If the vertices of a triangle have integer coordinates,then what type of triangle can it never be?

Let $L_1$ be a line passing through $(2,1)$ and $(3, \frac{5}{2})$. $L_2$ is a line perpendicular to $L_1$ and passing through $(4,-1)$. The area of the triangle formed by $L_1$,$L_2$ and the $y$-axis is

$A$ line $L$ passes through the points $(1, 1)$ and $(2, 0)$ and another line $L'$ passes through $\left( \frac{1}{2}, 0 \right)$ and is perpendicular to $L$. Then the area of the triangle formed by the lines $L, L'$ and the $y$-axis 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:

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

Let $A(1,2)$ and $C(-3,-6)$ be two diagonally opposite vertices of a rhombus, whose sides $AD$ and $BC$ are parallel to the line $7x-y=14$. If $B(\alpha, \beta)$ and $D(\gamma, \delta)$ are the other two vertices, then $|\alpha+\beta+\gamma+\delta|$ is equal to:

The points $(1,3)$ and $(5,1)$ are two opposite vertices of a rectangle. The other two vertices lie on the line $y = 2x + c$,where $c$ is a constant. The coordinates of the other two vertices are: