EN FR
← Course home

COM-2 Solutions: Coordinate Geometry and Lines

Solutions Reference · ← Back to Lesson COM-2

Section 5 — Guided Practice Solutions

▾

Problem 1 — Slope from Two Points (C3)

Variant 0: Through \( (2, 1) \) and \( (6, 9) \). Correct answer: m = 2.

\[ m = \frac{9 - 1}{6 - 2} = \frac{8}{4} = 2 \]

Variant 1: Through \( (-1, 4) \) and \( (3, -4) \). Correct answer: m = −2.

\[ m = \frac{-4 - 4}{3 - (-1)} = \frac{-8}{4} = -2 \]

Variant 2: Through \( (0, 5) \) and \( (4, 5) \). Correct answer: m = 0 (horizontal).

\[ m = \frac{5 - 5}{4 - 0} = \frac{0}{4} = 0 \]

A slope of 0 means the line is perfectly horizontal — both points have the same \( y \)-value, so there is no rise at all.

What to check if you got this wrong: Make sure \( \Delta y \) (difference in \( y \)-values) is in the numerator and \( \Delta x \) (difference in \( x \)-values) is in the denominator. The formula is "rise over run" — \( y \) on top, \( x \) on bottom.

Problem 2 — Point-Slope Form (C4)

Variant 0: \( m = 3 \), through \( (1, 2) \). Correct answer: y − 2 = 3(x − 1).

The point-slope template is \( y - y_1 = m(x - x_1) \). Here \( y_1 = 2 \), \( x_1 = 1 \), \( m = 3 \).

Variant 1: \( m = -2 \), through \( (-1, 3) \). Correct answer: y − 3 = −2(x + 1).

With \( x_1 = -1 \): \( x - x_1 = x - (-1) = x + 1 \). Don't forget: subtracting a negative becomes addition.

Variant 2: \( m = \frac{1}{2} \), through \( (4, 0) \). Correct answer: y = (1/2)(x − 4).

With \( y_1 = 0 \): \( y - 0 = \frac{1}{2}(x - 4) \), which simplifies to \( y = \frac{1}{2}(x-4) \). The \( y_1 = 0 \) term vanishes cleanly.

Common trap — swapping x₁ and y₁: If the point is \( (4, 0) \), then \( x_1 = 4 \) and \( y_1 = 0 \). Writing \( y - 4 = \frac12x \) treats 4 as a \( y \)-coordinate — that's wrong. Always identify which coordinate is \( x \) and which is \( y \) before plugging in.

Problem 3 — Parallel or Perpendicular? (C5)

Variant 0: \( m_1 = 3 \), \( m_2 = -\frac{1}{3} \). Correct answer: Perpendicular.

\[ m_1 \cdot m_2 = 3 \cdot \left(-\frac{1}{3}\right) = -1 \checkmark \]

Variant 1: \( m_1 = 5 \), \( m_2 = 5 \). Correct answer: Parallel.

Equal slopes, so the lines never intersect (assuming distinct lines — different \( y \)-intercepts).

Variant 2: \( m_1 = 2 \), \( m_2 = -2 \). Correct answer: Neither.

\[ m_1 \cdot m_2 = 2 \cdot (-2) = -4 \neq -1 \quad \text{(not perpendicular)} \]

\[ m_1 \neq m_2 \quad \text{(not parallel)} \]

The common error here is confusing "negative slope" with "perpendicular." The perpendicular slope to \( m = 2 \) is \( -\frac{1}{2} \), not \( -2 \).

Problem 4 — Distance Formula (C1) — Correct answer: 5

Points \( A(1, 1) \) and \( B(4, 5) \):

\[ d = \sqrt{(4-1)^2 + (5-1)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

This is the classic 3-4-5 right triangle. The exact answer is the integer 5 — not \( \sqrt25 \) (which is correct but not simplified) and not 7 (the Manhattan distance).

Section 6 — Independent Practice Solutions

▾

Problems 1 and 2 — Generator Problems

These problems are randomly generated, so the specific numbers depend on the problem you received. Use the patterns below to verify your work.

For distance and midpoint (Problem 1):

  • Compute \( \Delta x = x_2 - x_1 \) and \( \Delta y = y_2 - y_1 \).
  • Distance: \( \sqrt{(\Delta x)^2 + (\Delta y)^2} \). Simplify the radical if possible (check if the sum is a perfect square).
  • Midpoint: \( \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right) \). If the sum is odd, leave as a fraction — do not write a decimal.

For slope (Problem 2):

  • \( m = \dfrac{y_2 - y_1}{x_2 - x_1} \). The \( y \)-differences go on top.
  • Simplify the resulting fraction to lowest terms using GCD.
  • Check the sign: if the line falls from left to right, the slope is negative.

