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Year 8 Maths Graphing Linear Equations

Intersecting Lines

20 practice questions 1 video lesson Theory + worked examples
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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.

y = x + 1 and y = 5 βˆ’ x The lines y = x + 1 and y = 5 - x on one number plane. They cross at (2, 3). xy −2−112345−2−11234560 (2, 3) y = x + 1 y = 5 βˆ’ x
The lines cross at \((2, 3)\). Check: \(2 + 1 = 3\) and \(5 - 2 = 3\), so \(x = 2\) solves \(x + 1 = 5 - x\).

The two \(y\)-rows are equal only at \(x = 2\):

\(x\)\(0\)\(1\)\(2\)\(3\)
\(y = x + 1\)\(1\)\(2\)\(3\)\(4\)
\(y = 5 - x\)\(5\)\(4\)\(3\)\(2\)
One rule is not enough. Test a crossing point in both rules before you accept it.
The solution is the x-value. The point \((2, 3)\) means \(x = 2\) solves the equation; \(3\) is the value both sides take there.

Using intersecting lines

  1. Graph both rules on the same axes (or make a shared table).
  2. Read the crossing point \((x, y)\).
  3. Check the point in both rules.
  4. Answer: the \(x\)-value solves rule \(=\) rule; the \(y\)-value is the common value.
Example 1 — Reading the graphs
The graphs of \(y = x + 2\) and \(y = -2x - 1\) are drawn.
(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?
y = x + 2 and y = βˆ’2x βˆ’ 1 The lines y = x + 2 (blue) and y = -2x - 1 (red), crossing at (-1, 1). The blue line meets the x-axis at (-2, 0); the red line meets the y-axis at (0, -1). xy −3−2−1123−3−2−11230 (βˆ’1, 1)
Solution
\(-1 + 2\)\(=\)\(1\)
\(-2 \times (-1) - 1\)\(=\)\(1\)
x =−1(0,−1)(−2,0)

(i) The lines cross at \((-1, 1)\), so \(x = -1\).
(ii) \((0, -1)\)
(iii) \((-2, 0)\)

Example 2 — Read and check
\(y = 3x\) and \(y = 8 - x\) are drawn.
(i) Write the point of intersection.
(ii) Solve \(3x = 8 - x\).
y = 3x and y = 8 βˆ’ x The lines y = 3x and y = 8 - x crossing at (2, 6). xy −1123456−11234567890 (2, 6) y = 3x y = 8 βˆ’ x
Solution
\(3 \times 2\)\(=\)\(6\)
\(8 - 2\)\(=\)\(6\)
(2,6) x =2

(i) \((2, 6)\)
(ii) \(x = 2\)

Example 3 — From two tables
Find the point of intersection of \(y = 4x - 6\) and \(y = 2x\).
\(x\)\(0\)\(1\)\(2\)\(3\)\(4\)
\(4x - 6\)\(-6\)\(-2\)\(2\)\(6\)\(10\)
\(2x\)\(0\)\(2\)\(4\)\(6\)\(8\)
Solution

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)

\((3, 6)\), so \(x = 3\) solves \(4x - 6 = 2x\).

Example 4 — Two water tanks
Tank A holds \(18\) L and gains \(3\) L a minute. Tank B holds \(30\) L and leaks \(1\) L a minute.
(i) When do they hold the same?
(ii) How much is that?
(iii) After \(5\) minutes, which holds more?
Two water tanks Tank A (red) rises from 18 L by 3 L a minute; tank B (blue) falls from 30 L by 1 L a minute. The lines cross at 3 minutes and 27 L. 0 10 20 30 40 0 1 2 3 4 5 6 A B t (minutes) V (litres)
Solution

The rules are \(V = 18 + 3t\) and \(V = 30 - t\). The lines cross at \((3, 27)\).

\(18 + 3 \times 3\)\(=\)\(27\)
\(30 - 3\)\(=\)\(27\)
\(18 + 3 \times 5\)\(=\)\(33\)
\(30 - 5\)\(=\)\(25\)
3 min 27 L A by 8 L

(i) After \(3\) minutes
(ii) \(27\) L
(iii) Tank A, by \(8\) L

Example 5 — Between two columns
For \(y = 2x\) and \(y = 10 - 2x\): at \(x = 2\) the values are \(4\) and \(6\); at \(x = 3\) they are \(6\) and \(4\). Where do the lines cross?
Solution

\(y = 2x\) is below at \(x = 2\) and above at \(x = 3\), so they cross between, here halfway.

\(2 \times 2.5\)\(=\)\(5\)
\(10 - 2 \times 2.5\)\(=\)\(5\)
(2.5,5)

At \((2.5, 5)\).

Example 6 — Lines that never meet
Line \(A\) is \(y = x + 3\) and line \(B\) is \(y = x - 1\).
(i) Do they cross?
(ii) Which line is \((2, 1)\) on?
(iii) Solve \(x + 3 = x - 1\).
Solution
\(2 + 3\)\(=\)\(5\)
\(2 - 1\)\(=\)\(1\)
no B no solution

(i) No: both go up \(1\) for each \(1\) across, so they stay \(4\) apart (parallel).
(ii) Line \(B\) only
(iii) No solution

Example 7 — Three lines and a fraction
(i) Find the vertices of the triangle formed by \(y = x\), \(y = 4 - x\) and the \(x\)-axis.
(ii) Where does \(y = 3x - 2\) cut the \(x\)-axis?
A triangle from three lines The lines y = x, y = 4 - x and y = 0 (the x-axis) form a triangle with vertices (0, 0), (2, 2) and (4, 0). xy −112345−112340 (2, 2) (4, 0)
Solution

(i) \(x = 4 - x\) gives \(x = 2\), so \((2, 2)\). \(y = x\) meets \(y = 0\) at \((0, 0)\); \(4 - x = 0\) gives \((4, 0)\).

\(3x - 2\)\(=\)\(0\)
\(x\)\(=\)\(\dfrac{2}{3}\)
(0,0),(2,2),(4,0)(2/3,0)

(i) \((0, 0)\), \((2, 2)\) and \((4, 0)\)
(ii) \(\left(\dfrac{2}{3}, 0\right)\)

Common pitfalls

Checking only one rule. A point on one line is not the crossing point unless it fits the other rule too.
Giving the point as the solution. The solution of \(x + 1 = 5 - x\) is \(x = 2\).
Looking for a crossing of parallel lines. Same step means no point of intersection and no solution.
Rounding a fractional crossing. \(x = \dfrac{2}{3}\) is exact; do not write \(1\) or \(0.6\).

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 βˆ’ x with graphs?

Graph y = x + 1 and y = 5 βˆ’ 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.