Coordinates and graphs

Every time a delivery app drops a pin on a map, it's using the same idea you're about to learn: a pair of numbers that pins down an exact location. That's coordinate geometry , it's the bridge between algebra (equations) and geometry (shapes and space).

Analytic geometry (another name for coordinate geometry) lets you calculate things like distances, midpoints, and slopes using pure arithmetic, no ruler or protractor needed. Once you can describe a point with numbers, you can describe a line with an equation, and suddenly geometry becomes something you can solve.

The whole system runs on the Cartesian coordinate plane , a flat grid made by two number lines crossing at right angles. The horizontal one is the x-axis, the vertical one is the y-axis, and they meet at the origin, which sits at $(0,0)$. Every point on the plane gets a unique address written as an ordered pair $(x,y)$, where $x$ tells you how far left or right you are from the origin, and $y$ tells you how far up or down.

Here's something that surprises a lot of people: the formula for distance between two points is just Pythagoras in disguise. When you move from point $A$ to point $B$ on a grid, you're tracing the hypotenuse of a right triangle , the horizontal leg is the gap in x-values, and the vertical leg is the gap in y-values.

For two points $A(x_1,y_1)$ and $B(x_2,y_2)$, the horizontal change is $x_2-x_1$ and the vertical change is $y_2-y_1$. Pythagoras says the hypotenuse squared equals the sum of the two legs squared, so:

$AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

The squaring step is doing something clever: even if $x_2-x_1$ comes out negative (because you moved left), squaring it makes it positive. Distance can never be negative, and the squares guarantee that , so it genuinely doesn't matter which point you call $(x_1,y_1)$ and which you call $(x_2,y_2)$.

Sometimes you'll be asked to leave your answer as an exact surd like $\sqrt{n}$, and other times you'll round to a given number of significant figures. To simplify a surd, look for square factors inside the root. For example, $\sqrt{72}=\sqrt{36\times 2}=6\sqrt{2}$.