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Coordinate Geometry

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Coordinate Geometry

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Coordinate Geometry

🎯 Learning Objectives

By the end of this topic, you will be able to:

Locate points in the rectangular (Cartesian) coordinate system
Compute the distance between any two points using the distance formula
Apply the section formula to find a point dividing a line segment in a given ratio
Calculate the area of a triangle given three vertices
Determine whether three points are collinear

📋 Prerequisites

Sets and Set Operations — number systems, $\mathbb{R}^2$ as a set of ordered pairs
Basic arithmetic — squares, square roots, absolute value
Algebra — solving linear equations

📖 Core Content

1.1 The Cartesian Coordinate System

1.1.1 Intuition: The Grid of Mathematics

Imagine a city laid out on a perfect grid. Every location can be described by two numbers: how far east (x-coordinate) and how far north (y-coordinate) it is from a central reference point (the origin). This is the Cartesian coordinate system, named after René Descartes who famously developed the idea while watching a fly crawl on his ceiling! The plane is divided into 4 quadrants:

Quadrant	X-sign	Y-sign	Example
I	+	+	$(3, 2)$
II	−	+	$(-4, 1)$
III	−	−	$(-2, -5)$
IV	+	−	$(3, -1)$

The point where both axes meet is the origin

O = (0, 0)

💡 Why this matters: Every visualization in data science (scatter plots, line charts, histograms) uses the Cartesian coordinate system. Understanding coordinates is the first step to understanding any graph. (Diagram) Key convention: The ordered pair $(x, y)$ always lists the horizontal (x) coordinate first, then the vertical (y) coordinate. $(3, 5)$ means "3 units right, 5 units up."

1.1.2 Worked Examples

Example 1.1: Plot the points

A = (2, 3)

B = (-1, 4)

C = (-3, -2)

D = (4, -1)

and state which quadrant each lies in. Solution:

A(2,3)

: 2 right, 3 up → Quadrant I

B(-1,4)

: 1 left, 4 up → Quadrant II

C(-3,-2)

: 3 left, 2 down → Quadrant III

D(4,-1)

: 4 right, 1 down → Quadrant IV

\boxed{\text{I: }A,\ \text{II: }B,\ \text{III: }C,\ \text{IV: }D}

1.2 The Distance Formula

1.2.1 Intuition: Measuring Between Points

If you walk 3 blocks east and then 4 blocks north, how far are you from your starting point (as the crow flies)? The answer uses the Pythagorean theorem:

\sqrt{3^2 + 4^2} = \sqrt{25} = 5

blocks. The distance formula is just the Pythagorean theorem applied to the coordinate grid.

1.2.2 Formula

The distance

d

between two points

P_1 = (x_1, y_1)

and

P_2 = (x_2, y_2)

is:

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

Recipe:

Subtract the x-coordinates: $\Delta x = x_2 - x_1$
Subtract the y-coordinates: $\Delta y = y_2 - y_1$
Square both differences
Add the squares
Take the square root

1.2.3 Worked Examples

Example 2.1 (Easy): Find the distance between

(1, 2)

and

(4, 6)

. Step 1:

\Delta x = 4 - 1 = 3

\Delta y = 6 - 2 = 4

Step 2:

d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

\boxed{d = 5}

Example 2.2 (Medium): Find the distance between

(-3, 5)

and

(2, -7)

. Step 1:

\Delta x = 2 - (-3) = 5

\Delta y = -7 - 5 = -12

Step 2:

d = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13

\boxed{d = 13}

Example 2.3 (Exam-style): Show that the triangle with vertices

A(1, 1)

B(4, 5)

C(7, 1)

is isosceles. Step 1:

AB = \sqrt{(4-1)^2 + (5-1)^2} = \sqrt{3^2 + 4^2} = 5

Step 2:

BC = \sqrt{(7-4)^2 + (1-5)^2} = \sqrt{3^2 + (-4)^2} = 5

Step 3:

AC = \sqrt{(7-1)^2 + (1-1)^2} = \sqrt{6^2 + 0^2} = 6

Two sides (

AB

and

BC

) are equal → isosceles triangle.

\boxed{AB = BC = 5,\ \text{so isosceles}}

1.3 Section Formula

1.3.1 Intuition: Splitting a Segment

If you have a rope from point

A

to point

B

and you want to cut it at a point

P

that divides it in the ratio

m:n

(say 2:3), where is

P

? The section formula gives the coordinates of

P

using a weighted average of

A

and

B

1.3.2 Formula

P

divides the line segment

AB

(from

A(x_1, y_1)

B(x_2, y_2)

) in the ratio

m:n

(i.e.,

AP:PB = m:n

), then:

P = \left(\frac{m x_2 + n x_1}{m + n},\ \frac{m y_2 + n y_1}{m + n}\right)

💡 Memory aid: The coordinates of $P$ are the weighted average of $A$ and $B$ with weights $n$ and $m$ respectively — the weight for $B$ is $m$ (the ratio of $AP$ ) and the weight for $A$ is $n$ (the ratio of $PB$ ). Special case — midpoint: When $m = n = 1$ , $P$ is the midpoint:

P = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right)

1.3.3 Worked Examples

Example 3.1 (Easy): Find the midpoint of

(2, 3)

and

(6, 9)

. Step 1:

x = \frac{2 + 6}{2} = 4

y = \frac{3 + 9}{2} = 6

\boxed{(4, 6)}

Example 3.2 (Medium): Find the point

P

that divides the segment from

A(1, 2)

B(7, 8)

in the ratio

2:1

. Step 1:

m = 2

n = 1

(ratio

AP:PB = 2:1

) Step 2:

x = \frac{2(7) + 1(1)}{2 + 1} = \frac{14 + 1}{3} = 5

Step 3:

y = \frac{2(8) + 1(2)}{2 + 1} = \frac{16 + 2}{3} = 6

\boxed{P = (5, 6)}

Check:

AP = \sqrt{(5-1)^2 + (6-2)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}

PB = \sqrt{(7-5)^2 + (8-6)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}

AP:PB = 4\sqrt{2} : 2\sqrt{2} = 2:1

✓ Example 3.3 (Hard — external division): Find the point

Q

that divides

AB

externally in the ratio

3:1

where

A(2, 3)

and

B(4, 5)

. For external division, the formula becomes

x = \frac{m x_2 - n x_1}{m - n}

: Step 1:

x = \frac{3(4) - 1(2)}{3 - 1} = \frac{12 - 2}{2} = 5

Step 2:

y = \frac{3(5) - 1(3)}{3 - 1} = \frac{15 - 3}{2} = 6

\boxed{Q = (5, 6)}

1.4 Area of a Triangle

1.4.1 Intuition

Given three points on the coordinate plane, how do you find the area of the triangle they form without measuring side lengths? The formula uses the shoelace pattern — multiply and subtract coordinates in a cycle.

1.4.2 Formula

Area of triangle with vertices

A(x_1, y_1)

B(x_2, y_2)

C(x_3, y_3)

\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|

Shoelace method:

\text{Area} = \frac{1}{2} \left| (x_1 y_2 + x_2 y_3 + x_3 y_1) - (y_1 x_2 + y_2 x_3 + y_3 x_1) \right|

1.4.3 Collinearity Check

Three points are collinear (lie on the same straight line) if and only if the area of the triangle they form is

0

x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0

1.4.4 Worked Examples

Example 4.1 (Easy): Find the area of triangle with vertices

(0,0)

(4,0)

(0,3)

. Step 1:

x_1=0,y_1=0

x_2=4,y_2=0

x_3=0,y_3=3

Step 2:

\text{Area} = \frac{1}{2}|0(0-3) + 4(3-0) + 0(0-0)| = \frac{1}{2}|0 + 12 + 0| = 6

\boxed{6 \text{ square units}}

(Check: base=4, height=3, area =

\frac{1}{2} \times 4 \times 3 = 6

✓) Example 4.2 (Medium): Find the area of triangle with vertices

A(1,2)

B(5,3)

C(3,6)

. Step 1 — Using the formula:

\text{Area} = \frac{1}{2}|1(3-6) + 5(6-2) + 3(2-3)|

= \frac{1}{2}|1(-3) + 5(4) + 3(-1)|

= \frac{1}{2}|-3 + 20 - 3| = \frac{1}{2}|14| = 7

\boxed{7 \text{ square units}}

Step 2 — Using shoelace:

x_1y_2 + x_2y_3 + x_3y_1 = 1(3) + 5(6) + 3(2) = 3 + 30 + 6 = 39

y_1x_2 + y_2x_3 + y_3x_1 = 2(5) + 3(3) + 6(1) = 10 + 9 + 6 = 25

\text{Area} = \frac{1}{2}|39 - 25| = \frac{1}{2}|14| = 7

✓ Example 4.3 (Exam-style): Check whether points

(1,2)

(3,5)

(5,8)

are collinear. Step 1: Compute area:

\frac{1}{2}|1(5-8) + 3(8-2) + 5(2-5)|

= \frac{1}{2}|1(-3) + 3(6) + 5(-3)|

= \frac{1}{2}|-3 + 18 - 15| = \frac{1}{2}|0| = 0

\boxed{\text{Collinear — area is 0}}

1.5 Why This Matters

Coordinate geometry is the bridge between algebra and geometry. It allows us to:

Solve geometric problems using algebraic methods
Model real-world data as points on a plane
Fit lines and curves to data (regression analysis in statistics)
Visualize functions and their properties In data science, everything starts with coordinate geometry: scatter plots, the data space where each point is an observation with features as coordinates.

