Slopes of Perpendicular Lines Theorem

Since the theorem is a biconditional statement, the proof consists of two parts.

If two non-vertical lines are perpendicular, then the product of their slopes is $- 1 .$
If the product of the slopes of two non-vertical lines is $- 1,$ then the lines are perpendicular.

Part $1$

Let

ℓ_{1}

and

ℓ_{2}

be two perpendicular lines. Therefore, they intersect at one point. For simplicity, the lines will be translated so that the point of intersection is the origin.

Let

m_{1}

and

m_{2}

be the slopes of the lines

ℓ_{1}

and

ℓ_{2},

respectively. Next, consider the vertical line

x = 1 .

This line intersects both

ℓ_{1}

and

ℓ_{2} .

Since

ℓ_{1}

and

ℓ_{2}

are assumed to be perpendicular,

△ A O C

is a right triangle. Using the Distance Formula, the lengths of the sides of this triangle can be found.

Side	Points	$Distance Formula (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$	Length
$\overline{A O}$	$A (1, m_{1})$ $&$ $O (0, 0)$	$(0 - 1)^{2} + (0 - m_{1})^{2}$	$1 + m_{1}^{2}$
$\overline{C O}$	$C (1, m_{2})$ $&$ $O (0, 0)$	$(0 - 0)^{2} + (0 - m_{2})^{2}$	$1 + m_{2}^{2}$
$\overline{C A}$	$C (1, m_{2})$ $&$ $A (1, m_{1})$	$(1 - 1)^{2} + (m_{1} - m_{2})^{2}$	$m_{1} - m_{2}$

Since

△ A O C

is a right triangle, its side lengths satisfy the Pythagorean Equation.

A O^{2} + C O^{2} = C A^{2}

The next step is to substitute the lengths shown in the table.

A O^{2} + C O^{2} = C A^{2}

SubstituteExpressions

Substitute expressions

(1 + m_{1}^{2})^{2} + (1 + m_{2}^{2})^{2} = (m_{1} - m_{2})^{2}

▼

Simplify

PowSqrt

$(a)^{2} = a$

1 + m_{1}^{2} + 1 + m_{2}^{2} = (m_{1} - m_{2})^{2}

AddTerms

Add terms

2 + m_{1}^{2} + m_{2}^{2} = (m_{1} - m_{2})^{2}

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

2 + m_{1}^{2} + m_{2}^{2} = m_{1}^{2} - 2 m_{1} m_{2} + m_{2}^{2}

SubEqn

$LHS - m_{1}^{2} = RHS - m_{1}^{2}$

2 + m_{2}^{2} = - 2 m_{1} m_{2} + m_{2}^{2}

SubEqn

$LHS - m_{2}^{2} = RHS - m_{2}^{2}$

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

DivEqn

$LHS / (- 2) = RHS / (- 2)$

\frac{2}{- 2} = m_{1} m_{2}

MoveNegDenomToFrac

Put minus sign in front of fraction

- \frac{2}{2} = m_{1} m_{2}

QuotOne

$\frac{a}{a} = 1$

- 1 = m_{1} m_{2}

RearrangeEqn

Rearrange equation

m_{1} \cdot m_{2} = - 1

It has been proven that if two lines are perpendicular, then the product of their slopes is

- 1 .

$ℓ_{1} ⊥ ℓ_{2} \Rightarrow m_{1} \cdot m_{2} = - 1$

Part $2$

Here it is assumed that the slopes of two lines

ℓ_{1}

and

ℓ_{2}

are opposite reciprocals.

m_{1} \cdot m_{2} = - 1

Consider the steps taken in Part

1 .

This time, it should be found that

△ A O C

is a right triangle.

If the lengths of the sides of

△ A O C

satisfy the Pythagorean Theorem, then the triangle is a right triangle.

A O^{2} + C O^{2} = ? C A^{2}

The side lengths, which were previously found in Part

1,

can be substituted into the above equation.

A O^{2} + C O^{2} = ? C A^{2}

SubstituteExpressions

Substitute expressions

(1 + m_{1}^{2})^{2} + (1 + m_{2}^{2})^{2} = ? (m_{1} - m_{2})^{2}

▼

Simplify

PowSqrt

$(a)^{2} = a$

1 + m_{1}^{2} + 1 + m_{2}^{2} = ? (m_{1} - m_{2})^{2}

AddTerms

Add terms

2 + m_{1}^{2} + m_{2}^{2} = ? (m_{1} - m_{2})^{2}

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

2 + m_{1}^{2} + m_{2}^{2} = ? m_{1}^{2} - 2 m_{1} m_{2} + m_{2}^{2}

SubEqn

$LHS - m_{1}^{2} = RHS - m_{1}^{2}$

2 + m_{2}^{2} = ? - 2 m_{1} m_{2} + m_{2}^{2}

SubEqn

$LHS - m_{2}^{2} = RHS - m_{2}^{2}$

2 = ? - 2 m_{1} m_{2}

Substitute

$m_{1} m_{2} = - 1$

2 = ? - 2 (- 1)

MultNegNegOnePar

$- a (- b) = a \cdot b$

2 = 2 ✓

Since a true statement was obtained,

△ A O C

is a right triangle. Therefore,

ℓ_{1}

and

ℓ_{2}

are perpendicular lines. This completes the second part.

$m_{1} \cdot m_{2} = - 1 \Rightarrow ℓ_{1} ⊥ ℓ_{2}$

The biconditional statement has been proven.

$ℓ_{1} ⊥ ℓ_{2} \Leftrightarrow m_{1} \cdot m_{2} = - 1$

{{ article.displayTitle }}

Slopes of Perpendicular Lines Theorem

Proof

Part $1$

Part $2$

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount }}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount }}
	{{ 'ml-lesson-time-estimation' \| message }}

{{ article.displayTitle }}

Proof

Part 1

Part 2

Part $1$

Part $2$