Analytic geometry

Introduction

Analytic geometry describes geometric objects, points, lines and curves, by equations in coordinates, so that geometric questions become algebra. This chapter covers the distance of a point from a line, the bisectors of the angles between two lines, pairs of straight lines given by a single second-degree equation, and two conics: the circle and the parabola, including their tangents and normals. The algebra is routine once the standard forms are memorised, so the priority is knowing which formula fits the situation.

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Perpendicular distance from a point to a line

The perpendicular distance from a point $(x_1, y_1)$ to the line $ax + by + c = 0$ is

$$d = \frac{|a x_1 + b y_1 + c|}{\sqrt{a^2 + b^2}}.$$

The numerator is the value of the left-hand side at the point (its sign tells you which side of the line the point lies on), and the modulus keeps the distance non-negative. Putting $(x_1, y_1) = (0,0)$ gives the distance of the line from the origin as $\dfrac{|c|}{\sqrt{a^2+b^2}}$. The distance between two parallel lines $ax+by+c_1 = 0$ and $ax+by+c_2 = 0$ is $\dfrac{|c_1 - c_2|}{\sqrt{a^2+b^2}}$.

Worked example. Find the distance from $(2,3)$ to $3x - 4y + 5 = 0$:

$$ \begin{aligned} d &= \frac{|3(2) - 4(3) + 5|}{\sqrt{3^2 + (-4)^2}} \ &= \frac{|6 - 12 + 5|}{\sqrt{25}} = \frac{|-1|}{5} = \frac15. \end{aligned} $$

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Bisectors of angles between two lines

Given two lines $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$, a point lies on an angle bisector exactly when it is equidistant from the two lines. Equating the perpendicular distances gives the pair of bisectors

$$\frac{a_1 x + b_1 y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm,\frac{a_2 x + b_2 y + c_2}{\sqrt{a_2^2 + b_2^2}}.$$

The two signs give the two bisectors, which are always perpendicular to each other. After writing both constants $c_1, c_2$ positive, the $+$ sign gives the bisector of the angle that contains the origin, and the $-$ sign gives the bisector of the other (supplementary) angle.

Worked example. Find the bisectors of the angles between $3x - 4y + 1 = 0$ and $5x + 12y - 2 = 0$. Since $\sqrt{3^2 + 4^2} = 5$ and $\sqrt{5^2 + 12^2} = 13$,

$$\frac{3x - 4y + 1}{5} = \pm,\frac{5x + 12y - 2}{13}.$$

Cross-multiplying, $13(3x - 4y + 1) = \pm,5(5x + 12y - 2)$. Taking the two signs in turn:

$$ \begin{aligned} 39x - 52y + 13 &= 25x + 60y - 10 &&\Rightarrow&& 14x - 112y + 23 = 0, \ 39x - 52y + 13 &= -(25x + 60y - 10) &&\Rightarrow&& 64x + 8y + 3 = 0. \end{aligned} $$

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