Work with the gradient of a line as rise over run and as the tangent of its angle of inclination, write the equation of a line in gradient-intercept, point-gradient and general form, use the parallel and perpendicular gradient conditions, and find the length and midpoint of an interval

What this dot point is asking

Every straight line in the plane is captured by two numbers: how steep it is, and where it sits. This dot point is the toolkit for both. The gradient measures steepness as rise over run, and it is also the tangent of the line's angle to the horizontal, which links coordinate geometry to trigonometry. Once you have a gradient and a point, you can write the line's equation in whichever of three standard forms a question wants. Two lines are parallel when their gradients are equal and perpendicular when the product of their gradients is $-1$. And the length and midpoint of the interval joining two points come straight from Pythagoras and from averaging the coordinates.

None of these ideas is hard on its own, but the marks live in choosing the right tool and laying out the working cleanly. A "find the equation of the line through two points" question is really two steps (gradient, then point-gradient form); a "perpendicular" question hinges on the negative-reciprocal gradient; a "distance" question is Pythagoras in disguise. This page is also the launch pad for the rest of coordinate geometry: tangents and normals to curves in calculus are just lines through a point with a known (or perpendicular) gradient, so the technique you build here is used again and again.

The answer

Gradient: rise over run

The gradient of an interval (or of the line through it) measures its steepness as you walk from one point to the other. It is the rise (the vertical change) divided by the run (the horizontal change). For two points $P(x_1, y_1)$ and $Q(x_2, y_2)$,

The order does not matter as long as you subtract the coordinates the same way in the top and the bottom: swapping $P$ and $Q$ changes the sign of both the rise and the run, and the two sign changes cancel. A positive gradient slopes up to the right; a negative gradient slopes down to the right; a horizontal line has gradient $0$ (zero rise); and a vertical line has no gradient at all, because the run is zero and you cannot divide by zero.

The figure below shows the line through $(-2, -1)$ and $(2, 2)$. Walking from the first point to the second, the run is $4$ to the right and the rise is $3$ up, so the gradient is $\dfrac{3}{4}$. The right-angled triangle drawn under the line is the gradient triangle: its horizontal leg is the run, its vertical leg is the rise, and you can build it between any two points on the line and always get the same ratio.

Gradient as the tangent of the angle of inclination

There is a second, equivalent way to read a gradient. The angle of inclination $\theta$ is the angle the line makes with the positive $x$-axis, measured anticlockwise. Because the gradient triangle is a right-angled triangle with the rise opposite $\theta$ and the run adjacent to it,

So a gradient and an angle carry the same information. A line of gradient $1$ rises at $\tan^{-1} 1 = 45^\circ$; a steeper gradient gives a larger angle. A positive gradient gives an acute angle ($0^\circ < \theta < 90^\circ$); a negative gradient gives an obtuse angle ($90^\circ < \theta < 180^\circ$), because the line tilts back the other way. To go from a gradient to its angle you take $\tan^{-1}$ of the gradient; if that comes out negative (as it does for a downhill line, where the calculator returns a value between $-90^\circ$ and $0^\circ$), add $180^\circ$ to land in the correct $90^\circ$ to $180^\circ$ range.

The three forms of the equation of a line

NESA expects you to move fluently between three standard forms. Each is best for a different job.

• Gradient-intercept form $y = mx + b$. Here $m$ is the gradient and $b$ is the $y$-intercept (the value of $y$ when $x = 0$). This is the form to use when you can read the gradient and intercept directly, and the form most graphs are sketched from.
• Point-gradient form $y - y_1 = m(x - x_1)$. This builds the equation of the line with gradient $m$ that passes through a specific point $(x_1, y_1)$. It is the workhorse: whenever you know a gradient and one point, start here.
• General form $ax + by + c = 0$. Every term is on one side and equals zero, with integer coefficients where possible and the coefficient of $y$ non-zero for a genuine function. This is the tidy "final answer" form, and the one that also covers vertical lines, which the other two forms cannot (a vertical line $x = k$ has no gradient).

When a question does not specify a form, gradient-intercept or general form is acceptable; if you give general form, clear any fractions and divide out any common factor first. To convert $y = -\dfrac{3}{2}x + \dfrac{5}{2}$ to general form, multiply through by $2$ to get $2y = -3x + 5$, then collect: $3x + 2y - 5 = 0$.

The line through two given points

This is the most common equation question, and it is a clean two-step recipe. First find the gradient from the two points with the rise-over-run formula. Then substitute that gradient and either point into point-gradient form, and tidy up. It does not matter which of the two points you use in the second step; both give the same line, which is a useful self-check.

