Coordinate geometry

1.3 Coordinate geometry

Straight lines

For the points $(x_1, y_1)$ and $(x_2, y_2)$:

$\text{gradient } m = \frac{y_2 - y_1}{x_2 - x_1} \qquad \text{midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \qquad \text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

A line can be written in three forms:

$m = \dfrac{8 - 2}{3 - 1} = 3$, so $y - 2 = 3(x - 1)$, which is $y = 3x - 1$.

Parallel and perpendicular lines

The midpoint is $(2, 5)$ and the gradient of the line is 3, so the perpendicular gradient is $-\tfrac{1}{3}$:

$y - 5 = -\frac{1}{3}(x - 2) \quad\Rightarrow\quad x + 3y = 17$

The equation of a circle

A circle with centre $(a, b)$ and radius $r$ has equation

$(x - a)^2 + (y - b)^2 = r^2$

This is Pythagoras' theorem for the distance from the centre to any point $(x, y)$ on the circle.

If the equation is expanded, complete the square in $x$ and in $y$ to get back to this form.

$(x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0 \quad\Rightarrow\quad (x - 3)^2 + (y + 2)^2 = 25$

The centre is $(3, -2)$ and the radius is 5.

Circle facts to use

These come from circle geometry and turn a circle question into a straight-line question.

Tangents to a circle

$3^2 + 4^2 = 25$, so the point is on the circle. The radius from $(3, -2)$ to $(6, 2)$ has gradient $\tfrac{4}{3}$, so the tangent has gradient $-\tfrac{3}{4}$:

$y - 2 = -\frac{3}{4}(x - 6) \quad\Rightarrow\quad 3x + 4y = 26$

Lines and circles

Substitute the line into the circle to get a quadratic. Its discriminant says how they meet: two points if positive, a tangent if zero, no meeting if negative.

$x^2 + (2x + k)^2 = 5 \quad\Rightarrow\quad 5x^2 + 4kx + k^2 - 5 = 0$

For a tangent, $16k^2 - 20(k^2 - 5) = 0$, so $100 - 4k^2 = 0$ and $k = \pm 5$.

Finding a circle from given facts

To write down a circle you need its centre and radius.

The centre is the midpoint, $(4, 7)$. The radius is the distance from the centre to $A$: $\sqrt{3^2 + 4^2} = 5$. So

$(x - 4)^2 + (y - 7)^2 = 25$

For a circle through three given points, find the perpendicular bisectors of two of the chords. They cross at the centre.

Points of intersection of graphs

The points where two graphs meet are the solutions of their equations taken together.

• Set the two expressions for $y$ equal to find the $x$-coordinates.
• A repeated root means the graphs touch rather than cross.
• A question may ask you to use a graph to solve an equation: the roots of $\mathrm{f}(x) = \mathrm{g}(x)$ are the $x$-coordinates where $y = \mathrm{f}(x)$ and $y = \mathrm{g}(x)$ cross.