Parallel and perpendicular linesEdexcel International A Level Maths: Revision notes

Two lines are parallel if and only if they have the same gradient: $m_1=m_2$. To compare lines, rearrange each into the form $y=mx+c$ first. For example $2x+5y=10$ becomes $y=-\frac25x+2$, and $4x+10y=7$ becomes $y=-\frac25x+\frac{7}{10}$, so they are parallel (the same gradient, different intercepts). If the intercepts were also equal the lines would be the same line.

Two lines with gradients $m_1$ and $m_2$ are perpendicular if and only if $m_1m_2=-1,$ that is $m_2=-\frac{1}{m_1}$ (the negative reciprocal). A gradient of $2$ becomes $-\frac12$; $-\frac25$ becomes $\frac52$; $\frac34$ becomes $-\frac43$. Vertical and horizontal lines are perpendicular, but they are the one case where the product rule cannot be used because a vertical line has no gradient.

To find the line parallel to a given line through $(x_1,y_1)$: (1) find the gradient $m$ of the given line (rearrange if needed); (2) use the same $m$ in $y-y_1=m(x-x_1)$. Example: parallel to $2x+5y=10$ through $(5,-1)$: $m=-\frac25$, so $y+1=-\frac25(x-5)$, giving $2x+5y-5=0$.

To find the line perpendicular to a given line through $(x_1,y_1)$: (1) find the gradient $m$ of the given line; (2) take the negative reciprocal $-\frac1m$; (3) use $y-y_1=-\frac1m(x-x_1)$. Worked example from the specification: the line perpendicular to $3x+4y=18$ through $(2,3)$. The gradient is $-\frac34$, so the perpendicular gradient is $\frac43$ and $y-3=\frac43(x-2)$. Multiplying by $3$ gives $3y-9=4x-8$, so $4x-3y+1=0$.

To show that two lines are perpendicular, find both gradients and show $m_1m_2=-1$; state the conclusion. To prove a triangle is right-angled, show two sides have gradients with product $-1$. To find a fourth vertex of a rectangle, build the lines through known points parallel to the sides and solve them simultaneously. Always find intersection points by solving the two equations as simultaneous equations, and give exact fractions where the question asks for them.