📐 Key Formulas — Summary Table

Concept	Formula	When to Use
Distance	$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$	Between any two points
Midpoint	$\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$	Center of a segment
Section (internal)	$\left(\frac{m x_2 + n x_1}{m+n},\ \frac{m y_2 + n y_1}{m+n}\right)$	Point dividing segment in ratio $m:n$
Triangle area	$\frac{1}{2}\	x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)\
Collinearity	$x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) = 0$	Checking if 3 points are collinear
Shoelace	$\frac{1}{2}\	(x_1y_2 + x_2y_3 + x_3y_1) - (y_1x_2 + y_2x_3 + y_3x_1)\

⚠️ Common Pitfalls

Pitfall 1: Forgetting to Take the Absolute Value in Area

Mistake: Reporting a negative area. Why: The formula

\frac{1}{2}[x_1(y_2-y_3) + \ldots]

can be negative depending on vertex order (clockwise vs. counterclockwise). Area must be positive. Correct: Always take the absolute value:

\text{Area} = \frac{1}{2}|\ldots|

Pitfall 2: Reversing the Section Formula Ratio

Mistake: Using

m

and

n

in wrong order. Correct: If

AP:PB = m:n

, then

x = \frac{m x_2 + n x_1}{m+n}

. The coordinate of

B

(the far endpoint from

A

) gets the

m

weight. Memory aid: The point you're going towards (

B

) gets the first ratio part (

m

Pitfall 3: Confusing Collinearity with Parallelism

Mistake: Thinking collinear means parallel lines. Correct: Collinear means points lie on the same line. Three parallel lines have points on different lines that are parallel — these are NOT collinear.

Pitfall 4: Forgetting the Square Root in Distance

Mistake: Computing

d = (x_2-x_1)^2 + (y_2-y_1)^2

without the square root. Correct: Distance is the square root of the sum of squared differences.

📝 Practice Questions

Q1: Find the distance between $(-2, 3)$ and $(4, -5)$ .

Strategy Hint: Apply the distance formula directly.

Step 1: $\Delta x = 4 - (-2) = 6$ , $\Delta y = -5 - 3 = -8$ Step 2: $d = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

$\boxed{10}$ Q2: Find the midpoint of $(3, -2)$ and $(-5, 8)$ .

Strategy Hint: Average the x-coordinates and y-coordinates.

Step 1: $x = \frac{3 + (-5)}{2} = \frac{-2}{2} = -1$ Step 2: $y = \frac{-2 + 8}{2} = \frac{6}{2} = 3$

$\boxed{(-1, 3)}$ Q3: In what ratio does the point $P(3, 5)$ divide the segment joining $A(1, 3)$ and $B(5, 7)$ ?

Strategy Hint: Use the section formula and solve for $m:n$ .

Step 1: $x = \frac{m(5) + n(1)}{m+n} = 3 \implies 5m + n = 3m + 3n \implies 2m = 2n \implies m = n$

So the ratio is $1:1$ — P is the midpoint. Verify: $\frac{1+5}{2}=3$ , $\frac{3+7}{2}=5$ ✓

$\boxed{1:1}$ Q4: Find the area of triangle with vertices $(2,3)$ , $(5,7)$ , $(8,3)$ .

Strategy Hint: Use the formula with absolute value.

Step 1: $\text{Area} = \frac{1}{2}|2(7-3) + 5(3-3) + 8(3-7)|$ $= \frac{1}{2}|2(4) + 5(0) + 8(-4)|$ $= \frac{1}{2}|8 + 0 - 32| = \frac{1}{2}|-24| = 12$

$\boxed{12 \text{ square units}}$ Q5: Are the points $(0,0)$ , $(2,4)$ , $(4,8)$ collinear?

Strategy Hint: Compute the area — it should be 0 if collinear.

Step 1: $\text{Area} = \frac{1}{2}|0(4-8) + 2(8-0) + 4(0-4)|$ $= \frac{1}{2}|0 \cdot (-4) + 16 + 4(-4)| = \frac{1}{2}|0 + 16 - 16| = 0$

$\boxed{\text{Yes, collinear}}$ Q6: Find the point on the y-axis that is equidistant from $(2,3)$ and $(4,7)$ .

Strategy Hint: On the y-axis, $x = 0$ . So the point is $(0, y)$ . Use distance formula.

Step 1: Distance from $(0,y)$ to $(2,3)$ = distance to $(4,7)$ $\sqrt{(0-2)^2 + (y-3)^2} = \sqrt{(0-4)^2 + (y-7)^2}$

Step 2: Square both sides: $4 + (y-3)^2 = 16 + (y-7)^2$ $4 + y^2 - 6y + 9 = 16 + y^2 - 14y + 49$ $13 - 6y = 65 - 14y$ $8y = 52$ $y = 6.5$

$\boxed{(0, 6.5)}$ Q7: Find the area of quadrilateral with vertices $(0,0)$ , $(4,0)$ , $(6,3)$ , $(2,3)$ .

Strategy Hint: Divide into two triangles or use polygon area formula.

Step 1: Split into $\triangle ABC$ and $\triangle ACD$ (with $A(0,0)$ , $B(4,0)$ , $C(6,3)$ , $D(2,3)$ ). Step 2: Area $\triangle ABC = \frac{1}{2}|0(0-3) + 4(3-0) + 6(0-0)| = \frac{1}{2}|12| = 6$ Step 3: Area $\triangle ACD = \frac{1}{2}|0(3-3) + 6(3-0) + 2(0-3)| = \frac{1}{2}|18 - 6| = 6$ Step 4: Total = $6 + 6 = 12$

$\boxed{12 \text{ square units}}$ Q8: For what value of $k$ are $(2,3)$ , $(5,k)$ , $(8,9)$ collinear?

Strategy Hint: Set the area formula equal to 0 and solve for $k$ .

Step 1: $\frac{1}{2}|2(k-9) + 5(9-3) + 8(3-k)| = 0$ $|2k - 18 + 30 + 24 - 8k| = 0$ $|-6k + 36| = 0$ $-6k + 36 = 0 \implies k = 6$

$\boxed{k = 6}$ Q9: Find the centroid of the triangle with vertices $(2,4)$ , $(6,8)$ , $(10,3)$ .

Strategy Hint: Centroid = average of coordinates: $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$ .

Step 1: $x = \frac{2+6+10}{3} = \frac{18}{3} = 6$ Step 2: $y = \frac{4+8+3}{3} = \frac{15}{3} = 5$

$\boxed{(6,5)}$ Q10: Determine if points $(1,1)$ , $(2,4)$ , $(3,9)$ are collinear.

Strategy Hint: Compute area — a non-zero area means they form a triangle.

Step 1: $\text{Area} = \frac{1}{2}|1(4-9) + 2(9-1) + 3(1-4)|$ $= \frac{1}{2}|-5 + 16 - 9| = \frac{1}{2}|2| = 1$

Area = 1 (not 0), so they are NOT collinear (they lie on a parabola $y = x^2$ ).

$\boxed{\text{Not collinear}}$ Q11: The vertices of a triangle are $(3,4)$ , $(7,2)$ , $(5,6)$ . Find its area. What type of triangle is it?

Strategy Hint: Compute area, then side lengths to classify.

Step 1: $\text{Area} = \frac{1}{2}|3(2-6) + 7(6-4) + 5(4-2)|$ $= \frac{1}{2}|3(-4) + 7(2) + 5(2)| = \frac{1}{2}|-12 + 14 + 10| = \frac{1}{2}|12| = 6$

Step 2: Side lengths: $AB = \sqrt{(7-3)^2 + (2-4)^2} = \sqrt{16+4} = \sqrt{20}$ $BC = \sqrt{(5-7)^2 + (6-2)^2} = \sqrt{4+16} = \sqrt{20}$ $AC = \sqrt{(5-3)^2 + (6-4)^2} = \sqrt{4+4} = \sqrt{8}$

$AB = BC$ → isosceles triangle.

$\boxed{\text{Area} = 6,\ \text{isosceles}}$ Q12: A point $P$ divides the segment joining $A(2,3)$ and $B(6,7)$ in the ratio $3:2$ externally. Find $P$ .

Strategy Hint: External division: $x = \frac{m x_2 - n x_1}{m - n}$ .

Step 1: $m = 3$ , $n = 2$ , $A(2,3)$ , $B(6,7)$ Step 2: $x = \frac{3(6) - 2(2)}{3 - 2} = \frac{18 - 4}{1} = 14$ Step 3: $y = \frac{3(7) - 2(3)}{3 - 2} = \frac{21 - 6}{1} = 15$

$\boxed{P = (14, 15)}$

🔗 Cross-References

Next topic: Straight Lines — slope, intercepts, equations of lines
Related: Quadratic Functions — parabolas as curves
Across courses: BSMA1002 Stats 1 (scatter plots, linear regression); BSMA1003 Maths 2 (vectors in coordinate plane) Join Discord Previous1.3 FunctionsNext2.2 Straight Lines