The diagonals of a quadrilateral $ABCD$ are along the lines $x+3y=4$ and $6x-2y=7.$ Then, $ABCD$ must be a

Consider the following statements
$1$. The distance between the lines $y=mx+c_1$ and $y=mx+c_2$ is
$\frac{\left|c_1-c_2\right|}{\sqrt{1+m^2}}$

$2$. The distance between the lines $ax+by+c_1=0$
and $ax+by+c_2=0$ is $\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}.$.
$3$. The distance between the lines $x=c_1$ and $x=c_2$ is $\left|c_1-c_2\right|$

Which of the above statements are correct?