Equation of a line and area of a triangle

The equation of a line

A line is completely fixed once you know one point on it and its slope. The slope $m$ tells you how steep the line is. If $(x_1, y_1)$ is a known point and $(x, y)$ is any other point on the line, then the slope between them is $m$:

Multiplying across gives the form you will find in the Formulae and Tables booklet.

If you are given two points $(x_1, y_1)$ and $(x_2, y_2)$, first find the slope, then use either point:

You can leave the answer in any correct form: $y = mx + c$ or $ax + by + c = 0$. In the second form the numbers $a$, $b$, $c$ are usually whole numbers.

Use the sliders to see how $m$ and $c$ move the line $y = mx + c$. The number $c$ is the $y$-intercept, where the line cuts the $y$-axis.

Intercepts, parallel and perpendicular lines

Where does a line cut the axes?

• On the $x$-axis, $y = 0$. Put $y = 0$ into the equation and solve for $x$.

• On the $y$-axis, $x = 0$. Put $x = 0$ into the equation and solve for $y$.

Slope from $ax + by + c = 0$

Rearrange $ax + by + c = 0$ to $y = -\dfrac{a}{b}x - \dfrac{c}{b}$. So

Parallel and perpendicular

• Parallel lines have equal slopes: $m_1 = m_2$.

• Perpendicular lines have slopes whose product is $-1$: $m_1 m_2 = -1$. To get the perpendicular slope, turn the fraction upside down and change the sign.

Two lines with different slopes are intersecting lines and meet at exactly one point. Two lines with the same slope are parallel (no common point) unless they are the same line, in which case they are coincident (all points in common). To find the point where two lines meet, solve their equations as simultaneous equations.

Perpendicular bisector of a line segment

The perpendicular bisector of $[AB]$ is the line that passes through the midpoint of $[AB]$ and is perpendicular to it. The midpoint is $\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$.

Lines as models: fixed charge plus rate

Many real situations have a fixed charge plus a rate for each unit. The graph is a straight line. The slope is the rate, and the $y$-intercept is the fixed charge.

Area of a triangle

Suppose one vertex of the triangle is at the origin $O(0, 0)$ and the other two are $P(x_1, y_1)$ and $Q(x_2, y_2)$.

For a triangle with vertices anywhere, translate it so that one vertex lands on the origin. Moving a shape without turning it does not change its area. Subtract the coordinates of the chosen vertex from the other two.

Height from the area

If you know the area and the length of one side, you can find the perpendicular distance from the opposite vertex to that side. Use $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ and rearrange for the height, with the base from the distance formula $|AB| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Collinear points and triangles that touch the axes

Three points are collinear (lie on one line) exactly when the triangle they form has area $0$. Equivalently, the slope from the first to the second equals the slope from the first to the third.

When you are told the area, the absolute value gives two cases: $x_1 y_2 - x_2 y_1$ could be $+$ the number or $-$ the number. That is how you get two answers for an unknown coordinate.

A triangle formed by a line and the two axes has a simple area: if the line cuts the axes at $P(p, 0)$ and $Q(0, q)$, then triangle $OPQ$ has area $\frac{1}{2}|pq|$ (a right-angled triangle whose other two sides have lengths $|p|$ and $|q|$; remember that $p$ and $q$ can have opposite signs). The slope of the line $PQ$ is $-\frac{q}{p}$.

Back to the bike hire: the line was $C = 0.2t + 3$, so a 25-minute hire costs €8.

Summary

• The line through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$. Through two points, find $m = \frac{y_2 - y_1}{x_2 - x_1}$ first.

• Put $y = 0$ to find where a line cuts the $x$-axis, and $x = 0$ for the $y$-axis. For $ax + by + c = 0$, the slope is $-\frac{a}{b}$.

• Parallel lines have equal slopes. Perpendicular lines have $m_1 m_2 = -1$.

• In a fixed-charge-plus-rate model, the slope is the rate and the $y$-intercept is the fixed charge.

• A triangle with a vertex at the origin has area $\frac{1}{2}|x_1 y_2 - x_2 y_1|$. For any triangle, translate one vertex to the origin first. Area $0$ means the points are collinear.