Lines in the Coordinate Plane - Tenth Grade Geometry
Introduction to Lines in the Coordinate Plane
Coordinate Plane: Two-dimensional plane with x-axis (horizontal) and y-axis (vertical)
Linear Equation: Equation whose graph is a straight line
Slope: Measure of steepness of a line
Intercepts: Points where line crosses axes
Linear Equation: Equation whose graph is a straight line
Slope: Measure of steepness of a line
Intercepts: Points where line crosses axes
1. Slopes of Lines
Slope (m): Rate of change of y with respect to x
Rise: Vertical change (Δy)
Run: Horizontal change (Δx)
Notation: Usually represented by letter $m$
Rise: Vertical change (Δy)
Run: Horizontal change (Δx)
Notation: Usually represented by letter $m$
Slope Formula:
For two points $(x_1, y_1)$ and $(x_2, y_2)$:
$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}} = \frac{\Delta y}{\Delta x}$$
In words: Change in y divided by change in x
Important: Order of subtraction must be consistent!
For two points $(x_1, y_1)$ and $(x_2, y_2)$:
$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}} = \frac{\Delta y}{\Delta x}$$
In words: Change in y divided by change in x
Important: Order of subtraction must be consistent!
Example 1: Find slope between two points
Find slope of line through (2, 3) and (5, 11)
$$m = \frac{11 - 3}{5 - 2} = \frac{8}{3}$$
Answer: $m = \frac{8}{3}$
Find slope of line through (2, 3) and (5, 11)
$$m = \frac{11 - 3}{5 - 2} = \frac{8}{3}$$
Answer: $m = \frac{8}{3}$
Types of Slopes
Classification by Slope Value:
1. Positive Slope ($m > 0$):
• Line rises from left to right ↗
• As x increases, y increases
• Example: $m = 2$, $m = \frac{3}{4}$
2. Negative Slope ($m < 0$):
• Line falls from left to right ↘
• As x increases, y decreases
• Example: $m = -3$, $m = -\frac{1}{2}$
3. Zero Slope ($m = 0$):
• Horizontal line →
• No vertical change
• Form: $y = b$ (constant)
4. Undefined Slope:
• Vertical line ↑
• No horizontal change (division by zero)
• Form: $x = a$ (constant)
• Cannot be written in slope-intercept form
1. Positive Slope ($m > 0$):
• Line rises from left to right ↗
• As x increases, y increases
• Example: $m = 2$, $m = \frac{3}{4}$
2. Negative Slope ($m < 0$):
• Line falls from left to right ↘
• As x increases, y decreases
• Example: $m = -3$, $m = -\frac{1}{2}$
3. Zero Slope ($m = 0$):
• Horizontal line →
• No vertical change
• Form: $y = b$ (constant)
4. Undefined Slope:
• Vertical line ↑
• No horizontal change (division by zero)
• Form: $x = a$ (constant)
• Cannot be written in slope-intercept form
Example 2: Classify slopes
Line through (1, 2) and (4, 2):
$m = \frac{2-2}{4-1} = \frac{0}{3} = 0$ (Zero slope, horizontal)
Line through (3, 1) and (3, 5):
$m = \frac{5-1}{3-3} = \frac{4}{0}$ (Undefined slope, vertical)
Line through (0, 0) and (2, 6):
$m = \frac{6-0}{2-0} = \frac{6}{2} = 3$ (Positive slope, rises)
Line through (1, 2) and (4, 2):
$m = \frac{2-2}{4-1} = \frac{0}{3} = 0$ (Zero slope, horizontal)
Line through (3, 1) and (3, 5):
$m = \frac{5-1}{3-3} = \frac{4}{0}$ (Undefined slope, vertical)
Line through (0, 0) and (2, 6):
$m = \frac{6-0}{2-0} = \frac{6}{2} = 3$ (Positive slope, rises)
2. Graph a Linear Equation
Graphing: Plotting points and drawing line through them
x-intercept: Where line crosses x-axis (y = 0)
y-intercept: Where line crosses y-axis (x = 0)
Solution: Any point (x, y) that makes equation true
x-intercept: Where line crosses x-axis (y = 0)
y-intercept: Where line crosses y-axis (x = 0)
Solution: Any point (x, y) that makes equation true
Methods to Graph a Line
Method 1: Using Intercepts
Step 1: Find x-intercept (set y = 0, solve for x)
Step 2: Find y-intercept (set x = 0, solve for y)
Step 3: Plot both intercepts
Step 4: Draw line through them
Method 2: Using Slope and Y-Intercept
Step 1: Identify y-intercept (b) and plot point (0, b)
Step 2: Use slope m = rise/run to find second point
Step 3: Draw line through points
Method 3: Using Table of Values
Step 1: Choose several x-values
Step 2: Calculate corresponding y-values
Step 3: Plot all points
Step 4: Draw line through them
Step 1: Find x-intercept (set y = 0, solve for x)
Step 2: Find y-intercept (set x = 0, solve for y)
Step 3: Plot both intercepts
Step 4: Draw line through them
Method 2: Using Slope and Y-Intercept
Step 1: Identify y-intercept (b) and plot point (0, b)
Step 2: Use slope m = rise/run to find second point
Step 3: Draw line through points
Method 3: Using Table of Values
Step 1: Choose several x-values
Step 2: Calculate corresponding y-values
Step 3: Plot all points
Step 4: Draw line through them
Example 1: Graph using intercepts
Graph: $2x + 3y = 12$
x-intercept (y = 0):
$2x + 3(0) = 12$
$x = 6$ → Point (6, 0)
y-intercept (x = 0):
$2(0) + 3y = 12$
$y = 4$ → Point (0, 4)
Plot (6, 0) and (0, 4), draw line through them
Graph: $2x + 3y = 12$
x-intercept (y = 0):
$2x + 3(0) = 12$
$x = 6$ → Point (6, 0)
y-intercept (x = 0):
$2(0) + 3y = 12$
$y = 4$ → Point (0, 4)
Plot (6, 0) and (0, 4), draw line through them
Example 2: Graph using slope and y-intercept
Graph: $y = \frac{2}{3}x - 1$
y-intercept: b = -1 → Plot (0, -1)
Slope: $m = \frac{2}{3}$ → Rise 2, run 3
From (0, -1): move up 2 and right 3 → Plot (3, 1)
Draw line through (0, -1) and (3, 1)
Graph: $y = \frac{2}{3}x - 1$
y-intercept: b = -1 → Plot (0, -1)
Slope: $m = \frac{2}{3}$ → Rise 2, run 3
From (0, -1): move up 2 and right 3 → Plot (3, 1)
Draw line through (0, -1) and (3, 1)
3. Equations of Lines
Three Main Forms:
1. Slope-Intercept Form (most common)
2. Point-Slope Form (useful for writing equations)
3. Standard Form (integer coefficients)
1. Slope-Intercept Form (most common)
2. Point-Slope Form (useful for writing equations)
3. Standard Form (integer coefficients)
Slope-Intercept Form
Slope-Intercept Form:
$$y = mx + b$$
where:
• $m$ = slope
• $b$ = y-intercept
Advantages:
• Easy to identify slope and y-intercept
• Easy to graph
• Most useful for analyzing lines
$$y = mx + b$$
where:
• $m$ = slope
• $b$ = y-intercept
Advantages:
• Easy to identify slope and y-intercept
• Easy to graph
• Most useful for analyzing lines
Example 1: Write equation in slope-intercept form
Given: slope = 4, y-intercept = -3
$m = 4$, $b = -3$
$y = 4x - 3$
Answer: $y = 4x - 3$
