Find the equation of the line perpendicular to the line $x-7y+5=0$ and having $x$-intercept $3$.

The equation of a line with slope $-\frac{1}{\sqrt{2}}$ and making an intercept of $2 \sqrt{2}$ units on the negative direction of the $y$-axis is:

Match the items given in List-$I$ to the items given in List-$II$.

List-$I$	List-$II$
$A$. Line passing through $(-4, 3)$ and having intercepts in the ratio $5:3$	$1$. $2x - 5y + 4 = 0$
$B$. Line passing through $P(2, -5)$ such that $P$ bisects the part intercepted between the axes	$2$. $3x + 5y = 3$
$C$. Line parallel to $2x - 3y + 5 = 0$ with $x$-intercept $\frac{2}{5}$ is	$3$. $10x - 15y + 4 = 0$
$D$. Line perpendicular to $5x + 2y + 7 = 0$ with $y$-intercept $\frac{4}{5}$ is	$4$. $10x - 15y = 4$
	$5$. $5x - 2y - 20 = 0$

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

The equation of the line whose midpoint between the axes is $({x_1}, {y_1})$ is:

If a straight line is at a distance of $10$ units from the origin and the perpendicular drawn from the origin to it makes an angle of $\frac{\pi}{4}$ with the negative $X$-axis in the negative direction,then the equation of that line is: