CSCA Mathematics · Module 03 · Lesson 03.1

Coordinate Geometry

Understand the Cartesian plane, distance and midpoint formulas, coordinate reflections, and geometric applications through guided examples.

15 teaching sections   ·   12 worked examples   ·   Assessment in Tutor LMS quiz
Section 01

What is coordinate geometry?

Coordinate geometry connects algebra with geometry. Instead of describing a point only by its position in a diagram, we assign it two numbers. We can then use calculations to answer geometric questions:

  • Where is a point?
  • How far apart are two points?
  • What point lies halfway between them?
  • Do three points form an isosceles or right triangle?
  • What coordinates are missing from a geometric construction?

The main idea is:

Represent positions using coordinates, calculate using algebra, and interpret the result geometrically.

Section 02

The Cartesian coordinate plane

The coordinate plane contains two perpendicular number lines.

The x-axis

The x-axis is horizontal.

  • Positive x-values lie to the right.
  • Negative x-values lie to the left.

The y-axis

The y-axis is vertical.

  • Positive y-values lie above the x-axis.
  • Negative y-values lie below the x-axis.

The origin

The axes intersect at the origin: O = (0, 0).

Ordered pairs

A point is written as:

\[P=(x,y)\]

The first coordinate gives horizontal position. The second gives vertical position. For example: P = (3, −2) means:

  1. Start at the origin.
  2. Move 3 units right.
  3. Move 2 units down.

Order matters

The points (3, −2) and (−2, 3) are different. In general:

\[(x,y)\neq(y,x)\]

The two ordered pairs are equal only when the coordinates match in the same positions.

Section 03

Quadrants and points on the axes

The axes divide the plane into four quadrants.

Location Coordinate conditions Example
Quadrant I x > 0, y > 0 (3, 2)
Quadrant II x < 0, y > 0 (−3, 2)
Quadrant III x < 0, y < 0 (−3, −2)
Quadrant IV x > 0, y < 0 (3, −2)
x-axis y = 0 (4, 0)
y-axis x = 0 (0, −5)
Origin x = 0, y = 0 (0, 0)

Important: A point on an axis does not belong to any quadrant. The origin belongs to both axes.

Worked example 1: Classifying points

Classify:

  • A = (−4, 3).
  • B = (2, −5).
  • C = (0, 6).
  • D = (−7, 0).

Answers:

  • A is in Quadrant II.
  • B is in Quadrant IV.
  • C lies on the y-axis.
  • D lies on the x-axis.
Section 04

Distance from a point to an axis

For:

\[P=(x,y)\]

the perpendicular distance to the x-axis is:

\[|y|\]

The perpendicular distance to the y-axis is:

\[|x|\]

Absolute values are necessary because distance cannot be negative.

Worked example 2

Find the distances from P = (−5, 3) to the coordinate axes. Distance to the x-axis:

\[|3|=3\]

Distance to the y-axis:

\[|-5|=5\]

Distance to the x-axis: 3 units. Distance to the y-axis: 5 units. Remember: Distance to the x-axis uses the y-coordinate, and distance to the y-axis uses the x-coordinate.

Section 05

Coordinate changes and straight-line distance

Let:

\[A=(x_1,y_1),\qquad B=(x_2,y_2)\]

The signed horizontal change from A to B is:

\[\Delta x=x_2-x_1\]

The signed vertical change is:

\[\Delta y=y_2-y_1\]

These changes describe direction as well as size. The corresponding horizontal and vertical lengths are:

\[|\Delta x|,\qquad |\Delta y|\]

Horizontal segment

If the points have the same y-coordinate:

\[AB=|x_2-x_1|\]

Vertical segment

If the points have the same x-coordinate:

\[AB=|y_2-y_1|\]

Worked example 3

Find the distance between: A = (−3, 2), B = (5, 2). The segment is horizontal:

\[AB=|5-(-3)|=8\]

Distance: 8 units. For: C = (4, −6), D = (4, 1), the segment is vertical:

\[CD=|1-(-6)|=7\]

Distance: 7 units.

Section 06

Deriving the distance formula

For two general points, the horizontal and vertical changes form the legs of a right triangle. The straight segment between the points is the hypotenuse. By the Pythagorean theorem:

\[AB^2=(\Delta x)^2+(\Delta y)^2\]

Therefore:

\[AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]

This is the distance formula.

