Coordinates, Linear Graphs and Parallel Lines at Extended Level
From ax + by = c to the parallel line through a point
Quick recap. A point is written \((x, y)\): across first, then up or down. The axes split the grid into four quadrants, so either number can be negative — \((-3, 2)\) is 3 left and 2 up. To draw a straight line, make a table of values, plot at least three points and join them with a ruler.
Lines in the form \(ax + by = c\). Extended questions often give lines like \(2x + 3y = 12\). You have two good ways to deal with them:
- Intercepts (cover-up): put \(x = 0\) to find where it crosses the \(y\)-axis, and \(y = 0\) for the \(x\)-axis. For \(2x + 3y = 12\): \(x = 0 \Rightarrow y = 4\), and \(y = 0 \Rightarrow x = 6\). Plot \((0, 4)\) and \((6, 0)\).
- Rearrange to \(y = mx + c\): \(3y = -2x + 12 \Rightarrow y = -\dfrac{2}{3}x + 4\). Now you can read the gradient \(m = -\dfrac{2}{3}\) and the \(y\)-intercept 4.
Where two lines cross. Read the point from an accurate graph, or solve the two equations together (substitute one into the other). The crossing point fits BOTH equations — always check it in both.
Lines such as \(x = 3\) (vertical) and \(y = -2\) (horizontal) are straight lines too. Vertical lines are all parallel to each other, and so are horizontal lines.
Complete a table of values for \(y = 3 – 2x\) for \(x = -2\) to \(3\).
| \(x\) | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|
| \(y\) | 7 | 5 | 3 | 1 | −1 | −3 |
For \(x = -2\): \(y = 3 – 2(-2) = 3 + 4 = 7\). The \(y\) values go down by 2 each time, so the gradient is −2. If one value breaks the pattern, you have made an error.
Draw the line \(2x + 3y = 12\) and find its gradient.
Intercepts: \(x = 0 \Rightarrow y = 4\); \(y = 0 \Rightarrow x = 6\). Plot \((0, 4)\) and \((6, 0)\) and join them (the teal line below). A third point to check: \(x = 3 \Rightarrow 3y = 6 \Rightarrow y = 2\), so \((3, 2)\) is on it.
Gradient: \(y = -\dfrac{2}{3}x + 4\), so \(m = -\dfrac{2}{3}\).
The orange line on the grid above is \(y = x – 1\). Find where it crosses \(2x + 3y = 12\).
From the graph the lines meet at \((3, 2)\). Check by algebra: substitute \(y = x – 1\):
\(2x + 3(x – 1) = 12 \Rightarrow 5x – 3 = 12 \Rightarrow x = 3,\; y = 3 – 1 = 2\)
Check in both: \(2(3) + 3(2) = 12\) ✔ and \(3 – 1 = 2\) ✔. The crossing point is \((3, 2)\).
Find the equation of the line parallel to \(2x + 3y = 12\) that passes through \((-3, 5)\). Give your answer in the form \(ax + by = c\).
Gradient of the given line: \(-\dfrac{2}{3}\) (from Example 2). So the new line is \(y = -\dfrac{2}{3}x + c\).
\(5 = -\dfrac{2}{3}(-3) + c \Rightarrow 5 = 2 + c \Rightarrow c = 3\)
So \(y = -\dfrac{2}{3}x + 3\). Multiply by 3: \(3y = -2x + 9\), which is \(2x + 3y = 9\).
Shortcut: parallel lines in the form \(ax + by = c\) keep the same \(a\) and \(b\). Only \(c\) changes: \(2(-3) + 3(5) = 9\). Same answer.
Show that the line \(4x – 2y = 7\) is parallel to \(y = 2x + 5\).
\(-2y = -4x + 7 \Rightarrow y = 2x – 3.5\)
Both lines have gradient 2 (and different intercepts), so they are parallel. “Show that” needs this working, not just the word “yes”.
