The family of lines,forming an isosceles triangle with the lines $3x - 4y - 2 = 0$ and $12x - 5y + 6 = 0$,is

If the straight line $2x + 3y + 1 = 0$ bisects the angle between a pair of lines,one of which is $3x + 2y + 4 = 0$,then the equation of the other line in that pair is

The equations of two equal sides of an isosceles triangle are $7x - y + 3 = 0$ and $x + y - 3 = 0$. If the third side passes through the point $(1, -10)$,find the equation of the third side.

The vertices of a triangle are $A(1, 7)$,$B(-5, -1)$,and $C(-1, 2)$. Then,the equation of a bisector of the $\angle ABC$ is

The straight line $x+y+1=0$ bisects an angle between the pair of lines of which one is $2x+3y-4=0$. Then,the equation of the other line is

Let $A(1,0)$, $B(2,-1)$, and $C(\frac{7}{3},\frac{4}{3})$ be three points. If the equation of the internal angle bisector of $\angle ABC$ is $\alpha x+\beta y=5$, then the value of $\alpha^2+\beta^2$ is