Given: slope = 4, y-intercept = -3
$m = 4$, $b = -3$
$y = 4x - 3$
Answer: $y = 4x - 3$
Point-Slope Form
Point-Slope Form:
$$y - y_1 = m(x - x_1)$$
where:
• $m$ = slope
• $(x_1, y_1)$ = any point on the line
Use when: You know slope and one point
Advantage: Can use any point (not just y-intercept)
$$y - y_1 = m(x - x_1)$$
where:
• $m$ = slope
• $(x_1, y_1)$ = any point on the line
Use when: You know slope and one point
Advantage: Can use any point (not just y-intercept)
Example 2: Write equation using point-slope form
Line with slope 2 passing through (3, 5)
$m = 2$, $(x_1, y_1) = (3, 5)$
$y - 5 = 2(x - 3)$
Convert to slope-intercept:
$y - 5 = 2x - 6$
$y = 2x - 1$
Answer: $y - 5 = 2(x - 3)$ or $y = 2x - 1$
Line with slope 2 passing through (3, 5)
$m = 2$, $(x_1, y_1) = (3, 5)$
$y - 5 = 2(x - 3)$
Convert to slope-intercept:
$y - 5 = 2x - 6$
$y = 2x - 1$
Answer: $y - 5 = 2(x - 3)$ or $y = 2x - 1$
Standard Form
Standard Form:
$$Ax + By = C$$
where:
• $A$, $B$, $C$ are integers
• $A \geq 0$ (A should be positive)
• $A$, $B$, $C$ have no common factor (except 1)
To find slope from standard form:
$$m = -\frac{A}{B}$$
$$Ax + By = C$$
where:
• $A$, $B$, $C$ are integers
• $A \geq 0$ (A should be positive)
• $A$, $B$, $C$ have no common factor (except 1)
To find slope from standard form:
$$m = -\frac{A}{B}$$
Example 3: Convert to standard form
Given: $y = \frac{2}{3}x + 4$
Multiply by 3: $3y = 2x + 12$
Rearrange: $-2x + 3y = 12$
Make A positive: $2x - 3y = -12$
Or multiply by -1: $-2x + 3y = 12$
Answer: $2x - 3y = -12$ or $-2x + 3y = 12$
Given: $y = \frac{2}{3}x + 4$
Multiply by 3: $3y = 2x + 12$
Rearrange: $-2x + 3y = 12$
Make A positive: $2x - 3y = -12$
Or multiply by -1: $-2x + 3y = 12$
Answer: $2x - 3y = -12$ or $-2x + 3y = 12$
Example 4: Write equation given two points
Line through (1, 3) and (4, 9)
Step 1: Find slope
$m = \frac{9-3}{4-1} = \frac{6}{3} = 2$
Step 2: Use point-slope with either point
Using (1, 3): $y - 3 = 2(x - 1)$
Step 3: Convert to slope-intercept
$y - 3 = 2x - 2$
$y = 2x + 1$
Answer: $y = 2x + 1$
Line through (1, 3) and (4, 9)
Step 1: Find slope
$m = \frac{9-3}{4-1} = \frac{6}{3} = 2$
Step 2: Use point-slope with either point
Using (1, 3): $y - 3 = 2(x - 1)$
Step 3: Convert to slope-intercept
$y - 3 = 2x - 2$
$y = 2x + 1$
Answer: $y = 2x + 1$
4. Slopes of Parallel and Perpendicular Lines
Parallel Lines: Never intersect, same slope
Perpendicular Lines: Intersect at 90°, slopes are negative reciprocals
Reciprocal: Flip numerator and denominator
Negative Reciprocal: Flip and change sign
Perpendicular Lines: Intersect at 90°, slopes are negative reciprocals
Reciprocal: Flip numerator and denominator
Negative Reciprocal: Flip and change sign
Parallel Lines
Parallel Lines Theorem:
Two non-vertical lines are parallel if and only if they have the same slope.
$$\text{If } \ell_1 \parallel \ell_2, \text{ then } m_1 = m_2$$
Conversely: If $m_1 = m_2$, then $\ell_1 \parallel \ell_2$
Note: All vertical lines are parallel (undefined slope)
Note: All horizontal lines are parallel (slope = 0)
Two non-vertical lines are parallel if and only if they have the same slope.
