What is the normal form of the line $x + \sqrt{3}y - 4 = 0$?

In an isosceles triangle $ABC$,the coordinates of the points $B$ and $C$ on the base $BC$ are respectively $(1, 2)$ and $(2, 1)$. If the equation of the line $AB$ is $y = 2x$,then the equation of the line $AC$ is

Write the equation of the line passing through the points $(1, -1)$ and $(3, 5)$.

Let $u = \hat{i} - 2\hat{j}$ and $v = -3\hat{i} + 5\hat{j}$. Consider three points $P, Q,$ and $R$ having position vectors $-\frac{1}{7}\hat{i}$,$-\frac{1}{4}\hat{j}$,and $-2\hat{i} + 3\hat{j}$ respectively. Among these,the points on the line segment passing through $u$ and $v$ are

The equations of the lines which cut off an intercept of $-1$ from the $y$-axis and are equally inclined to the axes are

$A$ line cuts the $X$ and $Y$ axes at the points $A$ and $B$ respectively. If the point $(5, 6)$ divides the line segment $AB$ internally in the ratio $3: 1$,then the equation of the line is: