A point on a grid is written as an ordered pair $(x, y)$: the $x$-coordinate tells you how far to move horizontally from the origin $O(0,0)$, and the $y$-coordinate tells you how far to move vertically. Right and up are positive; left and down are negative.

The two axes divide the grid into four quadrants, numbered anticlockwise starting from the top-right. The signs of $x$ and $y$ tell you instantly which quadrant a point sits in, without needing to plot it:

• Quadrant 1 (top-right): $x > 0, \ y > 0$
• Quadrant 2 (top-left): $x < 0, \ y > 0$
• Quadrant 3 (bottom-left): $x < 0, \ y < 0$
• Quadrant 4 (bottom-right): $x > 0, \ y < 0$

A point sitting exactly on an axis (e.g. $(0, 5)$ or $(-3, 0)$) is not considered to be inside any quadrant — it's a boundary case worth recognising in a "describe the position" question.

You won't always be given a grid. If two points are $A(x_1, y_1)$ and $B(x_2, y_2)$, you can locate any point on segment $AB$ using the same left-right, up-down logic algebraically — this is the reasoning that later underpins the midpoint formula in E3.4. For now, treat coordinates as pure numbers you can add, subtract and average, not just marks on paper.

Every topic from here on (gradient, length, midpoint, equations of lines) is just arithmetic performed on coordinate pairs. If you can substitute correctly into $(x_2 - x_1)$ and $(y_2 - y_1)$, you have the one skill the rest of coordinate geometry is built from. Errors on later, harder topics almost always trace back to this stage — subtracting coordinates in the wrong order, or swapping $x$ and $y$ midway through a calculation.

A quadrilateral has vertices $A(-2, 1)$, $B(3, 1)$, $C(3, -4)$ and $D(-2, -4)$. Identify the shape and state its area.

1. Plot the points mentally: $A$ and $B$ share $y = 1$, so $AB$ is horizontal. $C$ and $D$ share $y = -4$, so $CD$ is horizontal. $A$ and $D$ share $x = -2$; $B$ and $C$ share $x = 3$. All four sides are horizontal or vertical, meeting at right angles — the shape is a rectangle.
2. Find the side lengths by subtracting coordinates: width $= 3 - (-2) = 5$, height $= 1 - (-4) = 5$.
3. Since width = height, it's actually a square of side $5$.
4. Area $= 5^2 = 25$ square units.

A point $P$ lies 4 units to the left of the $y$-axis and 7 units below the $x$-axis. State the coordinates of $P$ and the quadrant it lies in.

1. "4 units to the left" gives $x = -4$. "7 units below" gives $y = -7$. So $P = (-4, -7)$.
2. Both coordinates are negative, so $P$ lies in Quadrant 3.