Coordinate Geometry and Straight Lines

1. Core Concepts & Intuition

(Imagine a city map. To find a specific building, you use coordinates, like “3rd Avenue and 5th Street”. That’s the core idea of Coordinate Geometry: using numbers (coordinates) to describe positions and shapes. A straight line is the simplest path between two points. On our map, it’s like a perfectly straight road. We can describe this road with an equation, which acts as a universal address for every single point on that road.)

2. Formal Definitions, Jargon, and Nuances

The Rectangular Coordinate System (or Cartesian Plane) is a plane formed by two perpendicular number lines, the x-axis (horizontal) and the y-axis (vertical), which intersect at a point called the origin (0,0).

The slope (m) of a line is a measure of its steepness. It is the ratio of the vertical change (rise) to the horizontal change (run) between any two points on the line. $m=\frac{y_2-y_1}{x_2-x_1}$.

Forms of a Line’s Equation:

1. Slope-Intercept Form: $y=mx+c$, where m is the slope and c is the y-intercept (the point where the line crosses the y-axis).
2. Point-Slope Form: $y-y_1=m(x-x_1)$, where m is the slope and $(x_1,y_1)$ is a known point on the line.
3. General Form: $Ax+By+C=0$, where A, B, and C are constants.

Line Properties:

• Parallel Lines: Have the exact same slope ($m_1=m_2$).
• Perpendicular Lines: Have slopes that are negative reciprocals of each other ($m_1\cdot m_2=-1$).

3. Step-by-Step Procedures & Worked Examples

Procedure: Finding the Equation of a Line Given Two Points

1. Take the two points, $(x_1,y_1)$ and $(x_2,y_2)$.
2. Calculate the slope: $m=\frac{y_2-y_1}{x_2-x_1}$.
3. Choose one of the points (either one will work).
4. Substitute the slope m and the chosen point into the point-slope form: $y-y_1=m(x-x_1)$.
5. (Optional) Rearrange the equation into slope-intercept form ($y=mx+c$) for easier interpretation.

Example 1: Find the equation of the line passing through (2, 3) and (4, 7).

1. Points are (2, 3) and (4, 7).
2. Slope $m=\frac{7-3}{4-2}=\frac{4}{2}=2$.
3. Choose point (2, 3).
4. Point-slope form: $y-3=2(x-2)$.
5. Slope-intercept form: $y-3=2x-4⟹y=2x-1$.

Example 2: Find the equation of a line parallel to $y=-3x+5$ that passes through (1, -2).

1. The given line has a slope $m=-3$. A parallel line will have the same slope.
2. We have the slope $m=-3$ and a point (1, -2).
3. Using point-slope form: $y-(-2)=-3(x-1)$.
4. $y+2=-3x+3⟹y=-3x+1$.

Visual Representation

graph TD
    subgraph Cartesian Plane
        A((2, 3))
        B((4, 7))
        A -- Line (y=2x-1) --> B
    end

This represents the line from Example 1, passing through two specific points.

4. The “Exam Brain” Algorithm & Strategic Handbook

Pattern Recognition:

• Keywords to look for: “equation of a line”, “slope”, “intercept”, “parallel to”, “perpendicular to”, “passes through the point”.
• Question Formats: “Find the equation of the line…”, “What is the slope of the line perpendicular to…?”, “Do the lines intersect, are they parallel, or are they the same?“.

Mental Algorithm (The Approach):

1. Identify the Goal: What information do I need to find? An equation, a slope, or a relationship between lines?
2. Select the Tool: What information am I given?
   • Two points? Calculate slope first, then use point-slope form.
   • A point and a slope? Use point-slope form directly.
   • An equation and a relationship (parallel/perpendicular)? Extract the slope from the given equation, modify it based on the relationship, then use the given point.
3. Execute & Verify: Plug the values in carefully. Double-check the sign for perpendicular slopes (it must be flipped and inverted). Plug the original point(s) back into your final equation to ensure they satisfy it.

5. Common Pitfalls & Exam Traps

• Trap 1: Perpendicular Slope Error. Forgetting to take the negative reciprocal. The perpendicular slope to 2 is -1/2, not 1/2.
• Trap 2: Mixing up x and y in the slope formula. Always be consistent: $(y_2-y_1)$ is in the numerator (rise), and $(x_2-x_1)$ is in the denominator (run).

6. Practice Exercises (Scaffolded Difficulty)

Exercise 1 (Concept Check)

What is the slope of a horizontal line? What is the slope of a vertical line?

Exercise 2 (Application)

Find the equation of the line that is perpendicular to $2x+3y=6$ and passes through the point (4, 1).

Exercise 3 (Qualifier-Style Synthesis)

Two lines, L1 and L2, are described as follows:

• L1 passes through the points (0, 5) and (3, -1).
• L2 has an x-intercept of 2 and is perpendicular to L1.

At what point (x, y) do L1 and L2 intersect?

7. Comprehensive Solutions to Exercises

Solution 1:

• A horizontal line has no change in y (rise = 0), so its slope is $m=\frac{0}{\text{run}}=0$.
• A vertical line has no change in x (run = 0). Division by zero is undefined, so its slope is undefined.

Solution 2:

1. Identify the Goal: Find the equation of a perpendicular line through a point.
2. Select the Tool: Extract slope, find the perpendicular slope, use point-slope form.
3. Execute & Verify:
   • First, find the slope of the given line. Rearrange $2x+3y=6$ to slope-intercept form: $3y=-2x+6⟹y=-\frac{2}{3}x+2$. The slope is $m_1=-\frac{2}{3}$.
   • The slope of the perpendicular line, $m_2$, is the negative reciprocal: $m_2=-\frac{1}{(-2/3)}=\frac{3}{2}$.
   • Use the point (4, 1) and slope $m_2=3/2$ in point-slope form: $y-1=\frac{3}{2}(x-4)$.
   • $y-1=\frac{3}{2}x-6⟹y=\frac{3}{2}x-5$.

Solution 3:

1. Find the equation for L1:

   • Slope $m_1=\frac{-1-5}{3-0}=\frac{-6}{3}=-2$.
   • The y-intercept is given as (0, 5), so c = 5.
   • Equation for L1: $y=-2x+5$.

2. Find the equation for L2:

   • Slope $m_2$ is the negative reciprocal of $m_1=-2$. So, $m_2=\frac{1}{2}$.
   • L2 passes through the x-intercept (2, 0).
   • Using point-slope form: $y-0=\frac{1}{2}(x-2)⟹y=\frac{1}{2}x-1$.

3. Find the intersection:

   • Set the two equations equal to each other: $-2x+5=\frac{1}{2}x-1$.
   • Solve for x: $6=\frac{5}{2}x⟹12=5x⟹x=\frac{12}{5}=2.4$.
   • Substitute x back into either equation to find y. Using L1: $y=-2(\frac{12}{5})+5=-\frac{24}{5}+\frac{25}{5}=\frac{1}{5}=0.2$.
   • The intersection point is (12/5, 1/5) or (2.4, 0.2).

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8. Connections & Further Learning

• [[Relations and Functions]]
• [[Quadratic Functions]]
• [[Systems of Linear Equations]]

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