Question

The normal form of the equation $x-\sqrt{3}y+8=0$ is

• A. $xcos120^{\circ }-ysin120^{\circ }=4$
• B. $xcos120^{\circ }+ysin120^{\circ }=4$
• C. $xcos120^{\circ }+ysin120^{\circ }=8$
• D. None of these

Answer 1

Answer: (B)

Given equation of line is $x-\sqrt{3}y+8=0$
$\Rightarrow -x+\sqrt{3}y=8\dots \left(i\right)$
Now, divide $\left(i\right)$ by ${\sqrt{(coefficientofx{)}^2+(coefficientofy)}}^2$
$=\sqrt{{\left(-1\right)}^2+{\left(\sqrt{3}\right)}^2}=\sqrt{1+3}=2$, we get normal form as
$-\frac{1}{2}x+\frac{\sqrt{3}}{2}y=\frac{8}{2}$
$\Rightarrow \left(-cos60^{\circ }\right)x+\left(sin60^{\circ }\right)y=4$
$\Rightarrow xcos\left(180^{\circ }-60^{\circ }\right)+ysin\left(180^{\circ }-60^{\circ }\right)=4$
$\Rightarrow xcos120^{\circ }+ysin120^{\circ }=4$