E3.6 Parallel lines

Learning objectives

E3.6.1Find the gradient and equation of a straight line parallel to a given• Find the gradient and equation of a straight line parallel to a given line. e.g. find the equation of the line parallel to y = 4x – 1 that passes through (1, –3).

Distinct non-vertical parallel lines have the same gradient because they rise or fall at the same rate, but they have different intercepts because they occupy different positions.

$y=m_1x+c_1\ \parallel\ y=m_2x+c_2\quad\Longrightarrow\quad m_1=m_2$

Rearrange the given line into $y=mx+c$ if needed and identify $m$ .
Write the parallel line as $y=mx+k$ with the same gradient.
Substitute the point on the new line to calculate $k$ .
Simplify the equation and check both the gradient and the given point.

$y=4x-1,\ (1,-3):\quad -3=4(1)+k\Rightarrow k=-7\Rightarrow y=4x-7$

If a parallel line has gradient $1/7$ and crosses the $x$ -axis at $x=2$ , it passes through $(2,0)$ . Substitution gives $0=(1/7)(2)+k$ , so $k=-2/7$ and $y=(1/7)x-2/7$ , equivalently $7y=x-2$ .

The new line has the required gradient and its stated point satisfies the equation. In the syllabus example, substituting $(1,-3)$ into $y=4x-7$ gives $-3=4-7$ .

Do not copy the original intercept: that would reproduce the same line, not a distinct parallel one. Vertical lines are written $x=k$ ; two distinct vertical lines are parallel even though their gradients are undefined.

Next Topic

—E3.7 Perpendicular linesRead