Parallel and Perpendicular Lines Revision Notes for VCE SSCE Mathematical Methods

Parallel and Perpendicular Lines

Understanding the relationship between lines and their gradients is fundamental to coordinate geometry. This note explores how we can determine whether two lines are parallel or perpendicular by examining their gradients, and how to find equations of lines with specific relationships to other lines.

Parallel lines

When two non-vertical lines share identical gradients, they are parallel to each other. This means they will never intersect, no matter how far they are extended in either direction.

The reverse is also true: if two non-vertical lines are parallel, they must have the same gradient. This relationship can be proven by considering the angles of inclination that these lines make with the positive $x$ -axis.

Understanding the geometric basis

We can prove the parallel line rule using two key geometric principles:

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Principle 1: Two non-vertical lines are parallel if and only if they make equal angles ( $\theta_1$ and $\theta_2$ ) with the positive direction of the $x$ -axis.

Principle 2: When two angles $\theta_1$ and $\theta_2$ are acute, obtuse, or zero, the equation $\tan \theta_1 = \tan \theta_2$ means that $\theta_1 = \theta_2$ .

Since the gradient of a line equals $\tan \theta$ (where $\theta$ is the angle of inclination), parallel lines must have equal gradients.

Example of parallel lines

Consider these two linear equations:

y = 2x + 3

y = 2x - 4

Both equations have a gradient of 2. Since their gradients are equal, the lines are parallel to each other. Notice that they have different $y$ -intercepts ( $3$ and $-4$ ), which means they are shifted vertically but maintain the same steepness.

Perpendicular lines

Two lines are perpendicular when they meet at a right angle ( $90°$ ). For non-vertical lines, there is a special relationship between their gradients that allows us to determine perpendicularity.

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Key rule: Two lines with gradients $m_1$ and $m_2$ (both non-zero) are perpendicular if and only if $m_1m_2 = -1$ .

In other words, the product of the gradients of perpendicular lines equals $-1$ . This can also be expressed by saying that one gradient is the negative reciprocal of the other: if one line has gradient $m$ , a perpendicular line has gradient $-\frac{1}{m}$ .

Proof of the perpendicular gradient rule

The proof involves four steps, starting with lines through the origin and then generalising to any lines in the plane.

Step 1: Setting up the construction

We begin by drawing two lines through the origin $O$ , one with positive gradient $m_1$ and one with negative gradient $m_2$ . We construct right-angled triangles $OPQ$ and $OAB$ where $OQ = OB$ .

From the triangles, we can express the gradients:

Gradient $m_1 = \frac{AB}{BO}$

Gradient $m_2 = -\frac{OQ}{PQ}$

Therefore, the product of the gradients is:

m_1m_2 = -\frac{OQ}{PQ} \times \frac{AB}{BO}

Step 2: Proving perpendicular lines have product -1

If the two lines are perpendicular, then angle $POQ$ equals angle $AOB$ . This means triangles $OPQ$ and $OAB$ are congruent (they have the same shape and size).

Since the triangles are congruent, $PQ = AB$ . Substituting this into our product:

m_1m_2 = -\frac{AB}{PQ} = -\frac{AB}{AB} = -1

Step 3: Proving the converse

We now prove the reverse: if the product of gradients equals $-1$ , then the lines are perpendicular.

If $m_1m_2 = -1$ , then $AB = PQ$ . This means triangles $OAB$ and $OPQ$ are congruent. Therefore, angle $POQ$ equals angle $AOB$ , which means angle $AOP = 90°$ . The lines are perpendicular.

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Step 4: Generalising to any lines

The proof so far applies to lines passing through the origin. However, if we have two lines anywhere in the coordinate plane, we can draw parallel lines through the origin. Since parallel lines have the same gradients, the perpendicularity rule $m_1m_2 = -1$ applies to any pair of lines, not just those through the origin.

Worked example: Finding parallel and perpendicular lines

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Worked Example: Finding Parallel and Perpendicular Lines

Problem: Find the equation of the straight line which passes through $(1, 2)$ and is:

a) parallel to the line with equation $2x - y = 4$

b) perpendicular to the line with equation $2x - y = 4$

Solution:

First, we rearrange the given equation into gradient-intercept form: $2x - y = 4$ becomes $y = 2x - 4$ . This tells us the gradient is 2.

Part a) Finding the parallel line

A line parallel to $y = 2x - 4$ must have the same gradient, which is $2$ .

We use the point-slope form $y - y_1 = m(x - x_1)$ where the point is $(x_1, y_1) = (1, 2)$ and the gradient is $m = 2$ :

y - 2 = 2(x - 1)

y - 2 = 2x - 2

y = 2x

The equation of the parallel line is $y = 2x$ .

Part b) Finding the perpendicular line

A line perpendicular to a line with gradient $m = 2$ has gradient $-\frac{1}{m} = -\frac{1}{2}$ .

Using the point-slope form with $(x_1, y_1) = (1, 2)$ and $m = -\frac{1}{2}$ :

y - 2 = -\frac{1}{2}(x - 1)

Multiplying both sides by $2$ :

2y - 4 = -x + 1

2y + x = 5

The equation of the perpendicular line is $2y + x = 5$ (or equivalently $x + 2y = 5$ ).

Worked example: Proving perpendicularity in triangles

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Worked Example: Proving Perpendicularity in Triangles

Problem: The coordinates of the vertices of triangle $ABC$ are $A(0, -1)$ , $B(2, 3)$ and $C(3, -2\frac{1}{2})$ . Show that side $AB$ is perpendicular to side $AC$ .

Solution:

To prove perpendicularity, we need to show that the product of the gradients equals $-1$ .

Let $m_1$ be the gradient of line $AB$ :

m_1 = \frac{3 - (-1)}{2 - 0}

Let $m_2$ be the gradient of line $AC$ :

m_2 = \frac{-2\frac{1}{2} - (-1)}{3 - 0}

Now we check the product:

m_1 \times m_2 = 2 \times \left(-\frac{1}{2}\right) = -1

Since the product of the gradients equals $-1$ , the lines $AB$ and $AC$ are perpendicular to each other. This means angle $BAC$ is a right angle.

Exam tips

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Exam Tips:

Always rearrange equations into the form $y = mx + c$ to clearly identify the gradient
For perpendicular lines, remember the negative reciprocal: if $m_1 = 2$ , then $m_2 = -\frac{1}{2}$
When proving perpendicularity, calculate both gradients first, then check if their product equals $-1$
Use the point-slope form $y - y_1 = m(x - x_1)$ when you know a point and a gradient
Check your final answer by verifying it satisfies the given conditions

Remember!

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Key Points to Remember:

Parallel lines have identical gradients: if line 1 has gradient $m$ and line 2 has gradient $m$ , they are parallel
Perpendicular lines have gradients whose product is $-1$ : if $m_1 \times m_2 = -1$ , the lines are perpendicular
The perpendicular gradient is the negative reciprocal: if one line has gradient $m$ , the perpendicular line has gradient $-\frac{1}{m}$
Use the point-slope form $y - y_1 = m(x - x_1)$ when finding equations of lines through a specific point
To prove perpendicularity in geometric problems, calculate both gradients and verify their product equals $-1$

Parallel and Perpendicular Lines (VCE SSCE Mathematical Methods): Revision Notes