- Coordinate geometry (or analytic geometry) connects algebra and geometry by describing points and shapes using coordinates and equations.
- It is a foundation for building and validating mathematical models, for example when mapping locations, designing layouts, or interpreting linear data.
Coordinates Describe Location Precisely
A point on a plane is located using an ordered pair $(x,y)$.
Cartesian coordinate plane
A grid formed by two perpendicular number lines, the x-axis (horizontal) and y-axis (vertical), used to locate points with ordered pairs $(x,y)$.
- The x-coordinate tells you how far left or right to move from the origin.
- The y-coordinate tells you how far down or up to move.
- The origin is $(0,0)$.
Because every point has a unique coordinate pair, coordinate geometry lets you calculate geometric facts (like lengths and slopes) with algebra.
Distance Between Two Points Comes From Pythagoras
- Suppose you have two points $A(x_1,y_1)$ and $B(x_2,y_2)$.
- If you move from $A$ to $B$, the horizontal change is $\Delta x=x_2-x_1$ and the vertical change is $\Delta y=y_2-y_1$.
- Using Pythagoras' theorem on the right triangle formed by these changes gives the distance formula.
Distance formula
For points $A(x_1,y_1)$ and $B(x_2,y_2)$, the distance between them is
$$AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$
Why Squaring Matters
- The differences $x_2-x_1$ and $y_2-y_1$ can be negative, but distance cannot be negative.
- Squaring removes sign, so the formula works no matter which point you subtract first.
- A very common error is to forget the squares and write $\sqrt{(x_2-x_1)+(y_2-y_1)}$.
- The squares are essential because the distance is based on Pythagoras.
Find the distance between $(0,0)$ and $(5,12)$.
Solution
Here, $\Delta x=5-0=5$ and $\Delta y=12-0=12$.
$$AB=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$$
Exact Form Versus Decimal Form
Sometimes you are asked to leave answers as an exact surd, such as $\sqrt{n}$, and sometimes to round to a certain number of significant figures.
- To keep an exact answer, simplify inside the square root only when there is a square factor.
- For example, $\sqrt{72}=\sqrt{36\cdot 2}=6\sqrt{2}$
The Midpoint Is the Average of Coordinates
The midpoint of a line segment is the point exactly halfway between the endpoints.
Midpoint formula
For endpoints $A(x_1,y_1)$ and $B(x_2,y_2)$, the midpoint is
$$M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$
This works because moving halfway in the $x$ direction means averaging the $x$-coordinates, and similarly for $y$.
Find the midpoint of $A(1,4)$ and $B(4,8)$.
Solution
$$M\left(\frac{1+4}{2},\frac{4+8}{2}\right)=\left(\frac{5}{2},6\right)$$
A Useful Midpoint Property
- If $M$ is the midpoint of $AB$, then $AM=\tfrac12 AB$.
- One way to validate this is to:
- compute $AB$ using the distance formula,
- compute $AM$ using the distance formula,
- show $AM=\tfrac12 AB$.
- This kind of check is an example of using logic to validate a generalization.
For $A(a,b)$ and $B(c,d)$:
- Write down the midpoint $M$.
- Use the distance formula to write expressions for $AB$ and $AM$.
- What changes when you compare $(c-a)$ to $\left(\tfrac{c+a}{2}-a\right)$?
Gradient Measures Steepness And Direction
A straight line's gradient (also called slope) tells how steep it is and whether it rises or falls as you move left to right.
Gradient (slope)
For two distinct points $A(x_1,y_1)$ and $B(x_2,y_2)$ on a line, the gradient is
$$m=\frac{y_2-y_1}{x_2-x_1},\quad x_2\neq x_1$$
- If $m>0$, the line rises left to right.
- If $m<0$, the line falls left to right.
- If $m=0$, the line is horizontal.
- If $x_2=x_1$, the line is vertical and the gradient is undefined.
- Don't swap the order in the numerator and denominator inconsistently.
- If you use $(y_2-y_1)$ then you must use $(x_2-x_1)$ with the same point order.
Find the gradient of the line through $A(0,0)$ and $B(1,4)$.
Solution
$$m=\frac{4-0}{1-0}=4$$
Parallel And Perpendicular Lines Have Predictable Gradients
Gradients let you recognize angle relationships between lines without measuring angles.
Parallel lines
Two non-vertical lines are parallel if they have the same gradient.
Perpendicular lines
Two non-vertical, non-horizontal lines are perpendicular if their gradients multiply to $-1$. If one line has gradient $m\,(m\neq 0)$, then a perpendicular line has gradient $-\frac{1}{m}$.
Special Cases: Horizontal And Vertical
A horizontal line has gradient $0$ and a vertical line has undefined gradient, so the "negative reciprocal" rule cannot be applied directly in that form.
- A horizontal line is perpendicular to a vertical line.
- Horizontal lines are parallel to each other.
- Vertical lines are parallel to each other.
- When checking perpendicularity, first ask: "Is one line horizontal and the other vertical?"
- If yes, they are perpendicular even though gradients do not multiply neatly.
Equations Of Straight Lines Can Be Written In Different Forms
- A line can be represented by an equation in several equivalent forms.
- Choosing an efficient form is part of good problem solving.
Slope-Intercept Form Highlights Gradient
Slope-intercept form
The equation of a non-vertical line can be written as $$y=mx+c$$ where $m$ is the gradient and $c$ is the $y$-intercept.
The y-intercept is where the line crosses the $y$-axis, at $(0,c)$.
Point-Slope Form Is Fast When You Know A Point And The Gradient
Point-slope form
A line through $(x_1,y_1)$ with gradient $m$ can be written as $$y-y_1=m(x-x_1)$$
This is often the quickest way to build an equation from a point and a gradient.
Standard Form Is Useful For Intercepts And Some Exam Requirements
Standard form (linear)
A line can be written as $$ax+by+c=0$$ with constants $a,b,c$ (often integers, and sometimes with a required sign convention such as $a>0$).
To find intercepts quickly from standard form:
- y-intercept: set $x=0$ and solve for $y$.
- x-intercept: set $y=0$ and solve for $x$.
Find where the line $2y-3x-7=0$ intersects the axes.
Solution
- On the $y$-axis, $x=0$: $$2y-7=0\Rightarrow y=\frac{7}{2}$$
- So the y-intercept is $(0,\tfrac72)$
- On the $x$-axis, $y=0$: $$-3x-7=0\Rightarrow x=-\frac{7}{3}$$
- So the x-intercept is $(-\tfrac73,0)$
If an answer is required "in the form $ax+by+c=0$ with integers", avoid fractions by multiplying through by the lowest common multiple of denominators, then rearrange so one side is 0.
Finding A Line Equation In Common Situations
Through Two Points
- Find the gradient $m=\dfrac{y2-y1}{x2-x1}$.
- Use point-slope form with either point.
- Convert to the required form (slope-intercept or standard).
Parallel To A Given Line
- Parallel lines have the same gradient.
- Use the given point to find the intercept.
The line $y=px+q$ is parallel to $y=2x-6$ and passes through $(-1,7)$. Find $p$ and $q$.
Solution
- Parallel means $p=2$.
- Substitute the point into $y=2x+q$: $$7=2(-1)+q=-2+q\Rightarrow q=9$$
- So $p=2$ and $q=9$, and the line is $y=2x+9$
Perpendicular To A Given Line
- Find the gradient of the given line.
- Take the negative reciprocal to get the perpendicular gradient.
- Use point-slope form.
Find the equation of the line through $S(3,7)$ perpendicular to $l_2: 3x+2y+2=0$, and give it in the form $ax+by+c=0$ with $a>0$.
Solution
- First rewrite $l_2$ to see its gradient: $$3x+2y+2=0\Rightarrow 2y=-3x-2\Rightarrow y=-\frac{3}{2}x-1$$
- So $m_2=-\tfrac32$
- A perpendicular line has gradient $$m_1=-\frac{1}{m_2}=-\frac{1}{-\tfrac32}=\frac{2}{3}$$
- Use point-slope form through $(3,7)$: $$y-7=\frac{2}{3}(x-3)$$
- Expand: $$y-7=\frac{2}{3}x-2\Rightarrow y=\frac{2}{3}x+5$$
- Convert to standard form with integers: $$3y=2x+15\Rightarrow 2x-3y+15=0$$
Coordinate Geometry As A Modelling Tool
- Coordinate methods are especially useful for deciding whether a relationship is linear and interpreting what a linear model means.
- For data points $(n,C)$ (for example, number of units used $n$ and cost $C$), you can:
- plot points and check if they lie close to a straight line,
- compute gradients between pairs to see if the rate of change is roughly constant,
- use a linear equation $C=mn+c$ as a model.
- Interpretation is essential:
- the gradient $m$ represents cost per unit (rate of change),
- the y-intercept $c$ represents the fixed cost when $n=0$ (if $n=0$ is meaningful in context).
- A "treasure map" is a fun way to see modelling in action: if each grid unit represents one league, then distance and midpoint calculations become real navigation decisions.
- However, the model's validity depends on assumptions, for example whether the map scale is consistent and whether the terrain allows straight-line travel.
Rotations About The Origin Change Coordinates In Predictable Ways
- Coordinate geometry also supports geometric transformations.
- For rotations about the origin, there are simple coordinate rules.
Rotations about the origin:
For a point $P(a,b)$:
- $90^\circ$ clockwise: $P'(b,-a)$
- $90^\circ$ counterclockwise: $P'(-b,a)$
- $180^\circ$: $P'(-a,-b)$
These rules are useful for quickly finding images of points without drawing every time.
Be careful with the $180^\circ$ rule: both coordinates change sign, so $(a,b)$ becomes $(-a,-b)$.
Core toolkit of coordinate geometry:
- Distance: $\sqrt{(x2-x1)^2+(y2-y1)^2}$
- Midpoint: $\left(\tfrac{x1+x2}{2},\tfrac{y1+y2}{2}\right)$
- Gradient: $\tfrac{y2-y1}{x2-x1}$
- Parallel lines: same gradient
- Perpendicular lines: gradients multiply to $-1$ (negative reciprocals)
- Line equations: switch flexibly between $y=mx+c$, $y-y1=m(x-x1)$, and $ax+by+c=0$
