If three straight lines x-2y + 1 = 0, x + 3y-1 = 0 and ax + 2y-3 = 0 have two different intersections, then a=______ ．

If three straight lines x-2y + 1 = 0, x + 3y-1 = 0 and ax + 2y-3 = 0 have two different intersections, then a=______ ．

Three straight lines x-2y + 1 = 0, x + 3y-1 = 0 and ax + 2y-3 = 0 have two different intersections. The line ax + 2y-3 = 0 must be parallel to x-2y + 1 = 0 or x + 3y-1 = 0. When the line ax + 2y-3 = 0 is parallel to x-2y + 1 = 0, - 2a-2 × 1 = 0, the solution is a = - 1; when the line ax + 2y-3 = 0 is parallel to x + 3y-1 = 0, 3a-2 × 1 = 0

The line ax + 3y-6 = 0 and (a + 1) x-2y-1 = 0 are perpendicular to each other to find the possible value of A

The line ax + 3y-6 = 0 and (a + 1) x-2y-1 = 0 are perpendicular to each other
-a/3 *(a+1)/2=-1
a*(a+1)=6
a^2+a-6=0
(a-2)(a+3)=0
A = 2 or a = - 3

Two straight lines ax + by = 0, (A-1) x + y + B = 0, parallel to x + 2Y + 3 = 0, then the distance between them is

A line ax + by = 0, (A-1) x + y + B = 0, parallel to x + 2Y + 3 = 0, then the slopes of the three lines are equal, and the slope of X + 2Y + 3 = 0 is - 1 / 2, so the slope of (A-1) x + y + B = 0 is 1-A = - 1 / 2, so the slope of a = 3 / 2, ax + by = 0 is - A / b = - 1 / 2, because a = 3 / 2, so: B = 3, so the line is x / 2 + y + 3 = 0; 3x