3.1–3.3 - Gradients, perpendicular lines and distance

Gradient

The gradient of a non-vertical straight line is its change in $y$ divided by its change in $x$. A positive gradient rises from left to right; a negative gradient falls. For example, a rise of $3$ over a horizontal change of $2$ gives gradient $\frac32$, not $\frac23$.

Gradient between two points

For distinct points $(x_1,y_1)$ and $(x_2,y_2)$, with $x_2\ne x_1$,

A horizontal line has gradient $0$. A vertical line has no defined gradient because its horizontal change is $0$.
Not on the formulae sheet: learn it.

Worked example 1: a positive gradient

Question. Work out the gradient of the line through $P(-3,2)$ and $Q(5,6)$.

Plan. Subtract coordinates in the order $Q-P$, then divide the vertical change by the horizontal change.

Why this step? Subtracting a negative $x$-coordinate adds its distance to the left of zero: the horizontal gap is $8$, not $2$.

Answer. The gradient is $\frac12$.

Check. Moving $8$ units right at $\frac12$ unit up per unit across gives a rise of $4$, from $2$ to $6$.

Worked example 2: a negative gradient

Question. Work out the gradient of the line through $P(-3,2)$ and $R(5,-2)$.

Plan. Use the same point order for both signed changes.

Why this step? The negative numerator records a fall. Removing that sign would describe a different direction.

Answer. The gradient is $-\frac12$.

Check. Reversing both differences gives $\frac{2-(-2)}{-3-5}=\frac4{-8}=-\frac12$, the same gradient.

Watch out

Wrong: $m=\frac{\Delta x}{\Delta y}$, or subtracting in opposite orders above and below. Right: $m=\frac{y_2-y_1}{x_2-x_1}$. Give the resulting number as your gradient, rather than a term involving $x$.

Parallel and perpendicular lines

Distinct parallel lines never meet and have the same direction. Perpendicular lines meet at $90^\circ$. For non-vertical lines, equal gradients give parallel directions; equal gradients can also describe the same line, so check that the lines are distinct before calling them parallel.

Parallel and perpendicular directions

For non-vertical lines with gradients $m_1$ and $m_2$,

When $m_1\ne0$, the perpendicular gradient is $-\frac1{m_1}$. A horizontal line is perpendicular to a vertical line; the product rule cannot be used there because the vertical gradient is undefined.
Not on the formulae sheet: learn it.

Worked example 3: equal directions

Question. Line $L$ passes through $A(1,2)$ and $B(5,5)$. Line $M$ passes through $C(-2,-1)$ and $D(2,2)$. Show that the lines are parallel.

Plan. Calculate both gradients and confirm that these are distinct lines with the same gradient.

Why this step? A matching ratio of changes establishes the same direction, regardless of where the line sits.

Answer. Both gradients are $\frac34$. The lines are distinct: from $C$, moving $3$ units right along $M$ reaches height $-1+\frac94=\frac54$, not $A$'s height $2$. Therefore the lines are parallel.

Check. Both point pairs move $4$ units right and $3$ units up, so their directions agree.

Worked example 4: a right-angle turn

Question. Line $L$ passes through $A(1,2)$ and $B(5,5)$. Line $N$ passes through $C(-2,-1)$ and $E(1,-5)$. Show that the lines are perpendicular.

Plan. Calculate both gradients and test whether their product is $-1$.

Why this step? The sign changes and the fraction inverts. Either change on its own would not give a right-angle turn.

Answer. The lines are perpendicular because their gradients multiply to $-1$.

Check. The first direction moves $4$ right and $3$ up; the second moves $3$ right and $4$ down, a clockwise quarter-turn.

Exam tip

For “Show that”, finish with the required relationship and the reason. Two correct gradients leave your argument unfinished unless you explain why they establish parallel or perpendicular directions.