C3.6 Parallel lines

Learning objectives

C3.6.1Find 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 parallel straight lines have the same gradient but different intercepts. Keep the given line’s gradient, then use the new point to find its intercept.

Given line	Parallel-line form
$y=mx+c$	$y=mx+k$ , with a different intercept
vertical line $x=a$	$x=b$ , with $b
e a$

Write the new line as $y=mx+k$ using the known gradient $m$ ; substitute the coordinates of the point it passes through; solve for $k$ ; write the complete equation in fully simplified form; substitute the point once more to check it.

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

$\therefore\ y=4x-7$

If the given point lies on the $y$ -axis, its $x$ -coordinate is $0$ , so its $y$ -coordinate is immediately the new intercept. For example, a line parallel to $y=5x+6$ through $(0,-7)$ is $y=5x-7$ .

If the original line is shown on a grid or specified by two points, find its gradient first; only that gradient transfers to the parallel line. The original intercept does not.

Keeping both the same gradient and the same intercept reproduces the original line, not a distinct parallel line. Do not change the sign or take the reciprocal of the gradient.

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—Cambridge IGCSE Mathematics (0580)Read