$$\text{If } \ell_1 \parallel \ell_2, \text{ then } m_1 = m_2$$
Conversely: If $m_1 = m_2$, then $\ell_1 \parallel \ell_2$
Note: All vertical lines are parallel (undefined slope)
Note: All horizontal lines are parallel (slope = 0)
Example 1: Determine if lines are parallel
Line 1: $y = 3x + 2$
Line 2: $y = 3x - 5$
$m_1 = 3$, $m_2 = 3$
Since $m_1 = m_2$, lines are PARALLEL
Different y-intercepts ensure they don't coincide
Line 1: $y = 3x + 2$
Line 2: $y = 3x - 5$
$m_1 = 3$, $m_2 = 3$
Since $m_1 = m_2$, lines are PARALLEL
Different y-intercepts ensure they don't coincide
Perpendicular Lines
Perpendicular Lines Theorem:
Two non-vertical lines are perpendicular if and only if their slopes are negative reciprocals.
$$\text{If } \ell_1 \perp \ell_2, \text{ then } m_1 \cdot m_2 = -1$$
Or equivalently:
$$m_2 = -\frac{1}{m_1}$$
Special Cases:
• Horizontal line ($m = 0$) ⊥ Vertical line (undefined)
• Vertical line ⊥ Horizontal line
Two non-vertical lines are perpendicular if and only if their slopes are negative reciprocals.
$$\text{If } \ell_1 \perp \ell_2, \text{ then } m_1 \cdot m_2 = -1$$
Or equivalently:
$$m_2 = -\frac{1}{m_1}$$
Special Cases:
• Horizontal line ($m = 0$) ⊥ Vertical line (undefined)
• Vertical line ⊥ Horizontal line
Example 2: Find negative reciprocal
If $m_1 = \frac{2}{3}$, find $m_2$ for perpendicular line
$m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}$
Check: $\frac{2}{3} \cdot \left(-\frac{3}{2}\right) = -1$ ✓
Answer: $m_2 = -\frac{3}{2}$
If $m_1 = \frac{2}{3}$, find $m_2$ for perpendicular line
$m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}$
Check: $\frac{2}{3} \cdot \left(-\frac{3}{2}\right) = -1$ ✓
Answer: $m_2 = -\frac{3}{2}$
Example 3: Determine if lines are perpendicular
Line 1: $y = 4x + 1$
Line 2: $y = -\frac{1}{4}x + 3$
$m_1 = 4$, $m_2 = -\frac{1}{4}$
Check: $4 \cdot \left(-\frac{1}{4}\right) = -1$ ✓
Lines are PERPENDICULAR
Line 1: $y = 4x + 1$
Line 2: $y = -\frac{1}{4}x + 3$
$m_1 = 4$, $m_2 = -\frac{1}{4}$
Check: $4 \cdot \left(-\frac{1}{4}\right) = -1$ ✓
Lines are PERPENDICULAR
Quick Reference:
| Original Slope | Parallel Slope | Perpendicular Slope |
|---|---|---|
| $m = 2$ | $2$ | $-\frac{1}{2}$ |
| $m = -3$ | $-3$ | $\frac{1}{3}$ |
| $m = \frac{3}{4}$ | $\frac{3}{4}$ | $-\frac{4}{3}$ |
| $m = -\frac{2}{5}$ | $-\frac{2}{5}$ | $\frac{5}{2}$ |
5. Equations of Parallel and Perpendicular Lines
Strategy: Use known slope and given point
Parallel: Use same slope as given line
Perpendicular: Use negative reciprocal of given slope
Parallel: Use same slope as given line
Perpendicular: Use negative reciprocal of given slope
Writing Parallel Line Equations
Steps to Write Equation of Parallel Line:
Step 1: Find slope of given line
Step 2: Use same slope for parallel line
Step 3: Use point-slope form with given point
Step 4: Convert to desired form (usually slope-intercept)
Step 1: Find slope of given line
Step 2: Use same slope for parallel line
Step 3: Use point-slope form with given point
Step 4: Convert to desired form (usually slope-intercept)
Example 1: Parallel line through point
Write equation of line parallel to $y = 2x + 5$ through point (3, 4)
Step 1: Slope of given line: $m = 2$
Step 2: Parallel line has same slope: $m = 2$
Step 3: Point-slope with (3, 4):
$y - 4 = 2(x - 3)$
Step 4: Convert:
$y - 4 = 2x - 6$
$y = 2x - 2$
Answer: $y = 2x - 2$
Write equation of line parallel to $y = 2x + 5$ through point (3, 4)
Step 1: Slope of given line: $m = 2$
Step 2: Parallel line has same slope: $m = 2$
Step 3: Point-slope with (3, 4):
$y - 4 = 2(x - 3)$
Step 4: Convert:
$y - 4 = 2x - 6$
$y = 2x - 2$
Answer: $y = 2x - 2$
Writing Perpendicular Line Equations
Steps to Write Equation of Perpendicular Line:
Step 1: Find slope of given line
Step 2: Find negative reciprocal for perpendicular slope
Step 3: Use point-slope form with given point
Step 4: Convert to desired form
Step 1: Find slope of given line
Step 2: Find negative reciprocal for perpendicular slope
Step 3: Use point-slope form with given point
Step 4: Convert to desired form
Example 2: Perpendicular line through point
Write equation of line perpendicular to $y = -\frac{3}{4}x + 1$ through point (6, 2)
Step 1: Slope of given line: $m_1 = -\frac{3}{4}$
Step 2: Perpendicular slope: $m_2 = -\frac{1}{-3/4} = \frac{4}{3}$
Step 3: Point-slope with (6, 2):
$y - 2 = \frac{4}{3}(x - 6)$
Step 4: Convert:
$y - 2 = \frac{4}{3}x - 8$
$y = \frac{4}{3}x - 6$
Answer: $y = \frac{4}{3}x - 6$
Write equation of line perpendicular to $y = -\frac{3}{4}x + 1$ through point (6, 2)
Step 1: Slope of given line: $m_1 = -\frac{3}{4}$
Step 2: Perpendicular slope: $m_2 = -\frac{1}{-3/4} = \frac{4}{3}$
Step 3: Point-slope with (6, 2):
$y - 2 = \frac{4}{3}(x - 6)$
Step 4: Convert:
$y - 2 = \frac{4}{3}x - 8$
$y = \frac{4}{3}x - 6$
Answer: $y = \frac{4}{3}x - 6$
Example 3: Line perpendicular to standard form
Line perpendicular to $3x + 2y = 6$ through (-1, 5)
Step 1: Find slope of given line
$2y = -3x + 6$
$y = -\frac{3}{2}x + 3$
$m_1 = -\frac{3}{2}$
Step 2: Perpendicular slope
$m_2 = -\frac{1}{-3/2} = \frac{2}{3}$
Step 3: Point-slope form
$y - 5 = \frac{2}{3}(x - (-1))$
$y - 5 = \frac{2}{3}(x + 1)$
$y - 5 = \frac{2}{3}x + \frac{2}{3}$
$y = \frac{2}{3}x + \frac{17}{3}$
Answer: $y = \frac{2}{3}x + \frac{17}{3}$
Line perpendicular to $3x + 2y = 6$ through (-1, 5)
Step 1: Find slope of given line
$2y = -3x + 6$
$y = -\frac{3}{2}x + 3$
$m_1 = -\frac{3}{2}$
Step 2: Perpendicular slope
$m_2 = -\frac{1}{-3/2} = \frac{2}{3}$
Step 3: Point-slope form
$y - 5 = \frac{2}{3}(x - (-1))$
$y - 5 = \frac{2}{3}(x + 1)$
$y - 5 = \frac{2}{3}x + \frac{2}{3}$
$y = \frac{2}{3}x + \frac{17}{3}$
Answer: $y = \frac{2}{3}x + \frac{17}{3}$
6. Find the Distance Between a Point and a Line
Distance from Point to Line: Length of perpendicular segment from point to line
Shortest Distance: Always measured perpendicularly
Formula: Derived from geometry and algebra
Shortest Distance: Always measured perpendicularly
Formula: Derived from geometry and algebra
Distance from Point to Line Formula:
For line in standard form: $Ax + By + C = 0$
And point $(x_0, y_0)$:
$$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$
Key Points:
• Line must be in form $Ax + By + C = 0$ (move all terms to left)
• Use absolute value in numerator (distance is always positive)
• Square root in denominator
For line in standard form: $Ax + By + C = 0$
And point $(x_0, y_0)$:
$$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$
Key Points:
• Line must be in form $Ax + By + C = 0$ (move all terms to left)
• Use absolute value in numerator (distance is always positive)
• Square root in denominator
Steps to Find Distance:
Step 1: Convert line equation to standard form $Ax + By + C = 0$
Step 2: Identify values of A, B, and C
Step 3: Identify coordinates of point $(x_0, y_0)$
Step 4: Substitute into distance formula
Step 5: Simplify
Step 1: Convert line equation to standard form $Ax + By + C = 0$
Step 2: Identify values of A, B, and C
Step 3: Identify coordinates of point $(x_0, y_0)$
Step 4: Substitute into distance formula
Step 5: Simplify
Example 1: Find distance from point to line
Find distance from point (1, 2) to line $3x + 4y - 10 = 0$
$A = 3$, $B = 4$, $C = -10$
$(x_0, y_0) = (1, 2)$
$$d = \frac{|3(1) + 4(2) + (-10)|}{\sqrt{3^2 + 4^2}}$$
$$d = \frac{|3 + 8 - 10|}{\sqrt{9 + 16}} = \frac{|1|}{\sqrt{25}} = \frac{1}{5}$$
Answer: $d = \frac{1}{5}$ or 0.2 units
Find distance from point (1, 2) to line $3x + 4y - 10 = 0$
$A = 3$, $B = 4$, $C = -10$
$(x_0, y_0) = (1, 2)$
$$d = \frac{|3(1) + 4(2) + (-10)|}{\sqrt{3^2 + 4^2}}$$
$$d = \frac{|3 + 8 - 10|}{\sqrt{9 + 16}} = \frac{|1|}{\sqrt{25}} = \frac{1}{5}$$
Answer: $d = \frac{1}{5}$ or 0.2 units
Example 2: Convert equation first
Find distance from (2, 3) to line $y = 2x + 1$
Step 1: Convert to standard form
$y = 2x + 1$
$-2x + y - 1 = 0$ or $2x - y + 1 = 0$
Using $2x - y + 1 = 0$:
$A = 2$, $B = -1$, $C = 1$
$$d = \frac{|2(2) + (-1)(3) + 1|}{\sqrt{2^2 + (-1)^2}}$$
$$d = \frac{|4 - 3 + 1|}{\sqrt{4 + 1}} = \frac{|2|}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$$
Answer: $d = \frac{2\sqrt{5}}{5}$ units
Find distance from (2, 3) to line $y = 2x + 1$
Step 1: Convert to standard form
$y = 2x + 1$
$-2x + y - 1 = 0$ or $2x - y + 1 = 0$
Using $2x - y + 1 = 0$:
$A = 2$, $B = -1$, $C = 1$
$$d = \frac{|2(2) + (-1)(3) + 1|}{\sqrt{2^2 + (-1)^2}}$$
$$d = \frac{|4 - 3 + 1|}{\sqrt{4 + 1}} = \frac{|2|}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$$
Answer: $d = \frac{2\sqrt{5}}{5}$ units
7. Find the Distance Between Two Parallel Lines
Distance Between Parallel Lines: Perpendicular distance between them
Key Concept: Distance is same everywhere (parallel lines are equidistant)
Method: Pick any point on one line, find distance to other line
Key Concept: Distance is same everywhere (parallel lines are equidistant)
Method: Pick any point on one line, find distance to other line
Distance Between Parallel Lines:
For two parallel lines in form:
$Ax + By + C_1 = 0$
$Ax + By + C_2 = 0$
(Note: A and B are the same, only C differs)
$$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$$
Alternative Method:
1. Choose any point on first line
2. Find distance from that point to second line using point-to-line formula
For two parallel lines in form:
$Ax + By + C_1 = 0$
$Ax + By + C_2 = 0$
(Note: A and B are the same, only C differs)
$$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$$
Alternative Method:
1. Choose any point on first line
2. Find distance from that point to second line using point-to-line formula
Method 1: Using Parallel Lines Formula
Step 1: Convert both equations to form $Ax + By + C = 0$
Step 2: Verify lines are parallel (same A and B)
Step 3: Identify $C_1$ and $C_2$
Step 4: Apply formula $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$
Method 2: Using Point-to-Line Distance
Step 1: Find any point on first line (set x = 0 or y = 0)
Step 2: Use point-to-line distance formula with second line
Step 3: Calculate distance
Step 1: Convert both equations to form $Ax + By + C = 0$
Step 2: Verify lines are parallel (same A and B)
Step 3: Identify $C_1$ and $C_2$
Step 4: Apply formula $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$
Method 2: Using Point-to-Line Distance
Step 1: Find any point on first line (set x = 0 or y = 0)
Step 2: Use point-to-line distance formula with second line
Step 3: Calculate distance
Example 1: Using parallel lines formula
Find distance between $3x + 4y - 5 = 0$ and $3x + 4y + 10 = 0$
Verify parallel: $A = 3$, $B = 4$ for both ✓
$C_1 = -5$, $C_2 = 10$
$$d = \frac{|-5 - 10|}{\sqrt{3^2 + 4^2}} = \frac{|-15|}{\sqrt{25}} = \frac{15}{5} = 3$$
Answer: Distance = 3 units
Find distance between $3x + 4y - 5 = 0$ and $3x + 4y + 10 = 0$
Verify parallel: $A = 3$, $B = 4$ for both ✓
$C_1 = -5$, $C_2 = 10$
$$d = \frac{|-5 - 10|}{\sqrt{3^2 + 4^2}} = \frac{|-15|}{\sqrt{25}} = \frac{15}{5} = 3$$
Answer: Distance = 3 units
Example 2: Using point-to-line method
Find distance between $y = 2x + 1$ and $y = 2x + 5$
Step 1: Verify parallel
Both have slope $m = 2$ ✓
Step 2: Find point on first line
Let $x = 0$: $y = 2(0) + 1 = 1$
Point: (0, 1)
Step 3: Convert second line to standard form
$y = 2x + 5$
$-2x + y - 5 = 0$ or $2x - y + 5 = 0$
Step 4: Find distance from (0, 1) to $2x - y + 5 = 0$
$$d = \frac{|2(0) - 1 + 5|}{\sqrt{2^2 + (-1)^2}} = \frac{|4|}{\sqrt{5}} = \frac{4\sqrt{5}}{5}$$
Answer: $d = \frac{4\sqrt{5}}{5}$ units
Find distance between $y = 2x + 1$ and $y = 2x + 5$
Step 1: Verify parallel
Both have slope $m = 2$ ✓
Step 2: Find point on first line
Let $x = 0$: $y = 2(0) + 1 = 1$
Point: (0, 1)
Step 3: Convert second line to standard form
$y = 2x + 5$
$-2x + y - 5 = 0$ or $2x - y + 5 = 0$
Step 4: Find distance from (0, 1) to $2x - y + 5 = 0$
$$d = \frac{|2(0) - 1 + 5|}{\sqrt{2^2 + (-1)^2}} = \frac{|4|}{\sqrt{5}} = \frac{4\sqrt{5}}{5}$$
Answer: $d = \frac{4\sqrt{5}}{5}$ units
Example 3: Direct formula method
Distance between $2x - 3y + 6 = 0$ and $2x - 3y - 4 = 0$
$A = 2$, $B = -3$
$C_1 = 6$, $C_2 = -4$
$$d = \frac{|6 - (-4)|}{\sqrt{2^2 + (-3)^2}} = \frac{|10|}{\sqrt{13}} = \frac{10\sqrt{13}}{13}$$
Answer: $d = \frac{10\sqrt{13}}{13}$ units
Distance between $2x - 3y + 6 = 0$ and $2x - 3y - 4 = 0$
$A = 2$, $B = -3$
$C_1 = 6$, $C_2 = -4$
