What is the angle between the diagonals of the parallelogram formed by the lines $ℓx + my + n = 0$,$ℓx + my + n' = 0$,$mx + ℓy + n = 0$,and $mx + ℓy + n' = 0$?

All the vertices of a rectangle are of the form $(a, b)$ with $a, b$ integers satisfying the equation $(a-8)^2-(b-7)^2=5$. Then,the perimeter of the rectangle 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:

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

In an isosceles right-angled triangle,if the equation of the hypotenuse is $3x + 4y = 4$ and its opposite vertex is $(2, 2)$,then the slopes of the remaining two sides are:

Let $ABC$ be a triangle such that the equations of lines $AB$ and $AC$ are $3y-x=2$ and $x+y=2$,respectively,and the points $B$ and $C$ lie on the $x$-axis. If $P$ is the orthocentre of the triangle $ABC$,then the area of the triangle $PBC$ is equal to

Let the area of the triangle with vertices $A(1, \alpha)$,$B(\alpha, 0)$,and $C(0, \alpha)$ be $4 \text{ sq. units}$. If the points $(\alpha, -\alpha)$,$(-\alpha, \alpha)$,and $(\alpha^2, \beta)$ are collinear,then $\beta$ is equal to: