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

Find the value of $k$ for which the line $(k-3) x - (4-k^2) y + k^2 - 7k + 6 = 0$ passes through the origin.

If a line $L$ passes through the point of intersection of the lines $2x + 3y + 1 = 0$ and $x + y - 3 = 0$ and the angle made by the line $L$ with the $X$-axis measured in the positive direction from the positive $X$-axis is $\tan^{-1} \frac{2}{3}$,then the sum of the intercepts made by this line on the axes is

$A$ line passing through the point $P(1, 1)$ and parallel to the line $x - y = 5$ cuts the line $x + 3y - 2 = 0$ at $Q$. Then twice the length of the segment $PQ$ is

The Fahrenheit temperature $F$ and absolute temperature $K$ satisfy a linear equation. Given that $K=273$ when $F=32$ and that $K=373$ when $F=212$. Express $K$ in terms of $F$ and find the value of $F$,when $K =0$.

The equation of the line,where the length of the perpendicular segment from the origin to the line is $4$ and the inclination of this perpendicular segment with the positive direction of the $X$-axis is $30^{\circ}$,is: