Question

What is the slope of a line perpendicular to the line whose equation is 3x+6y=-183x+6y=−18. Fully reduce your answer.

Answer 1

Answer:

2

Step-by-step explanation:

The given equation of line is

$3x+6y=-18$

We need to find the slope of a line which is perpendicular to the given line.

The given equation can be rewritten as

$3x+6y+18=0$      ...(i)

If a line is defined as $ax+bx+c=0$, then the slope of the line is  

$m=-\dfrac{a}{b}$

In equation (i), a=3, b=6 and c=18. So, slope of the line is

$m_1=-\dfrac{3}{6}=-\dfrac{1}{2}$

Let $m_2$ be the slope of perpendicular line.

We know that product of two perpendicular line is -1.

$m_1\cdot m_2=-1$

$(-\dfrac{1}{2})\cdot m_2=-1$

Multiply both sides by -2.

$m_2=2$

Therefore, the slope of perpendicular line is 2.