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Math  /  Algebra

QuestionThe equations of three lines are given below. Line 1: 4y=3x+74 y=3 x+7 Line 2:6x+8y=62: 6 x+8 y=6 Line 3:y=43x−63: y=\frac{4}{3} x-6 For each pair of lines, determine whether they are parallel, perpendicular, or neither.
Line 1 and Line 2: Parallel Perpendicular Neither Line 1 and Line 3: Parallel Perpendicular Neither Line 2 and Line 3: Parallel Perpendicular Neither

Studdy Solution

STEP 1

What is this asking? We've got three lines, and we need to figure out if any of them are parallel, perpendicular, or neither! Watch out! Don't mix up parallel and perpendicular!
Parallel lines have the *same* slope, while perpendicular lines have *negative reciprocal* slopes.

STEP 2

1. Rewrite Line 1
2. Rewrite Line 2
3. Compare Line 1 and Line 2
4. Compare Line 1 and Line 3
5. Compare Line 2 and Line 3

STEP 3

We want to rewrite Line 1 in slope-intercept form (y=mx+by = mx + b), where mm is the **slope** and bb is the **y-intercept**.
Line 1 is given by 4y=3x+74y = 3x + 7.
To get yy by itself, we'll **divide both sides** by **4**: 4y4=3x+74 \frac{4y}{4} = \frac{3x + 7}{4} y=34x+74 y = \frac{3}{4}x + \frac{7}{4} So the **slope** of Line 1 is m1=34m_1 = \frac{3}{4}.

STEP 4

Line 2 is given by 6x+8y=66x + 8y = 6.
Let's **rewrite this** in slope-intercept form.
First, **subtract** 6x6x from **both sides**: 8y=−6x+6 8y = -6x + 6 Now, **divide both sides** by **8**: 8y8=−6x+68 \frac{8y}{8} = \frac{-6x + 6}{8} y=−68x+68 y = -\frac{6}{8}x + \frac{6}{8} Simplifying the fractions, we get: y=−34x+34 y = -\frac{3}{4}x + \frac{3}{4} So the **slope** of Line 2 is m2=−34m_2 = -\frac{3}{4}.

STEP 5

The slope of Line 1 is m1=34m_1 = \frac{3}{4} and the slope of Line 2 is m2=−34m_2 = -\frac{3}{4}.
Since the slopes are not equal, the lines are *not* parallel.
Since m1⋅m2=(34)(−34)=−916≠−1m_1 \cdot m_2 = (\frac{3}{4})(-\frac{3}{4}) = -\frac{9}{16} \neq -1, the lines are *not* perpendicular.
So, Line 1 and Line 2 are **neither** parallel nor perpendicular.

STEP 6

Line 3 is already in slope-intercept form: y=43x−6y = \frac{4}{3}x - 6.
The slope of Line 3 is m3=43m_3 = \frac{4}{3}.
The slope of Line 1 is m1=34m_1 = \frac{3}{4}.
Since m1⋅m3=(34)(43)=1212=1≠−1m_1 \cdot m_3 = (\frac{3}{4})(\frac{4}{3}) = \frac{12}{12} = 1 \neq -1, the lines are *not* perpendicular.
Since the slopes are not equal, the lines are *not* parallel.
So, Line 1 and Line 3 are **neither** parallel nor perpendicular.

STEP 7

The slope of Line 2 is m2=−34m_2 = -\frac{3}{4} and the slope of Line 3 is m3=43m_3 = \frac{4}{3}.
Since m2⋅m3=(−34)(43)=−1212=−1m_2 \cdot m_3 = (-\frac{3}{4})(\frac{4}{3}) = -\frac{12}{12} = -1, the lines *are* perpendicular!

STEP 8

Line 1 and Line 2: **Neither** Line 1 and Line 3: **Neither** Line 2 and Line 3: **Perpendicular**

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