Question

Find the length of the perpendicular drawn from the point $(4,-3)$ upon the straight line $6x+3y-10=0$, the angle between the axes being ${60}^o$.

Answer 1

When axes are inclined at an angle $\omega$ , the perpendicular distance of a point $P(h,k)$ from $ax+by+c=0$ is

$p=\frac{ah+bk+c}{\sqrt{a^2+b^2-2ab\cos \omega }}\sin \omega$

Here $P$ is $(4,-3)$ and line is $6x+3y-10=0$

$p=\frac{6\left(4\right)+3(-3)-10}{\sqrt{4^2+{(-3)}^2-2\left(4\right)(-3)\cos {60}^{\circ }}}\sin {60}^{\circ }p=\frac{24-9-10}{\sqrt{16+9+-2\left(4\right)(-3)\left(\frac{1}{2}\right)}}\frac{\sqrt{3}}{2}p=\frac{5}{\sqrt{37}}\times \frac{\sqrt{3}}{2}p=\frac{5}{2}\sqrt{\frac{3}{37}}$