are 2y=4x+6 and 3y+6x=12 perpendicular

Question

I really need to know this for my final 

Answer 1

In order for lines to be perpendicular, their slopes must be negative reciprocals.

 

We need to get these equations in y=mx+b form, where m is slope and b is y-intercept.

 

 

From the first equation.

 

Divide both sides of equation by 2.

 

y = 2x + 3

 

 

From the second equation.

 

Subtract 6x on both sides of equation.

 

3y = -6x + 12

 

 

Divide both sides of equation by 3.

 

y = -2x + 4

 

 

Compare the slopes.

 

2 and -2 are not negative reciprocals.  They are actually opposites.  Therefore, the equations are not perpendicular.

Answer 2

Put the two lines into the slope-intercept format, y = mx + b, where m is the slope:

 

2y = 4x + 6

y = 2x + 3  (slope = 2)

 

3y + 6x = 12

3y = -6x + 12

y = -2x + 4 (slope = -2)

 

Two lines are perpendicular if the their slopes are negative reciprocals.  Since the first line has a slope = 2, all lines perpendicular to it will have a slope = -1/2.  The second line has a slope of -2, not -1/2, so it is NOT perpendicular to the first line.

Answer 3

The lines of two equations are perpendicular if the slopes are negative reciprocals:

 

m₁ = -1/m₂

 

Put the two equations in slope-intercept form: y = mx + b

 

2y = 4x + 6             3y + 6x = 12

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

                                y = -2x + 4

 

slope = 2                  slope = -2

 

The lines are NOT perpendicular.

Answer 4

lines are perpendicular if their slopes are opposite reciprocals.

The slope of the first line is 4/2 = 2

The slope of the second line is -6/3 = -2

So they are not perpendicular.