Intersecting Lines
Two straight lines that are not parallel cross at one point, the point of intersection, and its coordinates fit both rules. In NSW Year 8 maths, students find this point from a graph or from a table of values (NESA outcome MA4‑LIN‑C‑01).
You will read intersection points from graphs and use them to solve equations such as \(2x + 1 = 7 - x\), find where two rows of a table give the same \(y\)-value, and check a point by substituting it into both rules. You will compare phone plans, bike hire shops and dance studios to see when one becomes cheaper, find when one walker catches another, and find the corners of a triangle formed by three lines. It is part of Year 8 Graphing Linear Equations.
Every question with a fully worked solution.
- Intersecting Lines - Video - How to find the point of intersection of two straight lines by graphing Watch
Theory
Two lines on one number plane meet at the point of intersection. That point is on both lines, so it fits both rules: \(y = x + 2\) and \(y = -2x - 1\) cross at \((-1, 1)\), and \(x = -1\) solves \(x + 2 = -2x - 1\). This Year 8 maths page shows how.
Read the point. From the crossing go straight down or up to the \(x\)-axis for \(x\), then across to the \(y\)-axis for \(y\). Write it as \((x, y)\).
Check it in both rules. Both must give the \(y\)-value of the point. A point that fits one rule is on that line only.
It solves an equation. Both rules give the same \(y\)-value there, so the \(x\)-value solves “first rule \(=\) second rule”, such as \(x + 1 = 5 - x\).
From two tables. Use a shared \(x\)-row. The column where the two \(y\)-values are equal gives the point. If one rule is below the other in one column and above it in the next, the lines cross between them, often halfway.
Each line crosses the axes. For the \(y\)-axis put \(x = 0\) into that rule; for the \(x\)-axis make that rule equal \(0\) and solve. The answer can be a fraction: \(3x - 2 = 0\) gives \(x = \dfrac{2}{3}\).
Parallel lines never meet. Rules with the same step stay the same distance apart, so no \(x\)-value makes them equal.
Three lines, no two parallel, usually form a triangle. Each vertex is where two of the lines cross.
Real life. Two charges, tanks or journeys are equal where the lines cross: the across-value says when, the up-value says how much.
The two \(y\)-rows are equal only at \(x = 2\):
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
| \(y = x + 1\) | \(1\) | \(2\) | \(3\) | \(4\) |
| \(y = 5 - x\) | \(5\) | \(4\) | \(3\) | \(2\) |
Using intersecting lines
- Graph both rules on the same axes (or make a shared table).
- Read the crossing point \((x, y)\).
- Check the point in both rules.
- Answer: the \(x\)-value solves rule \(=\) rule; the \(y\)-value is the common value.
(i) Solve \(x + 2 = -2x - 1\).
(ii) Where does \(y = -2x - 1\) cut the \(y\)-axis?
(iii) Where does \(y = x + 2\) cut the \(x\)-axis?
| \(-1 + 2\) | \(=\) | \(1\) |
| \(-2 \times (-1) - 1\) | \(=\) | \(1\) |
(i) The lines cross at \((-1, 1)\), so \(x = -1\).
(ii) \((0, -1)\)
(iii) \((-2, 0)\)
(i) Write the point of intersection.
(ii) Solve \(3x = 8 - x\).
| \(3 \times 2\) | \(=\) | \(6\) |
| \(8 - 2\) | \(=\) | \(6\) |
(i) \((2, 6)\)
(ii) \(x = 2\)
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|
| \(4x - 6\) | \(-6\) | \(-2\) | \(2\) | \(6\) | \(10\) |
| \(2x\) | \(0\) | \(2\) | \(4\) | \(6\) | \(8\) |
Compare the rows: at \(x = 2\), \(2 \ne 4\); at \(x = 3\), \(6 = 6\); at \(x = 4\), \(10 \ne 8\).
| \(4 \times 3 - 6\) | \(=\) | \(6\) |
| \(2 \times 3\) | \(=\) | \(6\) |
\((3, 6)\), so \(x = 3\) solves \(4x - 6 = 2x\).
(i) When do they hold the same?
(ii) How much is that?
(iii) After \(5\) minutes, which holds more?
The rules are \(V = 18 + 3t\) and \(V = 30 - t\). The lines cross at \((3, 27)\).
| \(18 + 3 \times 3\) | \(=\) | \(27\) |
| \(30 - 3\) | \(=\) | \(27\) |
| \(18 + 3 \times 5\) | \(=\) | \(33\) |
| \(30 - 5\) | \(=\) | \(25\) |
(i) After \(3\) minutes
(ii) \(27\) L
(iii) Tank A, by \(8\) L
\(y = 2x\) is below at \(x = 2\) and above at \(x = 3\), so they cross between, here halfway.
| \(2 \times 2.5\) | \(=\) | \(5\) |
| \(10 - 2 \times 2.5\) | \(=\) | \(5\) |
At \((2.5, 5)\).
(i) Do they cross?
(ii) Which line is \((2, 1)\) on?
(iii) Solve \(x + 3 = x - 1\).
| \(2 + 3\) | \(=\) | \(5\) |
| \(2 - 1\) | \(=\) | \(1\) |
(i) No: both go up \(1\) for each \(1\) across, so they stay \(4\) apart (parallel).
(ii) Line \(B\) only
(iii) No solution
(ii) Where does \(y = 3x - 2\) cut the \(x\)-axis?
(i) \(x = 4 - x\) gives \(x = 2\), so \((2, 2)\). \(y = x\) meets \(y = 0\) at \((0, 0)\); \(4 - x = 0\) gives \((4, 0)\).
| \(3x - 2\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\dfrac{2}{3}\) |
(i) \((0, 0)\), \((2, 2)\) and \((4, 0)\)
(ii) \(\left(\dfrac{2}{3}, 0\right)\)
Common pitfalls
Frequently asked questions
What is the point of intersection?
The point where two lines cross. It lies on both lines, so it fits both rules.
How do I solve x + 1 = 5 − x with graphs?
Graph y = x + 1 and y = 5 − x. The x-value of the point where they cross, 2, is the solution.
How do I find the intersection from tables?
Use the same x-values for both rules and find the column where the y-values are equal.
Can two lines have no point of intersection?
Yes. Parallel lines, which have the same step, never meet.
How do I find where a line cuts the x-axis?
Make its rule equal 0 and solve for x. The answer may be a fraction.