The area of a parallelogram formed by the lines $ax \pm by \pm c = 0$ is

The sides of a triangle are $3x + 4y$,$4x + 3y$,and $5x + 5y$ units,where $x, y > 0$. The triangle is:

If the vertices of a triangle $ABC$ are $A(1,7)$,$B(-5,-1)$,and $C(7,4)$,then the equation of the internal angle bisector of $\angle ABC$ is

If the incentre of the triangle formed by the lines $x-2=0$,$x+y-1=0$,and $x-y+3=0$ is $(\alpha, \beta)$,then $\beta=$

$A$ line $4x+y=1$ passes through the point $A(2,-7)$ and meets the line $BC$,whose equation is $3x-4y+1=0$,at the point $B$. The equation of the line $AC$ such that $AB=AC$ is

In the triangle with vertices at $A(6,3), B(-6,3)$ and $C(-6,-3)$,the median through $A$ meets $BC$ at $P$,the line $AC$ meets the $x$-axis at $Q$,while $R$ and $S$ respectively denote the orthocentre and centroid of the triangle. Then the correct matching of the coordinates of points in List-$I$ to List-$II$ is:

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$