For the line through $A(-3, 7)$ and $B(5, 1)$: the gradient is $\dfrac{1 - 7}{5 - (-3)} = \dfrac{-6}{8} = -\dfrac{3}{4}$, and then through $A(-3, 7)$, point-gradient form gives $y - 7 = -\dfrac{3}{4}(x + 3)$, which tidies to $y = -\dfrac{3}{4}x + \dfrac{19}{4}$ or, in general form, $3x + 4y - 19 = 0$. Substituting $B(5, 1)$ into the general form gives $15 + 4 - 19 = 0$, confirming $B$ is on the line.

Parallel and perpendicular lines

Two non-vertical lines are parallel exactly when they have the same gradient. That is the whole test: equal gradients means the lines never meet (or are the same line). Any two vertical lines are also parallel.

Two lines are perpendicular exactly when the product of their gradients is $-1$:

So the gradient perpendicular to a given one is its negative reciprocal: flip the fraction and change the sign. The gradient perpendicular to $\dfrac{2}{3}$ is $-\dfrac{3}{2}$; the gradient perpendicular to $-4$ is $\dfrac{1}{4}$. The one special case the product rule cannot see is a horizontal line (gradient $0$) with a vertical line (no gradient): these are perpendicular, but you cannot multiply $0$ by an undefined gradient, so handle that pair by inspection.

The figure below shows a perpendicular pair: $y = \dfrac{1}{2}x + 1$ has gradient $\dfrac{1}{2}$ and $y = -2x + 6$ has gradient $-2$, and their product is $\dfrac{1}{2} \times (-2) = -1$. They cross at $(2, 2)$, and the small square marks the right angle between them.

This same idea tests whether three points are collinear (all on one line): find the gradient between the first pair and the gradient between the second pair, and if the two gradients are equal, the points lie on a single straight line.

Length and midpoint of an interval

The interval joining $P(x_1, y_1)$ and $Q(x_2, y_2)$ has a length and a midpoint, and both come from the run and the rise.

The length (distance) is the hypotenuse of the gradient triangle, so by Pythagoras,

The squares make the order of subtraction irrelevant. Where the answer is not a whole number, leave it as a simplified surd unless a decimal is asked for, for example $\sqrt{45} = 3\sqrt{5}$.

The midpoint is the point halfway along, found by averaging the coordinates:

The figure below joins $C(2, 3)$ to $D(8, 11)$. The dashed triangle has run $6$ and rise $8$, so the length is $\sqrt{6^2 + 8^2} = \sqrt{100} = 10$, and the midpoint is the average $(5, 7)$, marked on the interval.

Lines in real contexts

A straight line is the natural model whenever a quantity changes at a constant rate. Suppose a Sydney plumber charges a fixed callout fee of $90 plus $75 for each hour on site. The total cost $C$ for $h$ hours is $C = 90 + 75h$. This is a line in disguise: the gradient $75$ is the rate (the cost per extra hour) and the intercept $90$ is the starting value (the cost before any hours are billed). For a three-hour job, $C = 90 + 75 \times 3 = 315$, so the bill is $315. Reading a worded rate problem as $y = mx + b$, with the rate as the gradient and the fixed amount as the intercept, turns it straight into the algebra above.

Gradient as $\tan\theta$ also has a physical reading: it is the slope of a ramp or road. A wheelchair ramp that rises $0.5$ m over a horizontal run of $6$ m has gradient $\dfrac{0.5}{6} = \dfrac{1}{12}$, and its angle to the horizontal is $\tan^{-1}\left( \dfrac{1}{12} \right) \approx 4.8^\circ$, which is why accessibility standards quote ramp slopes as a ratio like $1 : 12$.

How exam questions ask about lines

The wording tells you which tool to reach for:

• "Find the gradient of the line / interval through ... and ..." is the rise-over-run formula. State it as a fraction in lowest terms unless told otherwise.
• "Find the equation of the line through the two points ... and ..." is the two-step recipe: gradient first, then point-gradient form, then tidy. Either point works in step two.
• "Find the equation of the line through ... parallel to / perpendicular to ..." Read off the given line's gradient. Parallel keeps it; perpendicular takes the negative reciprocal. Then use point-gradient form through the given point.
• "Show that ... is perpendicular to ..." Compute both gradients and show their product is $-1$. "Show that ... is parallel to ..." shows the gradients are equal.
• "Show that the three points ... are collinear." Show the gradient of one pair equals the gradient of another pair sharing a point.
• "Find the midpoint / the length / the distance between ... and ..." Average the coordinates for the midpoint; use the distance formula for the length, leaving a surd in exact form.
• "Find the angle of inclination / the angle the line makes with the $x$-axis." Find the gradient, then take $\tan^{-1}$; add $180^\circ$ if the gradient is negative so the angle is obtuse.
• "Find where the line cuts the axes / find the intercepts." Put $y = 0$ for the $x$-intercept and $x = 0$ for the $y$-intercept.
• "Find the equation of the perpendicular bisector of ..." Combine two skills: the midpoint of the interval and the perpendicular gradient, then point-gradient form through the midpoint.