# A line has Cartesian equation `3x-2y+3=0`.Determine a direction vector for a line that is parallel to this line.

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### 1 Answer

You need to remember that the lines `3x - 2y + 3 = 0` and `ax + by + c = 0` are parallel if they follow the condition:

`3/a = -2/b =gt 3b=-2a =gt 3b + 2a = 0`

You need to remember that the direction vector of line `3x - 2y + 3 = 0` is P(-2,-3) and the orthogonal vector Q(3,-2). The vector Q is orthogonal to direction vector of parallel line such that:

`(3,-2)*(a,b) = 1 =gt 3a - 2b = 1`

Hence, you should solve the system of equatons:

`3b + 2a = 0`

`-2b + 3a = 1`

You should multiply by 2 the equation `3b + 2a = 0` and by 3 the equation `-2b + 3a = 1` and then you should add these equations such that:

`6b + 4a - 6b + 9a = 3 =gt 13a=3 =gt a=3/13`

`3b = 6/13 =gt b = 2/13`

**Hence, evaluating the direction vector of parallel line `3/13x + 2/13y + c = 0` yields `(-2/13,3/13).` **