Chapter 4 · Geometry and Equations · Concept 2
4-10 Parallel and Perpendicular Lines
Two lines never meeting, or meeting at a perfect right angle — both are decided by one number each. And both rules are about gradients, so the first move is always the same: get the gradient out.
Guiding question
What can two gradients tell you about how two lines sit?
Learning outcome. Apply the conditions for parallel and perpendicular lines to solve geometric problems.
Two rules, one preparation
parallel
m₁ = m₂
the gradients are equal
perpendicular
m₁ = −1m₂
negative reciprocals
“Negative reciprocal” is TWO changes.
Flip the fraction upside down and change its sign. So 2 becomes −1/2, and −3/5 becomes 5/3. Doing only one of the two is the mistake this lesson exists to prevent — and the quick test is that the two must multiply to −1.
And the step before either rule.
Most questions arrive as ax + by + c = 0, and you cannot read a gradient off that. Rearrange into y = mx + k first, every time. It is one line of work, and skipping it is where nearly all the lost marks in this lesson come from.
The blue line stays still. Turn the other one and watch the product of the two gradients — it hits exactly −1 at the moment the corner squares up, and nowhere else.
the blue line
m₁ = 2
the moving line
m₂ = -0.5
m₁ × m₂
-1.0
The product is exactly −1, and the corner is square. Notice what m₂ is: −0.5, which is 2 flipped over AND made negative. Both changes, not one.
3. Find the line through (−2, 5) perpendicular to 3x + 5y + 1 = 0.
- 1
Rearrange the given line so its gradient can be read.
3x + 5y + 1 = 0 → y = −35x − 15Take 3x and 1 across, then divide everything by 5. The gradient is the number multiplying x.
Work with a partner. For question 1, write out all six gradients before either of you says a word about pairs. Then compare lists — a disagreement is almost always a rearranging slip, not a comparing one.
A check that costs nothing.
Put the given point into your answer — it must give 0. And for a parallel question, the x and y coefficients of your answer should match the original's exactly; only the constant changes. Question 5 goes from 3x + 2y + 1 = 0 to 3x + 2y − 7 = 0, and nothing else moved.
L₁: 2x + 3y − 10 = 0
L₂: kx − 2y + 5 = 0
L₃: 4x + my − 1 = 0
L₁ is perpendicular to L₂, and L₁ is parallel to L₃. Find m ÷ k.
Reading a gradient off ax + by + c = 0.
You cannot. Rearrange into y = mx + k first — one line of work, and skipping it wrecks everything after.
Half-doing the negative reciprocal.
2 becomes −1/2: flipped AND negative. Test it by multiplying — the two must give −1.
Stopping at the first pair.
Exercise 2 has two perpendicular pairs and Try 1 has two parallel ones. Work out every gradient, then compare all of them.
Answering in the wrong form.
If the question used ax + by + c = 0, rearrange your answer to match. It is one more line.
Not checking with the point.
Substitute the given point into your finished equation — it must give 0. Five seconds, every time.
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