03Eparallellines

= Parallel & Perpendicular Lines =

Parallel Lines
Two lines are ** parallel ** if they have the __**same gradient**__.

For example, the following two equations produce parallel lines
 * y = 2x + 1
 * y = 2x – 4

** Example 1 **
Find if the following two equations are parallel.
 * 6x + 2y = 4
 * y = 5 – 3x


 * Solution: **

{Rearrange equation 1 to find the gradient} ... ... ... 6x + 2y = 4 ... ... ... {–6x} ... ... ... 2y = –6x + 4 ... ... .. {÷2} ... ... ... . y = –3x + 2

Both lines have a gradient of –3, hence they **__are__** parallel.

** Example 2 **
Show that AB is parallel to CD, given that
 * A (–1, –5)
 * B (5, 7)
 * C (–3, 1)
 * D (4, 15)


 * Solution:[[image:03Egradientrule.GIF align="right"]] **

math \\ \text{Gradient AB} = \dfrac{7 - (-5)}{5 - (-1)} \\. \\ . \qquad \qquad \quad \; = \dfrac{12}{6} \\. \\ . \qquad \qquad \quad \; = 2 math

math \\ \text{Gradient CD} = \dfrac{15-1}{4-(-3)} \\. \\ . \qquad \qquad \quad \; = \dfrac{14}{7} \\. \\ . \qquad \qquad \quad \; = 2 math

Both AB and CD have a gradient of 2, hence they __**are**__ parallel.



Collinear Points
Three points are ** collinear ** if they all lie on the __same__ straight line.

A, B, C will be collinear if the gradient of AB is the same as the gradient of AC.

** Example 3 **
Show that the following points are collinear.
 * A (2, 0)
 * B (4, 1)
 * C (10, 4)


 * Solution:[[image:03Egradientrule.GIF align="right"]] **

math \\ \text{Gradient AB} = \dfrac{1 - 0}{4 - 2} \\. \\ . \qquad \qquad \quad \; = \dfrac{1}{2} math

math \\ \text{Gradient AC} = \dfrac{4 - 0}{10 - 2} \\. \\ . \qquad \qquad \quad \; = \dfrac{4}{8} \\. \\ . \qquad \qquad \quad \; = \dfrac{1}{2} math

Both AB and AC have the same gradient hence A, B, C __**are**__ collinear.

Perpendicular Lines
Two lines are ** perpendicular ** if they meet at right angles (90º).

There is an important relationship between the gradients of two perpendicular lines.

Consider the two lines (1 and 2) shown on the right.

The two triangles drawn in to find their gradients are ** congruent ** (same size, same shape). {This is because the two angles at the intersection of lines 1 and 2 will be __**complementary**__ (add to 90 º) }

Gradient of line 1: math . \qquad m_1 = \dfrac{b}{a} math

Gradient of line 2: math . \qquad m_2 = \dfrac{-a}{b} math

Multiply the two gradients: math \\ . \qquad m_1 \times m_2 = \dfrac{b}{a} \times \dfrac{-a}{b} \\. \\ . \qquad \qquad \qquad = \dfrac{-ab}{ab} \\. \\ . \qquad \qquad \qquad = -1 math

So we get the rule: ... ... ... ** For any two perpendicular lines: m 1 × m 2 = –1 **

** Example 4 **
Show that the following two lines are perpendicular:
 * y = 4x –1
 * x + 4y = 3


 * Solution: **

{Rearrange equation 2 to find the gradient} math \\ . \qquad x + 4y = 3 \\. \\ . \qquad \quad 4y = -x + 3 \\. \\ . \qquad \quad y = \dfrac{-1}{4}x + \dfrac{3}{4} math

Gradient of equation 1 and 2 math . \qquad m_1 = 4 \qquad \qquad m_2 = -\dfrac{1}{4} math

Multiply the two gradients math . \qquad m_1 \times m_2 = 4 \times -\dfrac{1}{4} = -1 math

Since m 1 × m 2 = –1, these lines __are__ perpendicular.

** Example 5 **
Find the equation of the straight line that passes through (1, 2) and is perpendicular to the line y = 2x + 5


 * Solution: **

Gradient of original line math . \qquad m_1 = 2 math

Find the gradient of the perpendicular line math \\ . \qquad m_1 \times m_2 = -1 \\. \\ . \qquad 2 \times m_2 = -1 \\. \\ . \qquad m_2 = -\dfrac{1}{2} math

Substitute m 2 and (1, 2) into y = mx + c to find c. math \\ . \qquad y = mx + c \\. \\ . \qquad 2 = -\dfrac{1}{2} \big( 1 \big) + c \\. \\ . \qquad 2 = -\dfrac{1}{2} + c \\. \\ . \qquad 2\dfrac{1}{2} = c math

Substitute m and c into y = mx + c to get the equation of the line math . \qquad y = -\dfrac{1}{2}x + 2\dfrac{1}{2} math

Horizontal and Vertical Lines
Recall that:

A ** horizontal line **
 * has an equation in the form y = a ... ... {a is any real number}
 * has a gradient of zero (0)

A ** vertical line **
 * has an equation in the form x = b ... ... {b is any real number}
 * has an undefined gradient (or infinite gradient)

Any horizontal line and vertical line will be perpendicular. .