01 · Classify
Classify y = −2x + 5 and x − 2y + 8 = 0.
Hint
Rearrange the second equation.
Worked solution
y = x/2 + 4
(−2)(1/2) = −1
The lines are perpendicular.
Understand · explore · practise
Parallel and perpendicular lines
Find equations of parallel and perpendicular lines, explain the negative-reciprocal rule and prove geometric properties using exact gradients.
Before you startGradients and point-gradient equations
01 / Parallel lines
Equal finite gradients give the same direction.
Two distinct non-vertical lines are parallel when they have equal gradients. In y = mx + c, changing c moves the line without changing its direction.
y = 2x − 1 and y = 2x + 4
same gradient, different intercepts
If both gradient and intercept agree, the equations describe the same line. Do not conclude that there are two distinct parallel lines just because the gradients match.
Two distinct vertical lines are parallel too, although neither has a finite gradient. Select the vertical case in the model to see it.
y = 2x; y = 2x + 3Equal axis scales
Both gradients are 2. Their y-intercepts differ, so these are distinct parallel lines.
02 / Perpendicular lines
A quarter-turn swaps the changes and reverses one sign.
If a direction has horizontal change r and vertical change s, a 90° rotation has changes −s and r. The original gradient is s/r; the rotated gradient is −r/s.
m₂ = −1/m₁
so m₁m₂ = −1
for finite, non-zero gradients
Thus 2/3 pairs with −3/2; −4 pairs with 1/4. Negating the gradient alone works only in special cases such as 1 and −1.
When one line is horizontal, its perpendicular is vertical. Handle that case directly rather than dividing by zero.
The model uses equal unit scales on its axes. Otherwise, perpendicular coordinate directions could appear to meet at the wrong angle on screen.
Watch a quarter-turn explain the negative reciprocal
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Find an equation
First find the required gradient, then use the point.
Rearrange the given line before reading its gradient. Keep the parallel gradient unchanged, or take its negative reciprocal for a perpendicular line. Then substitute the required point into point-gradient form.
For a line parallel to 3x + 2y − 6 = 0 through (2, −1), the gradient is −3/2. Hence y + 1 = −3(x − 2)/2, or 3x + 2y − 4 = 0.
A perpendicular through a given pointWorked example
4x − 3y + 9 = 0
P = (2, −1)
Rearrange the original equation.
y = 4x/3 + 3
m₁ = 4/3 ⇒ m₂ = −3/4
Use the negative reciprocal.
y + 1 = −3(x − 2)/4
Anchor the new line at P.
3x + 4y − 2 = 0
Check P: 6 − 4 − 2 = 0, and (4/3)(−3/4) = −1.
04 / Vertical and horizontal
Use the coordinates instead of a gradient formula.
Through P(−2, 5), the line parallel to y = 3 is y = 5. The perpendicular is x = −2.
Through the same point, the line parallel to x = 7 is x = −2, and the perpendicular is y = 5.
A horizontal line y = c and a vertical line x = d meet at (d, c). The product-of-gradients test is unnecessary because one gradient is undefined.
Classify equations, not their appearance
2x − y + 3 = 0 and 4x − 2y + 6 = 0 describe the same line. But 4x − 2y + 5 = 0 has the same gradient with a different intercept, so it is a distinct parallel line. Non-matching gradients whose product is not −1 give neither relation.
05 / Prove a shape
Check all the conditions your conclusion needs.
For a rectangle, show that opposite sides are parallel and adjacent sides perpendicular, using vertices in their stated order. One right angle on its own is not enough to prove a general quadrilateral is a rectangle.
For A(−2, 1), B(4, 3), C(3, 6), D(−3, 4), the successive gradients are 1/3, −3, 1/3, −3. Opposite sides are parallel and adjacent products are −1, so ABCD is a rectangle.
For A(0, 0), B(6, 0), C(4, 3), D(1, 3), AB and CD are horizontal. BC has gradient −3/2 and DA has gradient 3, so the other pair is not parallel. This gives a trapezium with exactly one parallel pair.
Find a coordinate from a right angleWorked example
A = (−1, 1), B = (3, 3), C = (k, 7)
Angle ABC is 90°
The angle is at B, between BA and BC.
mAB = (3 − 1)/(3 − (−1)) = 1/2
The line BC must have gradient −2.
(7 − 3)/(k − 3) = −2
4 = −2k + 6 ⇒ k = 1
The denominator is nonzero at the solution.
BC: y − 3 = −2(x − 3)
y = −2x + 9
Check C(1, 7) lies on it.
06 / Your turn
Verify the gradient relation and the point.
A correct gradient with the wrong intercept describes the wrong line.
02 · Parallel through a point
Find the line through (−2, 3) parallel to 5x − 2y + 1 = 0.
Hint
The gradient is 5/2.
Worked solution
y − 3 = 5(x + 2)/2
5x − 2y + 16 = 0
03 · Perpendicular through a point
Find the line through (4, −1) perpendicular to 2x + 5y − 7 = 0.
Hint
The original gradient is −2/5.
Worked solution
m = 5/2
y + 1 = 5(x − 4)/2
5x − 2y − 22 = 0
04 · A vertical original
Find the line through (−3, 4) perpendicular to x = 6, and the line through it parallel to x = 6.
Hint
Use a constant y for the horizontal line.
Worked solution
Perpendicular: y = 4. Parallel: x = −3.
05 · Through an intercept
The line 3x + 4y − 12 = 0 meets the x-axis at A. Find its perpendicular through A.
Hint
A = (4, 0), and the original gradient is −3/4.
Worked solution
y = 4(x − 4)/3
4x − 3y − 16 = 0
06 · Prove a rectangle
A(0, 0), B(6, 2), C(5, 5), D(−1, 3) are successive vertices. Show ABCD is a rectangle.
Hint
Calculate four side gradients.
Worked solution
mAB = 1/3, mBC = −3
mCD = 1/3, mDA = −3
Opposite sides are parallel and every adjacent product is −1. The four vertices are distinct, so they form a rectangle.
07 · Unknown intercept
A(0, 6), B(−4, 0) and C(0, c) satisfy AB ⟂ BC. Find c.
Hint
AB has gradient 3/2; BC must have gradient −2/3.
Worked solution
c/4 = −2/3
c = −8/3
The negative answer places C below the origin, consistent with the falling line BC.
07 / Recap
Direction first, position second.
- Parallel non-vertical lines have equal gradients.
- Check whether equations describe the same line.
- Perpendicular finite, non-zero gradients multiply to −1.
- Horizontal and vertical lines form the exception to the formula.
- Use the required point to fix the new line’s position.
- Prove every geometric condition needed, not just one.
Section 1 of 7 · Parallel lines