$$d = \frac{|6 - (-4)|}{\sqrt{2^2 + (-3)^2}} = \frac{|10|}{\sqrt{13}} = \frac{10\sqrt{13}}{13}$$
Answer: $d = \frac{10\sqrt{13}}{13}$ units
Forms of Linear Equations Summary
| Form | Equation | When to Use | Key Features |
|---|---|---|---|
| Slope-Intercept | $y = mx + b$ | Graphing, analysis | m = slope, b = y-intercept |
| Point-Slope | $y - y_1 = m(x - x_1)$ | Writing equations | Uses point and slope |
| Standard | $Ax + By = C$ | Integer coefficients | A, B, C are integers |
| Horizontal Line | $y = k$ | Slope = 0 | Parallel to x-axis |
| Vertical Line | $x = h$ | Undefined slope | Parallel to y-axis |
Slope Classifications
| Slope Type | Value | Direction | Example |
|---|---|---|---|
| Positive | $m > 0$ | Rises left to right ↗ | $m = 2$, $m = \frac{3}{4}$ |
| Negative | $m < 0$ | Falls left to right ↘ | $m = -3$, $m = -\frac{1}{2}$ |
| Zero | $m = 0$ | Horizontal → | $y = 5$ |
| Undefined | Division by 0 | Vertical ↑ | $x = 3$ |
Parallel and Perpendicular Lines
| Relationship | Slope Condition | Example |
|---|---|---|
| Parallel | $m_1 = m_2$ | $y = 3x + 2$ and $y = 3x - 5$ |
| Perpendicular | $m_1 \cdot m_2 = -1$ or $m_2 = -\frac{1}{m_1}$ | $y = 2x + 1$ and $y = -\frac{1}{2}x + 3$ |
| Neither | Different slopes, product ≠ -1 | $y = 2x + 1$ and $y = 3x + 2$ |
Formula Quick Reference
| Formula Name | Formula | Use |
|---|---|---|
| Slope | $m = \frac{y_2 - y_1}{x_2 - x_1}$ | Find slope between two points |
| Slope-Intercept | $y = mx + b$ | Graph line, identify slope and intercept |
| Point-Slope | $y - y_1 = m(x - x_1)$ | Write equation from point and slope |
| Parallel Slope | $m_{\parallel} = m$ | Same slope as original |
| Perpendicular Slope | $m_{\perp} = -\frac{1}{m}$ | Negative reciprocal |
| Point to Line Distance | $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ | Shortest distance from point to line |
| Parallel Lines Distance | $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$ | Distance between parallel lines |
Success Tips for Lines in Coordinate Plane:
✓ Slope = rise/run = change in y / change in x
✓ Positive slope: line rises; Negative slope: line falls
✓ Horizontal line: slope = 0; Vertical line: undefined slope
✓ Parallel lines: SAME slope
✓ Perpendicular lines: slopes multiply to -1 (negative reciprocals)
✓ Slope-intercept form (y = mx + b) is easiest for graphing
✓ Point-slope form is best for writing equations
✓ Distance formulas require standard form: Ax + By + C = 0
✓ Always use absolute value for distances (positive)
✓ Check parallel lines have same A and B coefficients!
✓ Slope = rise/run = change in y / change in x
✓ Positive slope: line rises; Negative slope: line falls
✓ Horizontal line: slope = 0; Vertical line: undefined slope
✓ Parallel lines: SAME slope
✓ Perpendicular lines: slopes multiply to -1 (negative reciprocals)
✓ Slope-intercept form (y = mx + b) is easiest for graphing
✓ Point-slope form is best for writing equations
✓ Distance formulas require standard form: Ax + By + C = 0
✓ Always use absolute value for distances (positive)
✓ Check parallel lines have same A and B coefficients!