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Cambridge 0580 IGCSE  ·  Topic E3.6

E3.6 Parallel Lines

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01 — Explanation

What do parallel lines have in common algebraically?

Parallel lines never meet, so they must be equally steep in the same direction — parallel lines have equal gradients. That single fact is the whole topic; everything else is just applying y=mx+cy = mx+c with a gradient you already know.

02 — Explanation

How do I find the equation of a line parallel to a given line, through a given point?

  1. Identify the gradient of the given line (rearrange to y=mx+cy=mx+c if needed) — your new line has the same mm.
  2. Substitute the known point (x1,y1)(x_1, y_1) and this gradient into y−y1=m(x−x1)y - y_1 = m(x-x_1), or substitute directly into y=mx+cy = mx+c and solve for cc.

03 — Explanation

How do I check whether two lines given in different forms are actually parallel?

Rearrange both equations into y=mx+cy = mx + c form and compare their gradients directly — two lines are parallel if and only if their mm-values are identical (their cc-values can be anything). Two lines that look very different when written as, say, 2x+4y=82x + 4y = 8 and 3x+6y=53x + 6y = 5 can turn out to be parallel once both are simplified, since both rearrange to a gradient of −12-\tfrac12.

04 — Explanation

Can two "parallel" lines actually be the exact same line?

Yes — if both the gradient and the yy-intercept match, the two equations describe the identical line, not two separate parallel lines. Examiners sometimes test this by giving an equation that simplifies to be identical to another; always check both mm and cc before concluding two lines are genuinely distinct and parallel.

Exam tip

Don't confuse "parallel to" with "passes through the same yy-intercept" — two lines can be parallel and still cross the yy-axis at completely different points. Only the gradient carries over; cc must be recalculated using the new point every time.

05 — Explanation

How does this topic typically appear inside a longer geometry question?

Parallel-line reasoning is often the final step of a bigger problem rather than the whole question — for example, showing that one side of a quadrilateral is parallel to another to help classify the shape as a trapezium or parallelogram, or confirming that a constructed line matches a given direction. In these cases, calculate both gradients using the two-point formula from E3.3 first, since a "parallel or not" conclusion is only ever one short comparison once both gradients are known.

06 — Worked example

Worked example

Find the equation of the line parallel to y=4x−1y = 4x - 1 that passes through (1,−3)(1, -3).

  1. The given line has gradient m=4m = 4. A parallel line has the same gradient, so the new line is y=4x+cy = 4x + c for some cc.
  2. Substitute the point (1,−3)(1, -3) to find cc:
−3=4(1)+c  ⟹  −3=4+c  ⟹  c=−7-3 = 4(1) + c \implies -3 = 4 + c \implies c = -7
  1. The equation of the parallel line is:
y=4x−7y = 4x - 7

07 — Worked example

Worked example: testing whether two lines are parallel

Determine whether the lines 2x+4y=82x + 4y = 8 and 3x+6y=53x + 6y = 5 are parallel.

  1. Rearrange the first: 4y=8−2x  ⟹  y=2−12x4y = 8 - 2x \implies y = 2 - \tfrac12 x, so m1=−12m_1 = -\tfrac12.
  2. Rearrange the second: 6y=5−3x  ⟹  y=56−12x6y = 5 - 3x \implies y = \tfrac56 - \tfrac12 x, so m2=−12m_2 = -\tfrac12.
  3. Since m1=m2=−12m_1 = m_2 = -\tfrac12 but the yy-intercepts (22 and 56\tfrac56) are different, the lines are parallel and distinct (not the same line).

Both checks — comparing gradients, then comparing intercepts — are necessary together: matching gradients alone only proves the lines are either parallel or identical, and the intercept comparison is what tells them apart.