COM-2: Coordinate Geometry and Lines

Think about the last time you drove up a steep hill. Some hills feel gentle; others feel like you're launching into orbit. That steepness — how much you rise for every metre you travel forward — has a precise mathematical name: slope.

Or think about a car's speedometer. Speed is distance divided by time — change in position divided by change in time. That fraction, \( \Delta d / \Delta t \), is exactly the same idea as slope, just applied to motion instead of geometry.

Slope isn't a geometry concept. It's a universal concept. It measures how fast one quantity changes relative to another. That's why it appears everywhere: in physics, in economics, in biology, in engineering.

After this lesson, you will be able to:

• Compute the exact distance between two points using \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
• Find the midpoint of a segment using \( M = \left(\frac{x_1+x_2}{2},\, \frac{y_1+y_2}{2}\right) \)
• Compute slope as \( m = \frac{\Delta y}{\Delta x} \) and interpret its sign and magnitude
• Write equations of lines in point-slope form \( y - y_1 = m(x - x_1) \) — the primary form used in calculus
• Identify parallel and perpendicular lines using slope relationships
• Find the equation of the secant line through two points on a curve

This lesson builds directly on COM-1. The notation habits you established there — vertical work, exact values, implication arrows — are expected throughout. Here's what else you'll need:

If the Pythagorean theorem feels rusty, review it now — the distance formula is built on it, and it appears everywhere in this lesson.

C1 — The Distance Formula

Place two points on a coordinate plane: \( A(x_1, y_1) \) and \( B(x_2, y_2) \). To find the straight-line distance between them, draw a right triangle: the horizontal leg has length \( |x_2 - x_1| \) and the vertical leg has length \( |y_2 - y_1| \). The distance \( AB \) is the hypotenuse.

Quick example: distance from \( (1, 2) \) to \( (4, 6) \):

\[ d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

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C2 — The Midpoint Formula

The midpoint of a segment is the point exactly halfway between its two endpoints. You find it by averaging each coordinate separately.

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C3 — Slope

Slope measures steepness: how much a line rises (or falls) for every unit it moves to the right. It's the ratio of vertical change to horizontal change.

Special cases:

• Zero slope (\( m = 0 \)): the line is horizontal — it doesn't rise or fall at all
• Undefined slope: the line is vertical (\( x_2 = x_1 \), so the denominator is zero) — these are written as \( x = c \), not \( y = \ldots \)

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C4 — Equations of Lines

Once you know a slope and one point on a line, you can write its equation. There are two standard forms — and in calculus, one of them is far more useful than the other.

Example: slope \( m = 3 \), through \( (2, 5) \):

\[ y - 5 = 3(x - 2) \]

That's it — that's the equation. If you need slope-intercept:

\[ y - 5 = 3x - 6 \implies y = 3x - 1 \]

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C5 — Parallel and Perpendicular Lines

The perpendicular slope rule is sometimes stated as "negate and take the reciprocal." Both operations are required:

• If \( m_1 = 3 \), then \( m_2 = -\frac{1}{3} \) (not \( -3 \), which only negates; not \( \frac{1}{3} \), which only reciprocates)
• If \( m_1 = -\frac{2}{5} \), then \( m_2 = \frac{5}{2} \) (negate: \( \frac{2}{5} \), reciprocate: \( \frac{5}{2} \))

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C6 — Secant Lines

A secant line is a straight line that crosses a curve at exactly two distinct points. Its slope measures the average rate of change of the curve between those two points.

The secant line is the bridge between algebra and calculus. In the interactive visualization below, the fixed point P is anchored at \( (1, 1) \) on the parabola \( f(x) = x^2 \). Drag the slider to move the second point Q along the curve. Watch what happens to the slope of the secant as Q approaches P.

Example 1 — Distance and Midpoint (C1, C2)

Find the exact distance and midpoint between \( A(-3, 1) \) and \( B(1, 4) \).

Note: the midpoint \( y \)-coordinate is \( \frac{5}{2} \), not 2.5 — exact form from COM-1.

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Example 2 — Testing Perpendicularity with Slopes (C3, C5)

Points \( P(0,0) \), \( Q(2,4) \), \( R(4,-2) \). Is the angle at \( P \) a right angle? In other words, is \( PQ \perp PR \)?

Try computing both slopes before reading the solution.

Show Solution

Step 1 — Compute slope of PQ:

\[ m_{PQ} = \frac{4 - 0}{2 - 0} = \frac{4}{2} = 2 \]

Step 2 — Compute slope of PR:

\[ m_{PR} = \frac{-2 - 0}{4 - 0} = \frac{-2}{4} = -\frac{1}{2} \]

Step 3 — Check the product:

\[ m_{PQ} \cdot m_{PR} = 2 \cdot \left(-\frac{1}{2}\right) = -1 \checkmark \]

Since the product of slopes is \( -1 \), we conclude \( PQ \perp PR \) — the angle at \( P \) is exactly \( 90^\circ \).

\[ \therefore \triangle PQR \text{ is a right triangle with the right angle at } P. \]

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Example 3 — Equation of a Line Through Two Points (C3, C4)

Write the equation of the line through \( (-2, 5) \) and \( (3, -5) \).

