
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
What is the area of the triangle enclosed by the lines \(y=x\), \(y=6-2x\) and the \(x\)-axis?
- \(2\)
- \(\frac{9}{2}\)
- \(3\)
- \(4\)
- \(6\)
Show the answer and solution
Answer: C: \(3\)
The vertices are \((0,0)\), \((3,0)\) (where \(y=6-2x\) meets the axis) and \((2,2)\) (where \(x=6-2x\)). Base 3, height 2: area \(3\).
Question 2
The line \(L\) passes through \((-1,4)\) and is perpendicular to the line \(5x-2y=5\). At what value of \(x\) does \(L\) cross the \(x\)-axis?
- \(9\)
- \(\frac{18}{5}\)
- \(\frac{3}{5}\)
- \(-11\)
- \(3\)
Show the answer and solution
Answer: A: \(9\)
\(5x-2y=5\) has gradient \(\frac{5}{2}\), so \(L\) has gradient \(- \frac{2}{5}\): \(y-4=- \frac{2}{5}(x+1)\). Setting \(y=0\) gives \(x=9\).
Question 3
The line \(L\) passes through \((2,-2)\) and is perpendicular to the line \(2x+3y=1\). At what value of \(x\) does \(L\) cross the \(x\)-axis?
- \(-5\)
- \(0\)
- \(\frac{10}{3}\)
- \(5\)
- \(\frac{2}{3}\)
Show the answer and solution
Answer: C: \(\frac{10}{3}\)
\(2x+3y=1\) has gradient \(- \frac{2}{3}\), so \(L\) has gradient \(\frac{3}{2}\): \(y+2=\frac{3}{2}(x-2)\). Setting \(y=0\) gives \(x=\frac{10}{3}\).
Question 4
The line \(L\) passes through \((-1,4)\) and is perpendicular to the line \(3x-4y=10\). At what value of \(x\) does \(L\) cross the \(x\)-axis?
- \(2\)
- \(\frac{8}{3}\)
- \(-4\)
- \(\frac{13}{3}\)
- \(3\)
Show the answer and solution
Answer: A: \(2\)
\(3x-4y=10\) has gradient \(\frac{3}{4}\), so \(L\) has gradient \(- \frac{4}{3}\): \(y-4=- \frac{4}{3}(x+1)\). Setting \(y=0\) gives \(x=2\).
Question 5
The line \(L\) passes through \((2,-2)\) and is perpendicular to the line \(2x+3y=7\). At what value of \(x\) does \(L\) cross the \(x\)-axis?
- \(5\)
- \(0\)
- \(\frac{2}{3}\)
- \(-5\)
- \(\frac{10}{3}\)
Show the answer and solution
Answer: E: \(\frac{10}{3}\)
\(2x+3y=7\) has gradient \(- \frac{2}{3}\), so \(L\) has gradient \(\frac{3}{2}\): \(y+2=\frac{3}{2}(x-2)\). Setting \(y=0\) gives \(x=\frac{10}{3}\).
Question 6
The line \(L\) passes through \((2,3)\) and is perpendicular to the line \(x+2y=-3\). At what value of \(x\) does \(L\) cross the \(x\)-axis?
- \(-1\)
- \(5\)
- \(-4\)
- \(\frac{7}{2}\)
- \(\frac{1}{2}\)
Show the answer and solution
Answer: E: \(\frac{1}{2}\)
\(x+2y=-3\) has gradient \(- \frac{1}{2}\), so \(L\) has gradient \(2\): \(y-3=2(x-2)\). Setting \(y=0\) gives \(x=\frac{1}{2}\).
Question 7
The line \(L\) passes through \((2,-2)\) and is perpendicular to the line \(4x+3y=1\). At what value of \(x\) does \(L\) cross the \(x\)-axis?
- \(- \frac{2}{3}\)
- \(- \frac{7}{2}\)
- \(\frac{14}{3}\)
- \(\frac{7}{2}\)
- \(0\)
Show the answer and solution
Answer: C: \(\frac{14}{3}\)
\(4x+3y=1\) has gradient \(- \frac{4}{3}\), so \(L\) has gradient \(\frac{3}{4}\): \(y+2=\frac{3}{4}(x-2)\). Setting \(y=0\) gives \(x=\frac{14}{3}\).
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