Perpendicular lines — product of slopes = −1: Question bank

True or false — justify

True or false: If

m_{1} m_{2} = - 1

then the two lines are perpendicular, always.

True — the product equals

- 1

only when the inclination angles differ by

90°

, so it is an exact "if and only if" test (except when a slope is undefined, where the test simply doesn't apply).

True or false: Two perpendicular lines can both have positive slope.

False — if one line tilts uphill (positive slope) the negative reciprocal is negative, so the perpendicular must tilt downhill; you cannot have two positives multiplying to

- 1

True or false: A line perpendicular to

y = 5

has an undefined slope.

True —

y = 5

is horizontal (

m = 0

); the line at a right angle to it is vertical, and vertical lines have undefined slope, so the product rule can't be used here.

True or false: If

m_{1} = m_{2}

the lines could still be perpendicular.

False — equal slopes mean equal inclination angles, so the lines are parallel (or identical); a right angle needs the angles to differ by

90°

, which equal slopes can never do.

True or false: The condition

m_{1} m_{2} = - 1

works even if both lines are steep, like slopes

10

and

- 0.1

True — steepness doesn't matter;

10 \times (- 0.1) = - 1

, so these are perpendicular; the negative reciprocal of a steep line is a shallow line, which is geometrically exactly what a right-angle turn produces.

True or false: For perpendicular lines the angles with the x-axis add up to

90°

False — they differ by

90°

(

θ_{2} = θ_{1} + 90°

), they do not sum to

90°

; e.g. inclinations

30°

and

120°

are perpendicular but sum to

150°

True or false: If a line has slope

0

, no ordinary slope value can make a perpendicular to it.

True — its perpendicular is vertical (undefined slope), so there is no finite number

m_{2}

with

0 \cdot m_{2} = - 1

; you must handle it as the horizontal–vertical special case.

True or false: Swapping which line you call

L_{1}

and which

L_{2}

changes whether the product test passes.

False — multiplication is commutative, so

m_{1} m_{2} = m_{2} m_{1}

; the labelling never affects the outcome.

Spot the error

Spot the error: "Slope

\frac{2}{5}

, so perpendicular slope is

\frac{5}{2}

They flipped but forgot to negate;

\frac{2}{5} \times \frac{5}{2} = + 1

, not

- 1

. The correct perpendicular slope is

- \frac{5}{2}

(flip and sign-flip).

Spot the error: "Slope

2

, so perpendicular slope is

- 2

They negated but forgot to flip;

2 \times (- 2) = - 4 \neq = - 1

. Negating alone is not enough — the perpendicular slope is the negative reciprocal,

- \frac{1}{2}

Spot the error: "Lines

y = 3 x

and

y = 3 x + 4

are perpendicular because they never meet at the same point."

Never meeting means parallel, not perpendicular; equal slopes (

3 = 3

) give parallel lines. Perpendicular is about the angle at their crossing, and these two never even cross.

Spot the error: "The x-axis and y-axis fail the rule

0 \times undefined \neq = - 1

, so they aren't perpendicular."

The rule simply doesn't apply when one slope is undefined — you can't do arithmetic with "undefined". Geometrically the axes obviously meet at

90°

, so they are perpendicular; verify by picture, not by the product formula.

Spot the error: "Perpendicular of slope

- \frac{3}{4}

- \frac{4}{3}

, because I flipped the fraction."

They flipped but kept the original minus sign;

- \frac{3}{4} \times - \frac{4}{3} = + 1

. The negative reciprocal of a negative slope is positive: the answer is

+ \frac{4}{3}

Spot the error: "

tan (θ + 90°) = tan θ

, so the perpendicular has the same slope."

Tangent does not have period

90°

(its period is

180°

). Correctly,

tan (θ + 90°) = - cot θ = - 1/ tan θ

, which is exactly the negative reciprocal — the opposite of "same slope".

