The centre of a square of side $4$ units length is $(3,7)$ and one of the diagonals is parallel to the line $y=x$. If $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and $(x_4, y_4)$ are the vertices of this square,then $\frac{y_1 y_2 y_3 y_4}{x_1 x_2 x_3 x_4}=$

The vertices of a triangle are $(at_1t_2, a(t_1 + t_2))$,$(at_2t_3, a(t_2 + t_3))$,and $(at_3t_1, a(t_3 + t_1))$. Find the coordinates of its orthocentre.

The points $A(-4, -1)$,$B(-2, -4)$,$C(4, 0)$,and $D(2, 3)$ are the vertices of:

The incentre of the triangle formed by the lines $x = 0$,$y = 0$,and $3x + 4y = 12$ 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:

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

If the vertices of a triangle are $(0, 4)$,$(4, 1)$,and $(7, 5)$,find its perimeter.