2D Coordinate Geometry

IB Mathematics: Analysis and Approaches HL· Unit 3: Geometry & Trigonometry· 15 min read

1. Distance and Midpoint Formulas★☆☆☆☆⏱ 5 min

Any point on the 2D coordinate plane is described by an ordered pair $(x, y)$ , where $x$ is the horizontal coordinate and $y$ is the vertical coordinate. Using Pythagoras' theorem, we can derive the distance between any two points.

📘 Definition

Distance and Midpoint

For points $A (x_{1}, y_{1})$ , $B (x_{2}, y_{2})$

Distance: $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ Midpoint: $M (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$

Example:

Distance between $(1, 2)$ and $(4, 6)$ is 5 units, midpoint is $(2.5, 4)$

📐 Worked Example

Find the distance between $A (- 2, 3)$ and $B (4, - 5)$ , then find the midpoint $M$ of segment $A B$ .

1
Label coordinates: $x_{1} = - 2, y_{1} = 3, x_{2} = 4, y_{2} = - 5$
2
Substitute into the distance formula:
3
$d = \sqrt{(4 - (-2))^2 + (-5 - 3)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{100} = 10$
4
Substitute into the midpoint formula:
5
$M\left(\frac{-2 + 4}{2}, \frac{3 + (-5)}{2}\right) = M(1, -1)$
6
Final result: distance is 10 units, midpoint is $(1, - 1)$

Exam tip:

Always double check negative signs when subtracting coordinates, this is the most common error in this topic.

2. Straight Line Equations★★☆☆☆⏱ 6 min

A straight line is a set of points that satisfy a linear relationship between $x$ and $y$ . Its gradient (slope) describes steepness, and lines can be written in three common forms, all accepted in IB exams unless specified otherwise.

📘 Definition

Parallel and Perpendicular Lines

Parallel lines never intersect and have equal gradients: $m_{1} = m_{2}$ . Perpendicular lines intersect at $9 0^{\circ}$ and their gradients multiply to $- 1$ : $m_{1} m_{2} = - 1$ .

Common line forms: 1. Gradient-intercept: $y = m x + c$ ( $c$ = y-intercept), 2. Point-gradient: $y - y_{1} = m (x - x_{1})$ , 3. General: $a x + b y + c = 0$ .

📐 Worked Example

Find the equation of the line that passes through $(3, - 2)$ and is perpendicular to $2 x + 4 y - 5 = 0$ . Write your answer in general form.

1
Rearrange the given line to find its gradient:
2
$2x + 4y - 5 = 0 \implies y = -\frac{1}{2}x + \frac{5}{4}$
3
Gradient of given line is $- \frac{1}{2}$ . For perpendicular lines, find the required gradient:
4
$-\frac{1}{2}m = -1 \implies m = 2$
5
Use point-gradient form with the given point:
6
$y - (-2) = 2(x - 3) \implies y + 2 = 2x - 6$
7
Rearrange to general form:
8
$2x - y - 8 = 0$

Exam tip:

If the question asks for a specific form (e.g. general form), always give your answer in that form to earn full marks.

3. Intersections and Angle Between Two Lines★★★☆☆⏱ 7 min

Two distinct lines are either parallel (no intersection) or intersect at exactly one point. The intersection point satisfies both line equations, so we solve the system of linear equations to find its coordinates.

The acute angle $θ$ between two lines with gradients $m_{1}$ and $m_{2}$ is calculated using the formula derived from the tangent compound angle identity:

\tan\theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|, \quad 0^\circ \leq \theta \leq 90^\circ

📐 Worked Example

Find the acute angle between the lines $y = 3 x + 2$ and $y = - x + 5$ .

1
Identify the gradients from the equations: $m_{1} = 3$ , $m_{2} = - 1$
2
Substitute into the angle formula:
3
$\tan\theta = \left|\frac{-1 - 3}{1 + (3)(-1)}\right| = \left|\frac{-4}{-2}\right| = 2$
4
Calculate $θ$ using arctangent:
5
$\theta = \arctan(2) \approx 63.4^\circ$
6
This is already acute, so it is the required angle.

