Leave lesson

Algebra · Straight-line graphs

1 / 9

Parallel and perpendicular lines

Why parallel lines share a gradient (and nothing else), why perpendicular gradients multiply to −1 (the negative reciprocal), and how to build a line through a given point with either condition.

Algebra · Straight-line graphs

Parallel and perpendicular lines

Why parallel lines share a gradient (and nothing else), why perpendicular gradients multiply to −1 (the negative reciprocal), and how to build a line through a given point with either condition.

Why it works

Parallel = same steepness = same gradient. Two lines that never meet climb at the same rate: y=2x1y = 2x - 1 and y=2x+7y = 2x + 7 are parallel — same m=2m = 2, different cc. The intercept is irrelevant to parallelism; it just slides the line up or down. So "parallel to y=2x1y = 2x - 1 through (0,7)(0, 7)" is instant: keep m=2m = 2, read the new intercept: y=2x+7y = 2x + 7.

Perpendicular = negative reciprocal. Turning a line through 90° swaps its rise and run AND flips one sign: a gradient of 44 (right 1, up 4) becomes right 4, DOWN 1 — gradient 14-\frac{1}{4}. In general

m=1m,equivalentlym1×m2=1.m_\perp = -\frac{1}{m}, \qquad\text{equivalently}\qquad m_1 \times m_2 = -1.

Both moves are required: 14\frac{1}{4} (reciprocal, no flip) and 4-4 (flip, no reciprocal) are the two classic half-answers. The product test is the clean check: 4×(14)=14 \times (-\frac{1}{4}) = -1. ✓

"Show these lines are perpendicular" = show the product is −1. y=3x2y = 3x - 2 and x+3y=6x + 3y = 6: rearrange the second — y=2x3y = 2 - \frac{x}{3}, gradient 13-\frac{1}{3}. Product: 3×(13)=13 \times (-\frac{1}{3}) = -1 — perpendicular, and that one line of arithmetic IS the proof. (Never read a gradient off an un-rearranged form.)

Through a given point: condition first, then the usual recipe. For the line perpendicular to y=2x+3y = 2x + 3 through (4,1)(4, 1): the gradient must be 12-\frac{1}{2}, then cc from the point — 1=12(4)+c1 = -\frac{1}{2}(4) + c, so c=3c = 3:

y=12x+3.y = -\tfrac{1}{2}x + 3.

The given point is not decoration: the perpendicular gradient alone describes a whole family of lines; the point picks the one.