Question

The equations of three lines are given below. Line 1: y = - 3/4 x + 3 Line 2/8 x - 6y = 2 Line 3/- 3y = 4x + 7 For each pair of lines, determine whether they are parallel, perpendicular, or neither.

Answer 1

Answer:

No parallel lines

Lines 1 and 3 are perpendicular

Step-by-step explanation:

Given:

$Line\ 1:y = -\frac{3}{4}x + 3$

$Line\ 2: \frac{2}{8}x -6y = 2$

$Line\ 3: 3y = 4x + 7$

Required

Determine if they are parallel, perpendicular or not

The slope intercept of a line has the form:

$y = mx + b$

Where

$m = slope$

First, we calculate the slope of each lines

$Line\ 1:y = -\frac{3}{4}x + 3$

Compare the above to $y = mx + b$

$m_1 = -\frac{3}{4}$

$Line\ 2: \frac{2}{8}x -6y = 2$

$\frac{2}{8}x -6y = 2$

Make -6y the subject

$-6y = 2 - \frac{2}{8}x$

Divide through by -6

$y = -\frac{2}{6} + \frac{2}{8*6}x$

$y = -\frac{1}{3} + \frac{1}{8*3}x$

$y = -\frac{1}{3} + \frac{1}{24}x$

$y = \frac{1}{24}x-\frac{1}{3}$

Compare the above to $y = mx + b$

$m_2 = \frac{1}{24}$

$Line\ 3: 3y = 4x + 7$

$3y = 4x + 7$

Divide through by 3

$y = \frac{4}{3}x + \frac{7}{3}$

Compare the above to $y = mx + b$

$m_3 = \frac{4}{3}$

So, we have:

$m_1 = -\frac{3}{4}$

$m_2 = \frac{1}{24}$

$m_3 = \frac{4}{3}$

None of the slopes are the same, so none of the lines are parallel.

However, lines 1 and 3 are perpendicular.

This is shown below

When the slope of two lines satisfy the following condition, then they are  perpendicular.

$m_1 = -\frac{1}{m_3}$

This gives:

$-\frac{3}{4} = -\frac{1}{4/3}$

$-\frac{3}{4} = -1/\frac{4}{3}$

Convert / to *

$-\frac{3}{4} = -1*\frac{3}{4}$

$-\frac{3}{4} = -\frac{3}{4}$