Q1. \(ABCD\) is a parallelogram. \(A\) is \((-3, 2)\), \(B\) is \((1, 4)\) and \(C\) is \((3, -1)\). Find the coordinates of \(D\).
👀 Show answer
D = A moved 2 right, 5 down = (−1, −3) ✓ A1
Q2. Complete the table of values for \(3x – 2y = 6\).
| \(x\) | −2 | 0 | 2 | 4 |
|---|---|---|---|---|
| \(y\) |
👀 Show answer
y: −6, −3, 0, 3 ✓ A1
Q3. Find the coordinates of the points where the line \(5x – 2y = 20\) crosses the \(x\)-axis and the \(y\)-axis.
👀 Show answer
x = 0: −2y = 20 → (0, −10) ✓ B1
Q4. Find the gradient and the \(y\)-intercept of the line \(3x + 4y = 24\).
👀 Show answer
gradient −34, y-intercept 6 ✓ A1
Q5. Find the coordinates of the point where the lines \(y = 2x – 3\) and \(x + 2y = 9\) cross.
👀 Show answer
5x − 6 = 9 → x = 3 ✓ A1
y = 2(3) − 3 = 3 → (3, 3) ✓ A1
Q6. Find the equation of the line that is parallel to \(y = 4 – 3x\) and passes through \((2, -5)\).
👀 Show answer
−5 = −3(2) + c → c = 1 ✓ M1
y = −3x + 1 ✓ A1
Q7. Find the equation of the line parallel to \(2x – 5y = 10\) that passes through \((5, 1)\). Give your answer in the form \(ax + by = c\).
👀 Show answer
1 = 25(5) + c → c = −1 ✓ M1
y = 25x − 1 → 2x − 5y = 5 ✓ A1
Q8. Which two of these lines are parallel? Show your working.
A: \(y = 3x + 1\) B: \(6x – 2y = 5\) C: \(3y = x + 4\) D: \(x + 3y = 2\)
👀 Show answer
A and B ✓ A1
Q9. The diagram shows a straight line \(L\) passing through \((0, -2)\) and \((4, 4)\).
(a) Find the equation of \(L\).
(b) Find the equation of the line parallel to \(L\) that passes through \((2, 7)\).
👀 Show answer
y = 32x − 2 ✓ A1
(b) 7 = 32(2) + c → c = 4 ✓ M1
y = 32x + 4 ✓ A1
Q10. Solve to find where the lines \(0.4x + 1.5y = 6\) and \(y = 0.8x – 1\) cross. Give exact decimal values.
👀 Show answer
1.6x − 1.5 = 6 → x = 4.6875 ✓ A1
y = 0.8(4.6875) − 1 = 2.75 → (4.6875, 2.75) ✓ A1
- Reading the gradient straight from \(ax + by = c\): in \(3x + 4y = 24\) the gradient is \(-\dfrac{3}{4}\), not 3. Get \(y\) alone first.
- Coordinates the wrong way round: \((x, y)\) — across first, then up or down.
- Sign slips when rearranging: \(-2y = 6 – 3x\) means you divide EVERY term by −2.
- Copying the old \(c\) for the parallel line: parallel means same gradient, but the new line has its own \(c\). Use the point to find it.
- Only checking the crossing point in one equation: it must fit both.
- Plotting only two points: plot three. If they are not in a line, one of them is wrong.
- Paper 4: “Complete the table of values” then “On the grid, draw the graph” — usually 3–4 marks, often followed by “use your graph to solve…” (read where two lines cross).
- Paper 2: “Find the equation of the line parallel to … that passes through …” — typically 2–3 marks. Show the gradient and the substitution.
- Command words: “Write down the gradient” (no working needed), “Find the equation”, “Show that … is parallel to …” (working needed).
- This unit is the foundation for gradients, midpoints, lengths and perpendicular lines in the rest of the Coordinate geometry strand.