Parallel ⇒ equal gradients: Two lines are parallel when they have the same gradient (m₁ = m₂) — the same steepness, so they never meet. Only their y-intercepts differ.
IB-style question — are they parallel?
Show that y = 3x + 1 and 6x − 2y + 5 = 0 are parallel.
Step by step
- Rearrange the second line into y = mx + c.
- Compare gradients.
Final answer
Both gradients are 3, so the lines are parallel.
Same gradient AND same intercept = same line: Parallel means equal gradients but different intercepts. If the intercept matches too, it's the same line, not a parallel one.
Perpendicular ⇒ m₁ × m₂ = −1: Two lines are perpendicular (they meet at 90°) when their gradients multiply to −1. So the second gradient is m₂ = −1/m₁ — flip the fraction and change the sign (the negative reciprocal).
[Diagram: math-gradient-lines] - Available in full study mode
IB-style question — the perpendicular gradient
A line has gradient ⅔. Find the gradient of a line perpendicular to it.
Step by step
- Flip the fraction (reciprocal).
- Change the sign.
- Check the product is −1.
Final answer
Perpendicular gradient = −3/2.
Whole numbers too: Treat a whole number as a fraction over 1: a gradient of 5 = 5/1, so the perpendicular gradient is −1/5.
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Get the gradient, then point–gradient form: Same recipe both times: find the required gradient — the same one for parallel, the negative reciprocal for perpendicular — then drop it and the point into y − y₁ = m(x − x₁).
IB-style question — parallel through a point
Find the equation of the line through (1, 4) parallel to y = 2x + 7.
Step by step
- Parallel ⇒ same gradient.
- Point–gradient form with (1, 4).
- Tidy.
Final answer
y = 2x + 2.
IB-style question — perpendicular through a point
Find the equation of the line through (4, 1) perpendicular to y = 2x − 3.
Step by step
- Perpendicular ⇒ negative reciprocal of 2.
- Point–gradient form with (4, 1).
- Tidy.
Final answer
y = −½x + 3.
Bisector = midpoint + perpendicular: A perpendicular bisector cuts a segment in half at a right angle. So it passes through the midpoint of the two endpoints and has the negative-reciprocal gradient of the segment.
IB-style question — perpendicular bisector
Find the perpendicular bisector of A(1, 2) and B(5, 8).
Step by step
- Midpoint of AB.
- Gradient of AB.
- Negative reciprocal for the bisector, through (3, 5).
Final answer
y = −⅔x + 7 (the perpendicular bisector of AB).
Forward link — normals to curves: The normal to a curve at a point is the line perpendicular to the tangent there, so its gradient is −1/(tangent gradient). You'll use this exact step in calculus (Unit 5).