Spot the error: "Rotating direction

(Δ x, Δ y)

90°

gives

(Δ y, Δ x)

, so new slope

= \frac{Δ x}{Δ y}

90°

anticlockwise turn gives

(- Δ y, Δ x)

, not

(Δ y, Δ x)

; the missing minus is what makes the perpendicular slope

- \frac{Δ x}{Δ y}

and forces the product to be

- 1

Why questions

Why does the product of perpendicular slopes come out to exactly

- 1

and never some other constant?

Rotating the direction vector by

90°

swaps run and rise and injects one minus sign, so the two ratios are

\frac{Δ y}{Δ x}

and

- \frac{Δ x}{Δ y}

; the

Δ x

and

Δ y

cancel completely, leaving

- 1

with no room for any other value.

Why do we say "negative reciprocal" instead of just "reciprocal"?

The reciprocal alone gives product

+ 1

(that's a different, geometrically meaningless condition); the extra minus is needed so the product lands on

- 1

, encoding the sign flip that a quarter-turn produces.

Why does the parent note prove the rule three separate ways instead of once?

To show the result is inevitable and not a lucky coincidence — rotation, the tangent identity, and Pythagoras start from totally different ideas yet all collapse to

m_{1} m_{2} = - 1

, which builds real confidence in the rule.

Why can't the product rule be applied to a vertical line?

A vertical line rises without ever running, so

tan θ

(its slope) is undefined; you cannot multiply an undefined quantity, so that pair must be checked geometrically as the horizontal–vertical special case.

Why do a steep line and a shallow line end up perpendicular to each other?

The perpendicular slope is the negative reciprocal, and taking a reciprocal turns a large number into a small one; a right-angle turn literally trades "steep uphill" for "gentle downhill", which is exactly the reciprocal-with-a-sign-flip.

Why is

θ_{2} = θ_{1} + 90°

(adding, not subtracting) still fine even if it pushes past

180°

Direction angles are the same line whether you add or subtract

90°

(they point along the same perpendicular line, just opposite ways), and

tan

repeats every

180°

, so both choices give the same slope value.

Edge cases

Edge case: What is the perpendicular to a horizontal line

y = c

A vertical line

x = k

— its slope is undefined, so you describe it by an

x

-equation, not by a number; this is the one pair the product rule cannot handle numerically.

Edge case: Can a line be perpendicular to itself?

No — that would need

m \cdot m = m^{2} = - 1

, impossible for any real slope; no line meets itself at a right angle, so the equation has no real solution, matching the geometry.

Edge case: Is the line

y = x

perpendicular to

y = - x

Yes — slopes are

1

and

- 1

, and

1 \times (- 1) = - 1

; these are the diagonals of the plane meeting at a clean

90°

Edge case: Two lines both pass through the origin with slopes

m

and

- 1/ m

; are they perpendicular only at the origin?

They cross only at the origin (a shared point), and there they meet at

90°

; two distinct lines through one common point intersect exactly once, so the right angle occurs at that single crossing.

Edge case: If

m_{1}

is a huge number like

1000

, what happens to its perpendicular slope?

It becomes

- 1/1000

, a tiny negative number — as a line gets closer to vertical, its perpendicular gets closer to horizontal, the limiting behaviour that matches the vertical-⊥-horizontal special case.

Edge case: Does the perpendicular rule depend on where the lines are drawn (position/translation)?

No — slope depends only on direction, not position, so sliding either line around (translation) never changes

m_{1}

m_{2}

; perpendicularity is a property of directions alone.

Connections

Parent: the $m_1 m_2 = -1$ rule — every trap here tests it.
Slope of a Line — the $m = tan θ$ meaning behind "steep vs shallow" traps.
Parallel Lines — equal slopes — the twin rule people confuse with perpendicular.
Angle Between Two Lines — perpendicular as the $θ = 90°$ special case.
Rotation of Vectors by 90° — the source of the sign flip in "Why" questions.

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