Step 1 (slope) is fully shown. Try step 2 before reading the rest.

Show Steps 2 and 3

Step 2 — Write in point-slope form (using the point \( (3, -5) \)):

\[ y - (-5) = -2(x - 3) \]

\[ y + 5 = -2(x - 3) \]

Step 3 — Expand to slope-intercept if needed:

\[ y + 5 = -2x + 6 \implies y = -2x + 1 \]

Check: Using the other point \( (-2, 5) \) gives \( y - 5 = -2(x - (-2)) = -2(x + 2) \implies y = -2x + 1 \) — the same result. Either point works in step 2.

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Example 4 — The Secant Line (C3, C4, C6)

Let \( f(x) = x^2 \). Find the equation of the secant line through \( (1, f(1)) \) and \( (3, f(3)) \).

The steps are named. Try each one before reading.

Step 1: Evaluate f at both points

\[ f(1) = 1^2 = 1 \implies \text{point } (1, 1) \]

\[ f(3) = 3^2 = 9 \implies \text{point } (3, 9) \]

Step 2: Compute the slope of the secant

\[ m_{\text{sec}} = \frac{9 - 1}{3 - 1} = \frac{8}{2} = 4 \]

Step 3: Write the equation in point-slope form

Using \( (1, 1) \): \[ y - 1 = 4(x - 1) \]

Equivalent slope-intercept form: \( y = 4x - 3 \)

Notice: we evaluated \( f \) at two points on a curve, computed a slope, then wrote a line equation — using nothing but C1–C4 skills. This is exactly the difference quotient from BTC-2, before taking the limit.

Choose the correct answer from the dropdown. If your choice is wrong, you'll see an explanation of why — and why the correct answer is right.

Problem 1 — Slope from Two Points (C3)

Problem 2 — Point-Slope Form (C4)

Problem 3 — Parallel or Perpendicular? (C5)

Problem 4 — Distance Formula (C1)

Find the exact distance between \( A(1, 1) \) and \( B(4, 5) \).

Work through each problem independently. Show your steps using vertical work from COM-1. Use the "Show Solution" toggle to check your reasoning — not just your answer.

Problem 1 — Distance and Midpoint (C1, C2)

Problem 2 — Slope from Two Points (C3)

Problem 3 — Equation of a Line Through Two Points (C3, C4)

Problem 4 — Perpendicular Line Through a Point (C5, C4)

Problem 5 — Secant Line on a Curve (C3, C4, C6)

Let \( f(x) = x^2 - 2x \). Find the equation of the secant line through \( (1,\, f(1)) \) and \( (4,\, f(4)) \). Write your answer in point-slope form.

Show Full Solution

Step 1 — Evaluate f at both points:

\[ f(1) = 1^2 - 2(1) = 1 - 2 = -1 \implies \text{point } (1, -1) \]

\[ f(4) = 4^2 - 2(4) = 16 - 8 = 8 \implies \text{point } (4, 8) \]

Step 2 — Secant slope:

\[ m_{\text{sec}} = \frac{8 - (-1)}{4 - 1} = \frac{9}{3} = 3 \]

Step 3 — Point-slope form (using \( (1, -1) \)):

\[ y - (-1) = 3(x - 1) \]

\[ y + 1 = 3(x - 1) \]

Equivalent: \( y = 3x - 4 \)

Feynman Test — Explain It

In your own words, explain why point-slope form is more useful than slope-intercept form in calculus. Your answer should mention: (1) what information you typically know when working with curves, and (2) what extra step slope-intercept requires that point-slope avoids.

Show Model Answer

In calculus, when you find a tangent line to a curve at a specific point, you know two things: the coordinates of the point \( (x_1, y_1) \) and the slope \( m \) at that point (from the derivative). You almost never know the \( y \)-intercept directly.

Point-slope form \( y - y_1 = m(x - x_1) \) lets you write the equation immediately using exactly what you know. Slope-intercept form \( y = mx + b \) requires computing \( b = y_1 - m x_1 \) as an extra step — this isn't hard, but it's an additional arithmetic operation and an extra place to make an error.

More importantly, the point-slope structure mirrors how you find tangent lines in calculus: "I know a point, I have a slope, here's the line." Starting with slope-intercept hides that structure.

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Apply

Find the equation of the line perpendicular to \( y = -\dfrac{1}{3}x + 5 \) that passes through \( (2, 1) \). Write your answer in both point-slope and slope-intercept forms.

