Parallel and Perpendicular Lines in $y = mx + c$

Parallel Lines

• Parallel lines are like railroad tracks: they never meet, no matter how far they extend.
• In the equation $y = mx + c$, parallel lines have identical gradients ($m$) but different y-intercepts ($c$).
• This identical slope means the lines rise over the run at the same rate, keeping them an equal distance apart at all points.

Determining Parallel Lines

Two lines $y = m_1x + c_1$ and $y = m_2x + c_2$ are parallel if $m_1 = m_2$ and $c_1 \neq c_2$. The equality of $m$ ensures the lines have the same steepness, hence never intersecting.

Example

If we have the line $y = 2x + 7$, give the equation of the line that would be parallel to this and passes through the point $(3, 8)$.

Solution

Step 1: Determine Gradient 

Gradient of new line is $m_2 = 2$ as the lines are parallel

Step 2: Use the gradient and co-ordinate to solve for c 

$$y = mx + c \to m = 2 \therefore y = 2x + c  \\ \text{Substitute co-ordinate}  8 = 2(3) + c \to 8 = 6 + c \therefore c = 2 \\ y = 2x + 2$$

Perpendicular Lines

• Perpendicular lines intersect at right angles $(90\degree)$, forming perfect corners, like the edges of a square.
• For lines represented by $y = mx + c$, perpendicular lines have gradients that are negative reciprocals of each other.
• This means that the gradient of one is equal to the other 'flipped' and multiplied by $-1$ i.e $$m_1 = 2  \text{Gradient of perpendicular line is} \\ m_2 = \frac{1}{2} \times -1 = -\frac{1}{2}$$
• This relationship ensures the angles at their intersection point are right angles.

Determining Perpendicular Lines

Two lines $y = m_1x + c_1$ and $y = m_2x + c_2$ are perpendicular if $m_1 \cdot m_2 = -1$, or equivalently, $m_1 = -\frac{1}{m_2}$. This condition guarantees that one line’s slope is the negative inverse of the other’s.

Example

If we have the line $y = \frac{3}{4}x + 5$, give the equation of the line that would be parallel to this and passes through the point $(6, 11)$.

Solution

Step 1: Determine Gradient 

Gradient of new line is negative reciprocal $\therefore  m_2 = \frac{4}{3} \times -1 = -\frac{4}{3}$ as the lines are perpendicular.

Step 2: Use the gradient and co-ordinate to solve for c 

$$y = mx + c \to m = -\frac{4}{3} \therefore y = -\frac{4}{3}x + c  \\ \text{Substitute co-ordinate}  11 = -\frac{4}{3}(6) + c \to 11 = -8 + c \therefore c = 19 \\ y = -\frac{4}{3}x + 19$$