Chapter 3.4, Problem 49PPS

To determine

To determine whether lines are parallel, perpendicular or neither.

  $y-3=6(x+2)$

  $y+3=-\frac{1}{3}(x-4)$

Answer to Problem 49PPS

Lines are neither parallel nor perpendicular.

Explanation of Solution

Given information:

Lines

  $y-3=6(x+2)$

  $y+3=-\frac{1}{3}(x-4)$

Formula used:

For parallel lines

  $m_1=m_2$

For perpendicular lines

  $m_1\times m_2=-1$

Calculation:

Line 1

  $y-3=6(x+2)$

On solving

  $y-3=6x+12$

Add 3 both sides

  $y-3+3=6x+12+3$

On solving

  $y=6x+15$

Then slope of line 1

  $m_1=6$

Line 2

  $y+3=-\frac{1}{3}(x-4)$

On solving

  $y+3=-\frac{1}{3}x+\frac{4}{3}$

Subtract 3 both sides

  $y+3-3=-\frac{1}{3}x+\frac{4}{3}-3$

On solving

  $y=-\frac{1}{3}x-\frac{5}{3}$

Then slope of line 2

  $m_2=-\frac{1}{3}$

Here

  $m_1\ne m_2$

Then check for perpenticular

Using perperdicular lines formula

  $m_1\times m_2=-1$

Substitute values of slopes

  $6\times -\frac{1}{3}=-1$

On solving

  $-2\cancel{=}-1$

Hence, lines are neither parallel nor perpendicular.