# Decide the relative position of the lines 3x+3y-12=0 and 6x-3y+6=0

*print*Print*list*Cite

### 2 Answers

The lines we have are 3x + 3y - 12 = 0 and 6x - 3y + 6 = 0

3x + 3y - 12 = 0

=> 3y = 12 - 3x

=> y = 4 - x

This is in the slope intercept form and the slope is -1

Similarly 6x - 3y + 6 = 0

=> 3y = 6x + 6

=> y = 2x + 3

The slope is 2

The lines are not parallel, nor are they perpendicular.

Let us find the point of intersection.

3x + 3y - 12 = 0

=> -6x - 6y + 24 = 0

Add 6x - 3y + 6 = 0

=> -9y = -30

=> y = 10/3

3x = 12 - 3*(10/3)

=> x = 2/3

**The lines intersect each other at the point ( 2/3 , 10/3)**

To decide the relative position of the given lines, we'll solve the system formed of the given equations of the lines.

If the system has solutions, that means that the lines are intercepting each other in a certain point. If the system has no solutions, that means that the lines are not intercepting each other.

We'll divide the 1st equation by 3:

x + y - 4 = 0

We'll re-write the 1st equation:

x+y = 4

x = 4 - y (1)

We'll divide the 2nd equation by 3:

2x - y + 2 = 0 (2)

We'll substitute (1) in (2):2(4 - y) - y = -2

We'll remove the brackets:

8 - 2y - y = -2

We'll combine like terms and we'll subtract 8 both sides:

-3y = -2 - 8

-3y = -10

We'll divide by -3:

y = 10/3

We'll substitute y in (3):

x = 4 - 10/3

x = (12-10)/3

x = 2/3

**The solution of the system represents the coordinates of the intercepting point of the given lines: (2/3 ; 10/3).**