Topic 3: Coordinate geometry

The coordinate plane is one of the most powerful ideas in IGCSE Mathematics because it turns geometry into algebra. Once a point becomes a pair of numbers $(x, y)$, you can measure distances, find midpoints, and describe whole lines with a single equation. Topic 3 builds a toolkit you will reuse constantly: gradient tells you how steep a line is, the equation $y=mx+c$ packages a line into two numbers, and the parallel and perpendicular rules let you compare lines instantly. The skills here also feed directly into graphs of functions, transformations, and even calculus later on. Work through each section, and pay close attention to signs, because a single sign slip is the most common way to lose marks in this topic.

The Cartesian plane and plotting points

Every point on the plane is given by an ordered pair $(x, y)$, called its coordinates. The first number, $x$, is the horizontal position measured from the origin along the $x$-axis; the second number, $y$, is the vertical position measured along the $y$-axis. The origin itself is $(0, 0)$, where the two axes cross. The order matters: $(3, 5)$ is not the same point as $(5, 3)$. A handy memory aid is 'along the corridor, then up the stairs' meaning you move horizontally first, then vertically. Positive $x$ goes right, negative $x$ goes left; positive $y$ goes up, negative $y$ goes down. The axes divide the plane into four quadrants. In the first quadrant both coordinates are positive; in the second $x$ is negative and $y$ is positive; in the third both are negative; in the fourth $x$ is positive and $y$ is negative. Knowing the quadrant is a quick way to check a plotted point looks sensible. When plotting, always read the scale on each axis carefully, since the $x$ and $y$ scales are not always the same.

Gradient from two points

The gradient (or slope) measures how steep a line is. It is defined as the change in $y$ divided by the change in $x$, often remembered as 'rise over run'. For two points $(x_1, y_1)$ and $(x_2, y_2)$ the gradient is gradient $=\\frac{y_2-y_1}{x_2-x_1}$. For example, the gradient of the line through $(1, 2)$ and $(4, 11)$ is $\\frac{11-2}{4-1}=\\frac{9}{3}=3$. This means the line rises 3 units for every 1 unit moved to the right. A positive gradient means the line slopes upward from left to right; a negative gradient slopes downward. A horizontal line has gradient $0$ because $y$ never changes. A vertical line has an undefined gradient because the run, $x_2-x_1$, is zero and you cannot divide by zero. The most common mistake is subtracting the coordinates in a different order on the top and bottom. Always subtract in the same order: if you start with the second point's $y$ on top, start with the second point's $x$ on the bottom. Both orders, $\\frac{y_2-y_1}{x_2-x_1}$ and $\\frac{y_1-y_2}{x_1-x_2}$, give the same answer because the signs cancel.

Length of a line segment

The distance between two points is found using Pythagoras' theorem. The horizontal gap is $x_2-x_1$ and the vertical gap is $y_2-y_1$; these form the two shorter sides of a right-angled triangle, and the segment itself is the hypotenuse. So the length is length $=\\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. For example, the distance between $(1, 2)$ and $(4, 6)$ is $\\sqrt{(4-1)^2+(6-2)^2}=\\sqrt{3^2+4^2}=\\sqrt{9+16}=\\sqrt{25}=5$. Because each difference is squared, the sign of the difference does not matter, so you never have to worry about which point you call first. Just be sure to square each gap fully before adding. A frequent error is writing $\\sqrt{(x_2-x_1)^2}+\\sqrt{(y_2-y_1)^2}$, which is wrong; the square root applies to the whole sum, not to each term separately. If the answer is not a perfect square, leave it as a surd such as $\\sqrt{20}$ or simplify to $2\\sqrt{5}$, or give a rounded decimal if the question asks for it.

