Question

Are the lines `x-3y=-3` and `6x+2y=12` perpendicular, parallel, or neither?

Answer 1

The given lines are mutually perpendicular.

Explanation:

Let the given lines be `L_1 : x-3y=-3, and, L_2 : 6x+2y=12`.

Suppose that `m_1 and m_2` are their resp. slopes.

Then, `L_1 : x+3=3y, or, y=1/3x+1 rArr m_1=1/3`.

Similarly, `m_2=-3`

Now, `m_1*m_2=-1 rArr L_1 bot L_2`.

Answer 2

the lines are perpendicular because when you multiply the slopes you get a value of negative one

Explanation:

If lines are parallel then their slopes are the same.
If lines are perpendicular their slopes are opposites
This means that when you multiply the slopes the results is negative one. ( The slopes are multiplicative inverses)
One line has a positive value which means it goes to the right
One line has a negative value which means it goes to the left.
One slope must have a value greater than one (or one)
One slope must have a value less than one ( or one)

It helps to change the equations to the y = mx + b form
m the coefficient of x is the slope.

The first equation is x - 3y = -3

First subtract x from both sides of the equation.

x -x - 3y = -x - 3 = -3y = -1 x -3

Next divide both sides by - 3 to get y by itself.

-3y/3 = -1x/3 + -3/-3 = y = + 1/3 x + 1

The slope = 1/3

The Second equation is 6x + 2y = 12

First subtract -6x from both sides of the equation

• 6x - 6x + 2y = -6x + 12 = 2y = -6x + 12

Next divide both sides by + 2 to get y by itself.

2y/2 = -6x/2 + 12/2 = y = -3 x + 6

The slope is -3

Multiply -3 x 1/3 = -1

The slopes are inverses or opposites so the lines will be perpendicular.