✓ Quick check

Test your understanding:

What is the angle between two perpendicular lines?
- 30°
- 45°
- 90°
- 180°
Reveal answer
2 —
Perpendicular lines are defined as lines that intersect at a right angle (90°), which is consistent with the angle formula that gives undefined $tan θ$ , corresponding to 90°.

4. Equations of Circles★★★☆☆⏱ 6 min

A circle is defined as all points $(x, y)$ that are a fixed distance (radius $r$ ) from a fixed center point $(h, k)$ . We can derive the standard equation from the distance formula directly.

📘 Definition

Circle Equation Forms

Center $(h, k)$ , radius $r$

Standard form: $(x - h)^{2} + (y - k)^{2} = r^{2}$ . General form: $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , where center is $(- g, - f)$ and radius is $g^{2} + f^{2} - c$ .

📐 Worked Example

Find the center and radius of the circle with equation $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$ .

1
Group $x$ and $y$ terms, move the constant to the right-hand side:
2
$(x^2 - 4x) + (y^2 + 6y) = 12$
3
Complete the square for both variables:
4
$(x - 2)^2 - 4 + (y + 3)^2 - 9 = 12$
5
Simplify to get standard form:
6
$(x - 2)^2 + (y + 3)^2 = 25$
7
Compare to standard form to get center and radius:
8
Center = $(2, - 3)$ , Radius = $25 = 5$

Exam tip:

The tangent to a circle is always perpendicular to the radius at the point of contact. This is the key property used for all tangent to circle exam questions.

5. Common Pitfalls

Wrong move:

Writing $(x + h)^{2} + (y + k)^{2} = r^{2}$ for a circle with center $(h, k)$

Why:

Signs are flipped when moving from center coordinates to the standard equation

Correct move:

For center $(h, k)$ , use $(x - h)^{2} + (y - k)^{2} = r^{2}$ . A center at $(- 2, 3)$ becomes $(x + 2)^{2} + (y - 3)^{2} = r^{2}$

Wrong move:

Trying to use $m_{2} = - 1/ m_{1}$ for a horizontal (gradient 0) line to find its perpendicular

Why:

This leads to division by zero, which is undefined

Correct move:

A horizontal line ( $y = c$ ) has a vertical perpendicular ( $x = k$ ), and vice versa. Do not use the gradient product rule for these cases.

Wrong move:

Taking both positive and negative roots when calculating distance or radius

Why:

Distance and radius are scalar quantities, they are always positive

Correct move:

Only take the positive square root when calculating distance, radius or length of any segment.

Wrong move:

Leaving the negative result from the angle formula calculation, leading to an obtuse angle

Why:

The formula always requires an absolute value to get the acute angle between two lines, which is what exams ask for

Correct move:

Always take the absolute value of the fraction before calculating $arctan$ to get an angle between 0° and 90°.

6. Quick Reference Cheatsheet

Concept	Formula
Distance between $(x_{1}, y_{1}), (x_{2}, y_{2})$	$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Midpoint of segment	$(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
Gradient of line	$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Parallel lines	$m_{1} = m_{2}$
Perpendicular lines	$m_{1} m_{2} = - 1$
Circle (center $(h, k)$ , radius $r$ )	$(x - h)^{2} + (y - k)^{2} = r^{2}$
Acute angle between two lines	$tan θ = \frac{m _{2} - m _{1}}{1 + m _{1} m _{2}}$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Intersection of line and circle
2024 · 2
Angle between two lines
2023 · 1
Equation of tangent to circle

Going deeper

syllabusIB AA HL Official SyllabusFull topic specifications

What's Next

2D coordinate geometry is the foundation for all higher geometry topics in IB AA HL, including 3D coordinate geometry, 2D and 3D vectors, and conic sections. Mastering the basic formulas and problem solving patterns here will make it much easier to solve more complex problems involving intersections, tangents, and distances in higher dimensions. Many IB exam problems combine 2D coordinate geometry with calculus (for finding tangents to curves) and trigonometry, so a solid understanding of this topic is essential to achieve a high overall grade.