Chapter 3.4, Problem 48PPS

To determine

To determine whether lines are parallel, perpendicular or neither.

  $y-4=3(x+5)$

  $y+3=-\frac{1}{3}(x+1)$

Answer to Problem 48PPS

Lines are perpendicular.

Explanation of Solution

Given information:

Lines

  $y-4=3(x+5)$

  $y+3=-\frac{1}{3}(x+1)$

Formula used:

For parallel lines

  $m_1=m_2$

For perpendicular lines

  $m_1\times m_2=-1$

Calculation:

Line 1

  $y-4=3(x+5)$

On solving

  $y-4=3x+15$

Add 4 both sides

  $y-4+4=3x+15+4$

On solving

  $y=3x+19$

Then slope of line 1

  $m_1=3$

Line 2

  $y+3=-\frac{1}{3}(x+1)$

On solving

  $y+3=-\frac{1}{3}x-\frac{1}{3}$

Subtract 3 both sides

  $y+3-3=-\frac{1}{3}x-\frac{1}{3}-3$

On solving

  $y=-\frac{1}{3}x-\frac{10}{3}$

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

  $3\times -\frac{1}{3}=-1$

On solving

  $-1=-1$

Hence, lines are perpendicular.