Show Solution

Step 1 — Read the slope: \( m = -\dfrac{1}{3} \)

Step 2 — Perpendicular slope: \( m_{\perp} = 3 \) (negate and reciprocate \( -\frac{1}{3} \))

Step 3 — Point-slope form:

\[ y - 1 = 3(x - 2) \]

Step 4 — Slope-intercept form:

\[ y - 1 = 3x - 6 \implies y = 3x - 5 \]

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Analyze — Find the Error

A student is asked to find the slope of the line through \( (5, 2) \) and \( (1, 8) \). They write:

What is the error? What is the correct slope?

Show Analysis

The error: The student has computed \( \Delta x / \Delta y \) instead of \( \Delta y / \Delta x \). They put the \( x \)-differences in the numerator and the \( y \)-differences in the denominator — exactly backwards.

Correct slope:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 2}{1 - 5} = \frac{6}{-4} = -\frac{3}{2} \]

Note that the correct slope \( -\frac{3}{2} \) is the reciprocal of the student's answer \( -\frac{2}{3} \). This is the tell-tale sign of a reversed-formula error.

Self-Assessment

You've practiced individual concepts. Now combine them — under real constraints. Choose your path:

Challenge 1 — The Perpendicular Bisector (C1, C2, C4, C5)

The perpendicular bisector of a segment is the line that (1) passes through the midpoint of the segment and (2) is perpendicular to the segment.

Find the equation of the perpendicular bisector of the segment from \( A(1, 3) \) to \( B(5, 1) \).

Show Solution

Step 1 — Midpoint:

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

Step 2 — Slope of AB:

\[ m_{AB} = \frac{1-3}{5-1} = \frac{-2}{4} = -\frac{1}{2} \]

Step 3 — Perpendicular slope:

\[ m_{\perp} = 2 \]

Step 4 — Line through M with slope 2:

\[ y - 2 = 2(x - 3) \]

Equivalent: \( y = 2x - 4 \)

Perpendicular bisectors are used in calculus when finding the center of a circle tangent to a curve at a point — a concept related to curvature.

Challenge 2 — Three Equal Distances (C1)

Show that the three points \( A(0, 0) \), \( B(2, 0) \), \( C(1, \sqrt{3}) \) form an equilateral triangle — that is, all three side lengths are equal.

Show Solution

Side AB:

\[ AB = \sqrt{(2-0)^2 + (0-0)^2} = \sqrt{4} = 2 \]

Side BC:

\[ BC = \sqrt{(1-2)^2 + (\sqrt{3}-0)^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

Side AC:

\[ AC = \sqrt{(1-0)^2 + (\sqrt{3}-0)^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

Since \( AB = BC = AC = 2 \), triangle \( ABC \) is equilateral. \( \blacksquare \)

Notice how the exact form \( \sqrt{3} \) was essential — a decimal approximation of \( C \) would produce rounding errors that make the equilateral property impossible to verify cleanly.

Challenge 3 — The Secant Slope Pattern (C3, C6)

Let \( f(x) = x^2 \). Consider the secant line through the fixed point \( (2, 4) \) and the movable point \( (2+h,\; f(2+h)) \).

1. Compute the slope of this secant for \( h = 1,\; 0.5,\; 0.1,\; 0.01 \). Record the four values.
2. Without computing, what value does the slope appear to approach as \( h \to 0 \)?
3. Bonus: Simplify the general formula \( \dfrac{f(2+h) - f(2)}{h} \) algebraically (expand \( (2+h)^2 \)). What does the formula become after cancelling \( h \)?

Show Solution

Part (a) — Four slopes:

Secant slope \( = \dfrac{(2+h)^2 - 4}{h} \)

• \( h = 1 \): \( \dfrac{9 - 4}{1} = 5 \)
• \( h = 0.5 \): \( \dfrac{6.25 - 4}{0.5} = \dfrac{2.25}{0.5} = 4.5 \)
• \( h = 0.1 \): \( \dfrac{4.41 - 4}{0.1} = \dfrac{0.41}{0.1} = 4.1 \)
• \( h = 0.01 \): \( \dfrac{4.0401 - 4}{0.01} = \dfrac{0.0401}{0.01} = 4.01 \)

Part (b) — The pattern:

The slopes 5, 4.5, 4.1, 4.01, … are getting closer and closer to 4.

Part (c) — The algebraic simplification:

\[ \frac{(2+h)^2 - 4}{h} = \frac{4 + 4h + h^2 - 4}{h} = \frac{4h + h^2}{h} = \frac{h(4 + h)}{h} = 4 + h \]

As \( h \to 0 \), this expression \( 4 + h \to 4 \). So the slope approaches exactly \( 4 \) — which is the derivative of \( x^2 \) at \( x = 2 \). In BTC-2, you'll generalize this with the difference quotient \( \dfrac{f(x+h) - f(x)}{h} \).

You just computed your first derivative — without any calculus. The algebraic simplification in part (c) is exactly what the difference quotient formula does. Welcome to the threshold.

Complete, step-by-step solutions for all problems in Sections 5–9 are on the solutions page, including the generator problem patterns and Boss Fight full worked solutions.

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