Fundamentals of Analytic Geometry in the Plane

Analytic geometry turns geometric statements into algebra: a point becomes a pair of numbers, a line becomes an equation, and "do these two lines meet" becomes "solve this system." Every result in this chapter is a tool you will keep using for the rest of the course — the moment vectors appear in the next chapter, they are built on exactly the coordinate machinery developed here.

To follow the chapter you need the Cartesian plane, Pythagoras' theorem, and basic algebraic manipulation of linear equations. Nothing here is new content so much as a new, more powerful language for content you already know.

Coordinates fix a correspondence. The Cartesian plane assigns to every point $P$ an ordered pair $(x, y)$ — the signed distances from $P$ to two perpendicular axes. Once this correspondence is fixed, every geometric question about points and lines becomes a question about numbers.

1. Distance Between Two Points

Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be two points. Drop a horizontal line through $A$ and a vertical line through $B$; they meet at $C(x_2, y_1)$, forming a right triangle with legs $|AC| = |x_2 - x_1|$ and $|CB| = |y_2 - y_1|$ and hypotenuse $AB$. Pythagoras' theorem gives directly:

Worked example. Find the distance between $A(-1, 2)$ and $B(4, -3)$.

2. The Midpoint and the Section Formula

The midpoint $M$ of segment $AB$ has coordinates that are simply the averages of the endpoints:

This is a special case of a more general fact. If $P$ divides $AB$ in the ratio $AP:PB = m:n$ (measured along the segment from $A$ to $B$), then

which reduces to the midpoint formula when $m = n = 1$. The derivation follows from similar triangles: the horizontal run from $A$ to $P$ is the same fraction $\frac{m}{m+n}$ of the total run from $A$ to $B$ as the vertical rise is, so $x_P = x_1 + \frac{m}{m+n}(x_2 - x_1)$, which rearranges to the formula above (and likewise for $y_P$).

Worked example. Point $P$ divides $A(1, 1)$ to $B(13, 7)$ in the ratio $2:1$. Then

3. Gradient (Slope) of a Line

The gradient $m$ of the line through $A(x_1, y_1)$ and $B(x_2, y_2)$ measures steepness as a ratio of rise to run:

The gradient is independent of which two points on the line are chosen — any pair gives the same ratio, since all such pairs form similar right triangles. If the line makes angle $\theta$ with the positive $x$-axis, then $m = \tan\theta$, which follows directly from the right-triangle definition of tangent applied to the rise-over-run triangle. A horizontal line has $m = 0$; a vertical line has undefined gradient, since $x_2 - x_1 = 0$.

Steepness is not the same as angle. Doubling the gradient does not double the angle of inclination, because $\tan\theta$ is not linear in $\theta$. A gradient of $1$ corresponds to $45°$, but a gradient of $2$ corresponds to about $63.4°$, not $90°$.

4. Equations of a Straight Line

A line is the set of all points $(x,y)$ satisfying one linear equation. Three equivalent forms are used throughout the course:

• Point-gradient form, built directly from the gradient definition applied to a fixed point $(x_1, y_1)$ and a variable point $(x,y)$ on the line:

• Slope-intercept form, obtained by expanding and solving for $y$, where $c$ is the $y$-intercept:

• General form, useful when the gradient is undefined (vertical lines) or when symmetry between $x$ and $y$ is wanted:

Worked example. Find the equation of the line through $A(2, -1)$ and $B(6, 7)$.

First, $m = \dfrac{7-(-1)}{6-2} = \dfrac{8}{4} = 2$. Using point-gradient form with $A$:

5. Parallel and Perpendicular Lines

Two non-vertical lines with gradients $m_1$ and $m_2$ are parallel exactly when $m_1 = m_2$ — equal steepness means equal direction. They are perpendicular exactly when

To see why, rotate a line of gradient $m_1 = \tan\theta$ by $90°$; the new angle is $\theta + 90°$, and using the identity $\tan(\theta+90°) = -\dfrac{1}{\tan\theta}$, the new gradient is $-\dfrac{1}{m_1}$. So perpendicular gradients are negative reciprocals of one another.

Worked example — the harder case. Find the equation of the line through $C(3,4)$ perpendicular to the line $2x + 3y - 6 = 0$.

Rewriting the given line in slope-intercept form: $y = -\dfrac{2}{3}x + 2$, so $m_1 = -\dfrac{2}{3}$. The perpendicular gradient satisfies $m_1 m_2 = -1$, so $m_2 = \dfrac{3}{2}$. Through $C(3,4)$:

This case exposes the edge a careless student misses: forgetting to convert the given equation to slope-intercept form first, and instead reading off $m_1 = 2$ or $m_1 = 3$ directly from the general-form coefficients — the general form's coefficients are not the gradient.

6. Intersection of Two Lines

Two non-parallel lines meet at exactly one point, found by solving their equations simultaneously.

Worked example. Find the intersection of $y = 2x - 5$ and $3x - 2y - 1 = 0$ (the two lines from the previous examples).

Substitute the first into the second: $3x - 2(2x-5) - 1 = 0 \implies 3x - 4x + 10 - 1 = 0 \implies -x + 9 = 0 \implies x = 9$. Then $y = 2(9) - 5 = 13$. The lines meet at $(9, 13)$ — as expected, since their gradients ($2$ and $\tfrac32$) are unequal, so the lines are non-parallel and must cross at exactly one point (note: this gradient product is $2\times\tfrac32=3\neq-1$, so the two lines are not perpendicular — they simply happen to share a worked-example equation with the previous section).

7. The Angle Between Two Lines

If two lines have gradients $m_1$ and $m_2$ and meet at angle $\alpha$ (taken as the acute angle between them), the angle each makes with the $x$-axis gives, via the tangent subtraction formula,

The modulus is needed because swapping which line is "first" would otherwise flip the sign, while the acute angle between two lines is a single, well-defined, positive quantity.

Worked example. Find the acute angle between lines of gradient $m_1 = 3$ and $m_2 = -1$.

Common Exam Pitfalls

• Reading the gradient off the general form directly. In $ax+by+d=0$ the gradient is $-\dfrac{a}{b}$, not $a$ or $b$ — always rearrange to $y=mx+c$ first, or use $-a/b$ explicitly.
• Mixing up the ratio direction in the section formula. $AP:PB=m:n$ puts $m$ with point $B$'s coordinates and $n$ with point $A$'s in the weighted average — reversing them gives the point that divides $BA$ instead.
• Forgetting the vertical-line exception. A vertical line $x=k$ has no gradient (division by zero), and no amount of algebra on $y=mx+c$ can represent it; it must be written in general or point-coordinate form.
• Assuming equal gradient means equal line. Two lines can have $m_1=m_2$ and still be distinct, parallel lines — equality of gradient is necessary for coincidence, not sufficient; also check the intercept or a point.
• Sign errors in the perpendicularity condition. Perpendicular gradients are negative reciprocals ($m_1 m_2 = -1$), not simply reciprocals ($m_1 = 1/m_2$) — a common slip under exam pressure.
• Treating the angle-between-lines formula as signed. Without the absolute value bars, the raw expression can return a negative tangent corresponding to the obtuse angle; the acute angle is always the one intended unless stated otherwise.

Summary

• Distance: $|AB| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$, from Pythagoras.
• Midpoint: $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$; section formula for ratio $m:n$: $P = \left(\frac{nx_1+mx_2}{m+n}, \frac{ny_1+my_2}{m+n}\right)$.
• Gradient: $m = \dfrac{y_2-y_1}{x_2-x_1} = \tan\theta$, undefined for vertical lines.
• Line forms: point-gradient $y-y_1=m(x-x_1)$, slope-intercept $y=mx+c$, general $ax+by+d=0$ (gradient $-a/b$).
• Parallel: $m_1=m_2$. Perpendicular: $m_1 m_2 = -1$.
• Intersection of two lines: solve the two equations simultaneously.
• Angle between two lines: $\tan\alpha = \left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$. </content>