Important properties

  • Distance is nonnegative.
  • Reversing the endpoints gives the same distance.
  • Distance is zero exactly when the two points are identical.
  • The formula gives the straight-line distance, not the distance traveled along a chosen route.

Distance from the origin

For P = (x, y):

\[OP=\sqrt{x^2+y^2}\]
Section 07

Worked example 4: Distance with negative coordinates

Find the distance between: A = (−2, 3), B = (4, −5).

Step 1: Find the coordinate changes

\[\Delta x=4-(-2)=6\]
\[\Delta y=-5-3=-8\]

Step 2: Apply the formula

\[AB=\sqrt{6^2+(-8)^2}\]
\[AB=\sqrt{36+64}=10\]

Distance: 10 units.

Verification

The horizontal and vertical lengths are 6 and 8. These form a 6–8–10 right triangle. Sign warning:

\[(-8)^2=64\]

Use parentheses when squaring a negative difference.

Section 08

Exact distances and squared distances

A distance does not always simplify to an integer.

Worked example 5

Find the distance between: A = (1, −2), B = (5, 1).

\[AB=\sqrt{(5-1)^2+(1-(-2))^2}\]
\[AB=\sqrt{16+9}=5\]

Now consider: C = (1, 1), D = (5, 3).

\[CD=\sqrt{4^2+2^2} =\sqrt{20}=2\sqrt5\]

Exact distance: 2√5 units. Keep the exact form unless a decimal approximation is requested.

Comparing distances without square roots

Because distances are nonnegative, comparing their squares gives the same ordering. For example:

  • If AB² = 20.
  • If AC² = 25.

Then AB < AC. This is useful when:

  • Comparing side lengths.
  • Checking whether a triangle is isosceles.
  • Applying the Pythagorean theorem.
Section 09

Finding a missing coordinate from a distance

Worked example 6

Point P = (x, 2) is 5 units from A = (1, −1). Find all possible values of x.

Step 1: Write the distance equation

\[\sqrt{(x-1)^2+(2-(-1))^2}=5\]

Step 2: Square both sides

\[(x-1)^2+9=25\]
\[(x-1)^2=16\]

Step 3: Use both square-root signs

\[x-1=\pm4\]

Therefore:

\[x=5\quad\text{or}\quad x=-3\]

Possible points: (5, 2) and (−3, 2).

Verification

Each point has a horizontal separation of 4 and a vertical separation of 3 from A. Both distances are 5. Key lesson: A fixed-distance condition can produce more than one point.

Section 10

The midpoint formula

The midpoint lies halfway along a segment. For:

\[A=(x_1,y_1),\qquad B=(x_2,y_2)\]

the midpoint is:

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

Average the x-coordinates and y-coordinates separately.

Why the formula works

Starting from A, move halfway through each coordinate change:

\[x_M=x_1+\frac{x_2-x_1}{2} =\frac{x_1+x_2}{2}\]

Similarly:

\[y_M=\frac{y_1+y_2}{2}\]

Worked example 7

Find the midpoint of: A = (−4, 7), B = (6, −1).

\[M=\left(\frac{-4+6}{2},\frac{7+(-1)}{2}\right)\]

Midpoint: (1, 3).

Verification

From A to M, the coordinate changes are (5, −4). From M to B, the changes are also (5, −4). The two half-segments have the same length and direction.

Section 11

Finding a missing endpoint

If M is the midpoint of AB:

\[x_M=\frac{x_A+x_B}{2}, \qquad y_M=\frac{y_A+y_B}{2}\]

Rearrange:

\[x_B=2x_M-x_A\]
\[y_B=2y_M-y_A\]

Worked example 8

A = (−3, 5), and the midpoint of AB is M = (2, −1). Find B.

\[x_B=2(2)-(-3)=7\]
\[y_B=2(-1)-5=-7\]

B = (7, −7).

Verification

\[\left(\frac{-3+7}{2},\frac{5+(-7)}{2}\right) =(2,-1)\]

The midpoint matches the given information.

A midpoint warning

Being equally distant from A and B does not, by itself, make a point their midpoint. A midpoint must also lie on segment AB. For example, (0, 2) is equally distant from (−1, 0) and (1, 0), but their midpoint is (0, 0).

Section 12

Basic coordinate reflections

Reflections provide a useful way to interpret signs and symmetry.

Reflection Image of (x, y) Coordinate change
Across the x-axis (x, −y) Change the sign of y.
Across the y-axis (−x, y) Change the sign of x.
Through the origin (−x, −y) Change both signs; equivalent to a 180° rotation.
Across the line y = x (y, x) Interchange the coordinates.