Midpoint of a line segment

The midpoint is the point exactly halfway between two endpoints. You find it by averaging the $x$-coordinates and averaging the $y$-coordinates separately: midpoint $=\\left(\\frac{x_1+x_2}{2}, \\frac{y_1+y_2}{2}\\right)$. For example, the midpoint of $(2, 3)$ and $(8, 7)$ is $\\left(\\frac{2+8}{2}, \\frac{3+7}{2}\\right)=(5, 5)$. Notice the contrast with gradient and length: for the midpoint you add the coordinates, whereas for gradient and length you subtract them. Mixing these up is a very common slip, so pause and ask yourself which operation the question needs. Midpoints appear often in problems about the centre of a circle when you know the ends of a diameter, or when you need the point where the diagonals of a parallelogram cross. A useful reverse skill is finding one endpoint when given the midpoint and the other endpoint: set the midpoint formula equal to the known midpoint and solve for the unknown coordinates.

The equation of a straight line: y = mx + c

Every non-vertical straight line can be written in the form $y=mx+c$, where $m$ is the gradient and $c$ is the $y$-intercept, the value of $y$ where the line crosses the $y$-axis (that is, where $x=0$). For example, in $y=2x+3$ the gradient is $2$ and the line crosses the $y$-axis at $(0, 3)$. In $y=-\\frac{1}{2}x+4$ the gradient is $-\\frac{1}{2}$ and the intercept is $4$. To read $m$ and $c$ from an equation, first make sure it is rearranged into the form $y=mx+c$ with $y$ alone on the left. For instance, $2y=6x+8$ is not yet in this form; dividing by $2$ gives $y=3x+4$, so $m=3$ and $c=4$. Likewise $x+y=5$ rearranges to $y=-x+5$, giving gradient $-1$ and intercept $5$. Given the equation, you can plot the line quickly: mark the intercept on the $y$-axis, then use the gradient as rise over run to step to a second point, and draw the line through both. Horizontal lines have the special form $y=c$ and vertical lines have the form $x=k$, where $k$ is a constant.

Parallel and perpendicular gradients

Two lines are parallel when they have exactly the same gradient. So $y=3x+1$ and $y=3x-4$ are parallel because both have $m=3$; they never meet, no matter how far they are extended. The intercepts differ, which simply shifts one line above the other. Two lines are perpendicular when they cross at a right angle. This happens when the product of their gradients is $-1$. In other words, if one line has gradient $m$, a line perpendicular to it has gradient $-\\frac{1}{m}$, the negative reciprocal. To take a negative reciprocal you flip the fraction and change the sign. For example, a line perpendicular to $y=2x+5$ has gradient $-\\frac{1}{2}$, because $2 \\times -\\frac{1}{2}=-1$. A line perpendicular to one with gradient $-\\frac{3}{4}$ has gradient $\\frac{4}{3}$. A special case: a horizontal line (gradient $0$) is perpendicular to a vertical line (undefined gradient); the negative reciprocal rule does not apply directly here because you cannot divide by zero, so just remember they meet at $90$ degrees.

Finding the equation of a line

There are two common situations. First, suppose you know the gradient $m$ and one point $(x_1, y_1)$ on the line. Substitute the gradient into $y=mx+c$, then substitute the point's coordinates to find $c$. For example, a line with gradient $2$ passing through $(3, 11)$ gives $11=2(3)+c$, so $c=5$ and the equation is $y=2x+5$. Second, suppose you know two points. First find the gradient using $\\frac{y_2-y_1}{x_2-x_1}$, then use either point to find $c$ as above. For the points $(1, 4)$ and $(3, 10)$, the gradient is $\\frac{10-4}{3-1}=3$. Substituting $(1, 4)$ gives $4=3(1)+c$, so $c=1$ and the equation is $y=3x+1$. For parallel or perpendicular problems, work out the required gradient first using the rules from the previous section, then proceed exactly as above with the given point. For instance, the line through $(2, 7)$ parallel to $y=4x-1$ has gradient $4$, so $7=4(2)+c$ gives $c=-1$ and the equation is $y=4x-1$. Always present your final answer in the clean form $y=mx+c$ unless the question asks otherwise.