Slope and Equations of Lines

Finding and using slope in geometry

Slope and Equations of Lines

Slope Formula

The slope between points $(x_1, y_1)$ and $(x_2, y_2)$ :

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}$

Types of Slope

Positive slope: Line rises (goes up from left to right)

Negative slope: Line falls (goes down from left to right)

Zero slope: Horizontal line ( $y = k$ )

Undefined slope: Vertical line ( $x = k$ )

Parallel Lines

Parallel lines have equal slopes: $m_1 = m_2$

Perpendicular Lines

Perpendicular lines have negative reciprocal slopes: $m_1 \cdot m_2 = -1$

Or: $m_2 = -\frac{1}{m_1}$

Equation Forms

Slope-intercept form: $y = mx + b$

Point-slope form: $y - y_1 = m(x - x_1)$

Standard form: $Ax + By = C$

Applications

Proving lines are parallel or perpendicular
Finding equations of medians, altitudes, perpendicular bisectors
Coordinate proofs

📚 Practice Problems

1Problem 1easy

❓ Question:

Find the slope of the line passing through points (2, 5) and (6, 13).

💡 Show Solution

Step 1: Recall the slope formula: m = (y₂ - y₁)/(x₂ - x₁)

Step 2: Identify the coordinates: Point 1: (x₁, y₁) = (2, 5) Point 2: (x₂, y₂) = (6, 13)

Step 3: Substitute into the formula: m = (13 - 5)/(6 - 2) m = 8/4 m = 2

Answer: The slope is 2

2Problem 2easy

❓ Question:

Find the slope of the line through $(2, 3)$ and $(6, 11)$ .

💡 Show Solution

Use the slope formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{11 - 3}{6 - 2}$

$m = \frac{8}{4} = 2$

Answer: Slope = $2$

3Problem 3easy

❓ Question:

Write the equation of a line with slope 3 passing through point (4, 7).

💡 Show Solution

Step 1: Use point-slope form: y - y₁ = m(x - x₁)

Step 2: Substitute m = 3 and point (4, 7): y - 7 = 3(x - 4)

Step 3: Convert to slope-intercept form: y - 7 = 3x - 12 y = 3x - 12 + 7 y = 3x - 5

Step 4: Verify with the given point: When x = 4: y = 3(4) - 5 = 12 - 5 = 7 ✓

Answer: y = 3x - 5

4Problem 4medium

❓ Question:

Are the lines through $(1, 2)$ , $(3, 6)$ and $(0, 5)$ , $(2, 9)$ parallel, perpendicular, or neither?

💡 Show Solution

Find slope of first line: $m_1 = \frac{6 - 2}{3 - 1} = \frac{4}{2} = 2$

Find slope of second line: $m_2 = \frac{9 - 5}{2 - 0} = \frac{4}{2} = 2$

Since $m_1 = m_2 = 2$ , the lines are parallel.

Answer: Parallel

5Problem 5medium

❓ Question:

Find the equation of the line passing through points (-1, 4) and (3, -2).

💡 Show Solution

Step 1: Find the slope: m = (y₂ - y₁)/(x₂ - x₁) m = (-2 - 4)/(3 - (-1)) m = -6/4 m = -3/2

Step 2: Use point-slope form with point (-1, 4): y - 4 = (-3/2)(x - (-1)) y - 4 = (-3/2)(x + 1)

Step 3: Convert to slope-intercept form: y - 4 = (-3/2)x - 3/2 y = (-3/2)x - 3/2 + 4 y = (-3/2)x - 3/2 + 8/2 y = (-3/2)x + 5/2

Step 4: Verify with both points: Point (-1, 4): y = (-3/2)(-1) + 5/2 = 3/2 + 5/2 = 8/2 = 4 ✓ Point (3, -2): y = (-3/2)(3) + 5/2 = -9/2 + 5/2 = -4/2 = -2 ✓

Answer: y = (-3/2)x + 5/2 or y = -1.5x + 2.5

6Problem 6medium

❓ Question:

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

💡 Show Solution

Step 1: Find the slope of the given line: y = 2x + 3 has slope m₁ = 2

Step 2: Find the perpendicular slope: Perpendicular slopes are negative reciprocals m₂ = -1/m₁ = -1/2

Step 3: Use point-slope form: y - 1 = (-1/2)(x - 4)

Step 4: Simplify: y - 1 = (-1/2)x + 2 y = (-1/2)x + 2 + 1 y = (-1/2)x + 3

Step 5: Verify perpendicularity: Product of slopes: 2 × (-1/2) = -1 ✓ (Perpendicular lines have slopes whose product is -1)

Answer: y = (-1/2)x + 3

7Problem 7hard

❓ Question:

Find the equation of the line perpendicular to $y = 3x - 2$ that passes through $(6, 1)$ .

💡 Show Solution

Step 1: Find the perpendicular slope

Original slope: $m_1 = 3$

Perpendicular slope: $m_2 = -\frac{1}{3}$

Step 2: Use point-slope form with $(6, 1)$

$y - y_1 = m(x - x_1)$

$y - 1 = -\frac{1}{3}(x - 6)$

Step 3: Simplify to slope-intercept form

$y - 1 = -\frac{1}{3}x + 2$

$y = -\frac{1}{3}x + 3$

Answer: $y = -\frac{1}{3}x + 3$

8Problem 8hard

❓ Question:

Three points A(1, 2), B(4, k), and C(7, 14) are collinear (on the same line). Find the value of k.

💡 Show Solution

Step 1: Understand collinearity: If three points are collinear, they all lie on the same line Therefore, the slope between any two pairs must be equal

Step 2: Find slope from A to C: m_AC = (14 - 2)/(7 - 1) m_AC = 12/6 m_AC = 2

Step 3: Find slope from A to B: m_AB = (k - 2)/(4 - 1) m_AB = (k - 2)/3

Step 4: Set the slopes equal: m_AB = m_AC (k - 2)/3 = 2

Step 5: Solve for k: k - 2 = 6 k = 8

Step 6: Verify using slope from B to C: m_BC = (14 - 8)/(7 - 4) = 6/3 = 2 ✓ All three pairs have slope 2, confirming collinearity

Step 7: Find the equation (optional): Using A(1, 2) and slope 2: y - 2 = 2(x - 1) y = 2x

Check all points: A(1, 2): 2 = 2(1) ✓ B(4, 8): 8 = 2(4) ✓ C(7, 14): 14 = 2(7) ✓

Answer: k = 8

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