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

Variant 0: Through \( (2, 3) \) and \( (5, 9) \).

\[ m = \frac{9-3}{5-2} = \frac{6}{3} = 2 \]

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

Slope-intercept: \( y = 2x - 1 \)

Variant 1: Through \( (-1, 5) \) and \( (3, -3) \).

\[ m = \frac{-3-5}{3-(-1)} = \frac{-8}{4} = -2 \]

\[ y - 5 = -2(x + 1) \]

Slope-intercept: \( y = -2x + 3 \)

Variant 2: Through \( (0, 4) \) and \( (6, 1) \).

\[ m = \frac{1-4}{6-0} = \frac{-3}{6} = -\frac{1}{2} \]

\[ y - 4 = -\frac{1}{2}(x - 0) \implies y = -\frac{1}{2}x + 4 \]

Since the known point is the \( y \)-intercept, the slope-intercept form emerges directly here.

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

Variant 0: Original slope \( m = 3 \), through \( (6, 2) \).

Perpendicular slope: \( m_\perp = -\dfrac{1}{3} \)

\[ y - 2 = -\frac{1}{3}(x - 6) \]

Slope-intercept: \( y = -\dfrac{1}{3}x + 4 \)

Variant 1: Line \( y = -2x + 1 \), through \( (0, 4) \).

Original slope \( m = -2 \). Perpendicular slope: \( m_\perp = \dfrac{1}{2} \)

\[ y - 4 = \frac{1}{2}(x - 0) \implies y = \frac{1}{2}x + 4 \]

Variant 2: Line \( y = \frac12x - 3 \), through \( (2, -1) \).

Original slope \( m = \dfrac{1}{2} \). Perpendicular slope: \( m_\perp = -2 \)

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

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

Slope-intercept: \( y = -2x + 3 \)

Problem 5 — Secant Line on \( f(x) = x^2 - 2x \)

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

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

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

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

Slope-intercept: \( y = 3x - 4 \)

Common mistakes on secant lines:

  • Forgetting to evaluate \( f \) at both points first (using the \( x \)-values directly as if they were the \( y \)-values)
  • Plugging in \( x_1 \) and \( x_2 \) rather than \( f(x_1) \) and \( f(x_2) \) into the slope formula
  • Forgetting the minus sign when \( f(1) = -1 \) appears in the numerator: \( 8 - (-1) = 9 \), not \( 8 - 1 = 7 \)

Section 7 — Mastery Check Solutions

▾

Feynman Test — Model Answer

A strong answer addresses both parts:

In calculus, when studying curves, you typically know a specific point on the curve and the slope at that point (from the derivative). You rarely know the \( y \)-intercept directly — it would require finding where the curve's tangent line crosses the \( y \)-axis, which is an extra calculation.

Point-slope form \( y - y_1 = m(x - x_1) \) lets you write the tangent line equation immediately from what you have. Slope-intercept form \( y = mx + b \) requires computing \( b = y_1 - m x_1 \) first — an extra step and an extra chance for error. More importantly, point-slope explicitly shows the "anchor point" \( (x_1, y_1) \), which keeps the geometry visible.

Apply — Perpendicular to \( y = -\frac{1}{3}x + 5 \) through \( (2, 1) \)

Original slope: \( m = -\dfrac{1}{3} \)

Perpendicular slope: \( m_\perp = 3 \) (negate and reciprocate)

Point-slope form: \[ y - 1 = 3(x - 2) \]

Slope-intercept form: \[ y = 3x - 5 \]

Verify: at \( (2, 1) \): \( 3(2) - 5 = 1 \checkmark \)

Analyze — Find the Error

The student computed \( \dfrac{5-1}{2-8} = \dfrac{4}{-6} = -\dfrac{2}{3} \).

The error: The numerator should contain \( \Delta y \) (difference of \( y \)-values) and the denominator should contain \( \Delta x \) (difference of \( x \)-values). The student put the \( x \)-differences on top and the \( y \)-differences on bottom — exactly reversed.

Correct calculation:

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

Notice: the student's answer \( -\dfrac{2}{3} \) is the reciprocal of the correct answer \( -\dfrac{3}{2} \). A reversed-formula error always produces the reciprocal — this is the diagnostic sign to look for when checking your own work.

Section 8 — Boss Fight Solutions

▾

Path A — The Surveyor: Intersections A(0,0), B(4,3), C(7,−1)

Part 1 — Is AB ⊥ BC?

\[ m_{AB} = \frac{3-0}{4-0} = \frac{3}{4} \]

\[ m_{BC} = \frac{-1-3}{7-4} = \frac{-4}{3} \]

\[ m_{AB} \cdot m_{BC} = \frac{3}{4} \cdot \frac{-4}{3} = -1 \checkmark \]

Yes — road AB is perpendicular to road BC. The angle at B is exactly 90°.

Part 2 — Midpoint M of AC:

\[ M = \left(\frac{0+7}{2},\; \frac{0+(-1)}{2}\right) = \left(\frac{7}{2},\; -\frac{1}{2}\right) \]

Part 3 — New road through M, parallel to AB:

Parallel slope: \( m = \dfrac{3}{4} \) (same as \( m_{AB} \))

\[ y - \left(-\frac{1}{2}\right) = \frac{3}{4}\left(x - \frac{7}{2}\right) \]

\[ y + \frac{1}{2} = \frac{3}{4}x - \frac{21}{8} \]

\[ y = \frac{3}{4}x - \frac{21}{8} - \frac{4}{8} = \frac{3}{4}x - \frac{25}{8} \]

Key checks: Is the product of AB and BC slopes exactly \( -1 \)? Is the midpoint left as exact fractions? Is the new road's equation written with point-slope as the starting form?

Path B — The Analyst: \( f(x) = x^2 + 1 \)

Part 1 — Secant through (1, f(1)) and (3, f(3)):

\[ f(1) = 1 + 1 = 2, \quad f(3) = 9 + 1 = 10 \]

\[ m_1 = \frac{10-2}{3-1} = \frac{8}{2} = 4 \]

Equation: \( y - 2 = 4(x - 1) \), i.e., \( y = 4x - 2 \)

Part 2 — Secant through (1, f(1)) and (2, f(2)):

\[ f(2) = 4 + 1 = 5 \]

\[ m_2 = \frac{5-2}{2-1} = \frac{3}{1} = 3 \]

Equation: \( y - 2 = 3(x - 1) \), i.e., \( y = 3x - 1 \)

Part 3 — Geometric description:

As the second point moves from \( x = 3 \) to \( x = 2 \), the secant slope drops from 4 to 3. As the second point continues sliding toward \( (1, 2) \), the secant rotates around the anchor point, and its slope decreases further.

In the limit — as the second point merges with the first — the secant becomes the tangent line to the curve at \( (1, 2) \). The slope it approaches is the derivative of \( f \) at \( x = 1 \). For this function, that limiting slope is \( 2 \) (which you'll compute rigorously in BTC-2).

Section 9 — Challenge Problem Solutions

▾

Challenge 1 — Perpendicular Bisector of A(1,3) to B(5,1)

Midpoint: \( M = \left(\dfrac{1+5}{2}, \dfrac{3+1}{2}\right) = (3, 2) \)

Slope of AB: \( m_{AB} = \dfrac{1-3}{5-1} = \dfrac{-2}{4} = -\dfrac{1}{2} \)

Perpendicular slope: \( m_\perp = 2 \)

Perpendicular bisector: \( y - 2 = 2(x - 3) \), i.e., \( y = 2x - 4 \)

A point \( P \) lies on the perpendicular bisector of \( AB \) if and only if it is equidistant from \( A \) and \( B \). Perpendicular bisectors appear in geometry (circumscribed circles), physics (perpendicular fields), and calculus (curvature).

Challenge 2 — Equilateral Triangle: A(0,0), B(2,0), C(1,\sqrt{3})

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

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

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

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

The exact coordinate \( \sqrt{3} \) is essential here. Using a decimal approximation (e.g., 1.732) would make \( BC \) and \( AC \) come out to \( \sqrt{1 + 2.999...} \approx 1.9999... \) instead of exactly 2 — the equilateral property would be invisible.

Challenge 3 — Secant Slope Pattern for f(x) = x²

Part (a) — Four slopes:

General secant slope through \( (2, 4) \) and \( (2+h, (2+h)^2) \):

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

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

Part (b) — Pattern: The slopes 5, 4.5, 4.1, 4.01 are approaching 4.

Part (c) — 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 \( h \to 0 \): \( 4 + h \to 4 \).

The expression \( 4 + h \) confirms the pattern exactly. In BTC-2, you'll apply this same algebraic process to a general function \( f(x) \) with the difference quotient \( \dfrac{f(x+h)-f(x)}{h} \). This is the derivative — and you've already been doing it.

If you made it through all three challenges, you're well prepared for the rest of this course — and for the first weeks of calculus. The pattern from Challenge 3 is exactly the limit definition of the derivative, which you'll formalize in BTC-2 and BTC-3.