Question

the perpendicular form of the line √3x-y+4=0 is​

Answer 1

$\textbf{Given:}$

$\textsf{Equation of line is}$

$\mathsf{\sqrt{3}x-y+4=0}$

$\textbf{To find:}$

$\textsf{Perpendicular form of the given line}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{\sqrt{3}x-y+4=0}$

$\textsf{This can be written as}$

$\mathsf{-\sqrt{3}x+y=4}$

$\textsf{Divide bothsides by 2}$

$\mathsf{\dfrac{-\sqrt{3}}{2}x+\dfrac{1}{2}y=2}$

$\boxed{\mathsf{x\,cos150^\circ+y\,sin150^\circ=2}}$

$\mathsf{This\;equation\;is\;of\;the\;form\;x\,cos\alpha+y\,sin\alpha=p}$

$\textbf{Find more:}$

Reduce the equation root 3x+y+2=0 to 1)slope intercept form 2)intercept form and 3) normal form

https://brainly.in/question/2754095

Answer 2

$\textbf\red{Given:}$

$\textsf{Equation of line is}$

$\mathsf{\sqrt{3}x-y+4=0}$

$\textbf\red{To find:}$

$\textsf{Perpendicular form of the given line}$

$\textbf\red{Solution:}$

$\textsf\red{Consider,}$

$\mathsf{\sqrt{3}x-y+4=0}$

$\textsf{This can be written as}$

$\mathsf{-\sqrt{3}x+y=4}$

$\textsf{Divide bothsides by 2}$

$\mathsf{\dfrac{-\sqrt{3}}{2}x+\dfrac{1}{2}y=2}$

$\boxed{\mathsf{x\,cos150^\circ+y\,sin150^\circ=2}}$

$\mathsf{This\;equation\;is\;of\;the\;form\;x\,cos\alpha+y\,sin\alpha=p}$