Question

In what follows, p denotes the distance of the straight line from the origin and a denotes the angle
made by the normal ray drawn from the origin to the straight line with Ox measured in the
anti-clockwise sense. Find the equations of the straight lines with the following values of
(1) p = 5, alpha = 60°
​

Answer 1

$\textbf{Given:}$

$\mathsf{p=5\;\;and\;\;\alpha=60^\circ}$

$\textbf{To find:}$

$\textsf{The equation the given line}$

$\textbf{Solution:}$

$\textbf{Formula used:}$

$\boxed{\begin{minipage}{8cm} \\\underline{\mathsf{Normal\;form\;of\;st.line:}}\\\\\mathsf{x\;cos\alpha+y\;sin\alpha=p}\\\\\mathsf{where,}\\\mathsf{p\;is\;the\;perpendicular\;distance\;of\;the\;line\;from\;origin}\\\mathsf{\alpha\;is\;the\;angle\;made\;by\;the\;perpendicular\;with\;x-axis}\\ \end{minipage}}$

$\mathsf{Here,\;p=5,\;\alpha=60^\circ}$

$\textsf{The required line is}$

$\mathsf{x\;cos\alpha+y\;sin\;\alpha=p}$

$\mathsf{x\;cos60^\circ+y\;sin60^\circ=5}$

$\mathsf{x\left(\dfrac{1}{2}\right)+y\;sin\left(\dfrac{\sqrt3}{2}\right)=5}$

$\mathsf{\dfrac{x}{2}+\dfrac{\sqrt3\,y}{2}=5}$

$\mathsf{\dfrac{x+\sqrt3\,y}{2}=5}$

$\mathsf{x+\sqrt3\,y=10}$

$\implies\boxed{\mathsf{x+\sqrt3\,y-10=0}}$

$\textbf{Find more:}$

Transform the equation x+y- 2 = 0) into slope intercept form, intercept form and

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transform the following equation √3x+y+10=0 into 1)normal form,2)intercept form and 3)slope intercept form​

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

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Answer 2

Answer:

1) p=5 and alpha=60°

Step-by-step explanation:

x cos alpha+ysin alpha=5

x cos 60°+y sin 60°=5

x (1/2)+y sin(√3/2)=5

x/2+√3/2y=5

x+√3y/2=5

x+√3y=10

x+√3-10=0