Worked example 9

Reflect P = (−3, 4):

  • Across the x-axis: (−3, −4).
  • Across the y-axis: (3, 4).
  • Through the origin: (3, −4).
  • Across y = x: (4, −3).

Reflections preserve distances.

Reflection through a given midpoint

If A and B are reflections of one another through M, then M is their midpoint. Thus the missing-endpoint formula also finds the reflection of a point through a specified center.

Section 13

Applications to triangles and rectangles

Worked example 10: Classifying a triangle

Let:

  • A = (0, 0).
  • B = (4, 0).
  • C = (0, 3).

Find the side lengths and classify the triangle.

\[AB=4,\qquad AC=3\]
\[BC=\sqrt{(0-4)^2+(3-0)^2}=5\]

Since:

\[3^2+4^2=5^2\]

the triangle is right-angled. The right angle is at A, where the sides of lengths 3 and 4 meet. Its area is:

\[\text{Area}=\frac12(4)(3)=6\]

Area: 6 square units.

Worked example 11: An isosceles triangle

Let:

  • A = (−2, 0).
  • B = (2, 0).
  • C = (0, 3).

Calculate squared lengths:

\[AC^2=2^2+3^2=13\]
\[BC^2=(-2)^2+3^2=13\]

Therefore AC = BC. The triangle is isosceles, with equal sides AC and BC. The midpoint of AB is (0, 0), directly below C. The base length is 4 and the perpendicular height is 3. Area: 6 square units.

Worked example 12: An axis-aligned rectangle

Three consecutive vertices of a rectangle are:

  • A = (−1, 2).
  • B = (5, 2).
  • C = (5, −2).

Find the fourth vertex, center, and diagonal length. Because the sides are parallel to the coordinate axes: Fourth vertex: D = (−1, −2). The center is the midpoint of diagonal AC:

\[M=\left(\frac{-1+5}{2},\frac{2+(-2)}{2}\right)\]

Center: (2, 0). The diagonal length is:

\[AC=\sqrt{6^2+(-4)^2} =\sqrt{52}=2\sqrt{13}\]

Diagonal length: 2√13 units.

Section 14

Common mistakes and verification

Mistake 1: Reversing the coordinate order

In (x, y), horizontal position comes first.

Mistake 2: Assigning an axis point to a quadrant

If either coordinate is zero, the point is on an axis, not in a quadrant.

Mistake 3: Giving a negative distance

Coordinate changes may be negative. Distances cannot be negative.

Mistake 4: Adding coordinate differences

The expression:

\[|x_2-x_1|+|y_2-y_1|\]

measures the length of a horizontal-and-vertical route, not generally the straight-line distance.

Mistake 5: Forgetting parentheses

For a difference of −7:

\[(-7)^2=49\]

Mistake 6: Using differences in the midpoint formula

Midpoints use coordinate sums divided by 2.

Mistake 7: Taking only one square-root sign

From:

\[(x-1)^2=16\]

both x − 1 = 4 and x − 1 = −4 must be considered.

Mistake 8: Relying on the appearance of a diagram

Unequal axis scales can distort lengths and angles visually. Use the coordinate values and formulas. For a geometrically faithful sketch, use equal unit lengths on both axes.

Verification checklist

  • Have I used the same coordinate order throughout?
  • Have I handled negative signs correctly?
  • Is the distance nonnegative?
  • Is the straight-line distance at least the absolute change in either coordinate?
  • Does the midpoint lie between the endpoint coordinates?
  • Does a recovered endpoint reproduce the given midpoint?
  • Have I retained all valid solutions?
  • Are lengths in units and areas in square units?
Section 15

Lesson review

You should now be able to:

  • Interpret and plot ordered pairs.
  • Identify quadrants and points on the coordinate axes.
  • Find distances to the axes and origin.
  • Calculate horizontal, vertical, and general distances.
  • Work with exact square-root answers.
  • Use squared distances to compare lengths.
  • Find missing coordinates from a distance condition.
  • Calculate midpoints and recover missing endpoints.
  • Apply basic coordinate reflections.
  • Solve elementary triangle and rectangle problems.
  • Verify results using coordinates, distances, and units.

The central formulas are:

\[AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]

and:

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

Completion target: Score at least 8/10 on the lesson quiz and independently correct every missed or guessed answer.