Parallel and perpendicular linesIB MYP Maths Extended: Revision notes

Parallel lines never meet because they slope at the same rate. They have the same gradient: $m_1=m_2$. The lines $y=3x-2$, $y=3x+5$ and $y=3x$ are all parallel to each other, because each has gradient $3$. They differ only in their $y$-intercept.

Perpendicular lines meet at a right angle ($90^\circ$). Their gradients multiply to $-1$: $m_1\times m_2=-1$. To find a perpendicular gradient, flip the fraction and change the sign: $3\to-\frac13$, $-\frac12\to2$ and $\frac23\to-\frac32$. Example: $y=2x+7$ and $y=-\frac12x+1$ are perpendicular because $2\times\left(-\frac12\right)=-1$.

You must have the equation in the form $y=mx+c$ before reading off the gradient. Example: $2x+4y=8$. Subtract $2x$: $4y=-2x+8$. Divide every term by $4$: $y=-\frac12x+2$. The gradient is $-\frac12$, not $2$.

1. Find the gradient of the given line. 2. Use the same gradient in $y=mx+c$. 3. Substitute the point to find $c$. Example: parallel to $y=3x-2$ through $(2,1)$. $m=3$, so $1=3(2)+c$, giving $c=-5$ and $y=3x-5$.

1. Find the gradient of the given line. 2. Take the negative reciprocal. 3. Substitute the point to find $c$. Example: perpendicular to $y=\frac12x+1$ through $P(4,5)$. $m=-2$, so $5=-2(4)+c$, giving $c=13$ and $y=-2x+13$. The perpendicular from a point to a line gives the shortest distance between them. Find where the two lines meet, then use Pythagoras to find the distance.