Analytic Geometry
Cheat sheet.

The address system

$\mathbb{R} \times \mathbb{R} = \mathbb{R}^2$: every point is one ordered pair $P(x, y)$. $x$ = abscissa (horizontal), $y$ = ordinate (vertical). Order matters.

Distances and locations

Sign quick-read

$x > 0$ right, $x < 0$ left; $y > 0$ above, $y < 0$ below.

Quadrant sign table

Problem patterns

• Known quadrant → known signs of $x$ and $y$ → sign of any product or sum by sign rules.
• "Point in quadrant N" → two inequalities → intersect the solution sets.
• "Point on $y$-axis" → set abscissa $= 0$; "on $x$-axis" → set ordinate $= 0$.

The formula

$|AB| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ - Pythagoras with legs $|x_2 - x_1|$ and $|y_2 - y_1|$. Subtraction order never matters.

Special cases

Unknown coordinate procedure

Set formula equal to the given distance, square both sides, isolate the squared binomial, take $\pm$ roots - expect two answers. Classic integer triangles to recognize: 3-4-5, 5-12-13, 8-15-17.

The formula

$M\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$ - averages of the coordinates.

Problem patterns

The definition

$m = \dfrac{\text{rise}}{\text{run}}$ - vertical change over horizontal change, between ANY two points of the line.

Reading the number

The key fact

A straight line has ONE slope everywhere - any two measuring triangles are similar, so rise/run always reduces to the same value.

From two points

$m=\dfrac{y_2-y_1}{x_2-x_1}$. Keep the subtraction order consistent.

From an equation

Rewrite as $y=mx+b$, or use $m=-\dfrac{a}{b}$ for $ax+by+c=0$.

From inclination

$m=\tan(\alpha)$, where $\alpha$ is measured counterclockwise from the positive $x$-axis.

Special lines

Horizontal lines have slope $0$. Vertical lines have undefined slope.

The one formula

$y - y_0 = m(x - x_0)$ - slope $m$, any known point $(x_0, y_0)$. Standard form: $ax + by + c = 0$. Slope-intercept: $y = mx + n$.

Finding $m$ first

Traps

The two families

Reading the wording

Traps

Intercept form

$\dfrac{x}{a} + \dfrac{y}{b} = 1$ - where $a$ is the $x$-intercept and $b$ the $y$-intercept.

Moves

When it does not apply

The test

$P(x_0, y_0)$ is on $ax + by + c = 0$ when $ax_0 + by_0 + c = 0$. Substitute; zero means yes.

Three uses of one substitution

Notes

The idea

The crossing point satisfies both equations. Solve the system; check the answer in both.

Methods

What the collapse means

Two points, then a ruler

Checks

The three cases

What solving reveals

Trap

Compare ratios in a fixed order, always the same line on top. $\dfrac{a_1}{a_2}$ against $\dfrac{b_2}{b_1}$ is the classic slip.

The condition

$m_1 = m_2$ with $n_1 \ne n_2$ - equal slopes, different intercepts. In standard form: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$.

Building a parallel through a point

Shortcut

Parallel to $ax + by + c = 0$ is $ax + by + k = 0$: keep both coefficients, substitute the point for $k$.

Note

Two vertical lines are parallel although neither has a slope - the ratio test covers them, the slope test does not.

The condition

$m_1 \cdot m_2 = -1$, that is $m_2 = -\dfrac{1}{m_1}$ - the negative reciprocal. In standard form: $a_1a_2 + b_1b_2 = 0$.

Flipping slopes

Shortcut

Perpendicular to $ax + by + c = 0$ is $bx - ay + k = 0$: swap the coefficients, negate one, substitute the point for $k$.

Trap

Horizontal and vertical lines are perpendicular, but the product rule cannot show it - $0 \times$ undefined means nothing. Use the coefficient form.

The condition

$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ - all three ratios agree, so one equation is a multiple of the other.

Fast test

Divide the first coefficients for the multiplier, then check it carries the other two terms. Failing on the constant alone means parallel.

In a system

Infinitely many solutions means coincident: the second equation adds no information, so no unique point can be found.