Question

the perpendicular distance of a line from the origin is 4 units . if the perpendicular makes 60 degree with the x axis , find the equation of the line . also prove the line passes through (5/root under 3)

Answer 1

EXPLANATION.

The perpendicular distance from the line from

the origin is 4 units.

perpendicular makes 60° with the x- axis

To find the equation of the line.

$\sf : \implies \: by \: using \: parametric \: form \\ \\\sf : \implies \: x \cos( \alpha ) + y \sin( \alpha ) = p \: \\ \\ \sf : \implies \: p \: = 4 \: \: \: and \: \: \alpha = 60 \degree$

$\sf : \implies \: \cos(60 \degree) = \cos(90 \degree - 30 \degree) \\ \\ \sf : \implies \: \cos(a - b) = \cos(a) \cos(b) + \sin(a) \sin(b) \\ \\ \sf : \implies \: \cos(90 \degree) \cos(30 \degree) + \: \sin(90 \degree) \sin(30 \degree) \\ \\ \sf : \implies \: 0 \times \frac{ \sqrt{3} }{2} + 1 \times \frac{1}{2} \\ \\ \sf : \implies \: \cos(60 \degree) = \frac{1}{2}$

$\sf : \implies \: \sin( \theta) = \sqrt{1 - \cos {}^{2} ( \theta) } \\ \\ \sf : \implies \: \sin(60 \degree) = \sqrt{1 - (\frac{1}{2} ) {}^{2} } \\ \\ \sf : \implies \: \sin(60 \degree) = \sqrt{1 - \frac{1}{4} } \\ \\ \sf : \implies \: \sin(60 \degree) = \frac{ \sqrt{3} }{2}$

$\sf : \implies \: put \: the \: value \: of \: \sin( \theta) \: \: and \: \: \cos( \theta) \: \: and \: \: p \: \: \: in \: \: \: equation \\ \\ \sf : \implies \: x( \cos60 \degree ) + y( \sin 60 \degree ) = 4 \\ \\ \sf : \implies \: \frac{x}{2} + \frac{ \sqrt{3}y }{2} = 4 \\ \\ \sf : \implies \: x \: + \sqrt{3} y = 8$

$\sf : \implies \: \green{{ \underline{equation \: of \: the \: line \: = x + \sqrt{3} y = 8}}}$

Answer 2

Answer:To find the equation of the line.

\begin{gathered}\sf : \implies \: by \: using \: parametric \: form \\ \\\sf : \implies \: x \cos( \alpha ) + y \sin( \alpha ) = p \: \\ \\ \sf : \implies \: p \: = 4 \: \: \: and \: \: \alpha = 60 \degree\end{gathered}

:⟹byusingparametricform

:⟹xcos(α)+ysin(α)=p

:⟹p=4andα=60°

​

\begin{gathered}\sf : \implies \: \cos(60 \degree) = \cos(90 \degree - 30 \degree) \\ \\ \sf : \implies \: \cos(a - b) = \cos(a) \cos(b) + \sin(a) \sin(b) \\ \\ \sf : \implies \: \cos(90 \degree) \cos(30 \degree) + \: \sin(90 \degree) \sin(30 \degree) \\ \\ \sf : \implies \: 0 \times \frac{ \sqrt{3} }{2} + 1 \times \frac{1}{2} \\ \\ \sf : \implies \: \cos(60 \degree) = \frac{1}{2}\end{gathered}

:⟹cos(60°)=cos(90°−30°)

:⟹cos(a−b)=cos(a)cos(b)+sin(a)sin(b)

:⟹cos(90°)cos(30°)+sin(90°)sin(30°)

:⟹0×

2

3

​

​

+1×

2

1

​

:⟹cos(60°)=

2

1

​

​

\begin{gathered}\sf : \implies \: \sin( \theta) = \sqrt{1 - \cos {}^{2} ( \theta) } \\ \\ \sf : \implies \: \sin(60 \degree) = \sqrt{1 - (\frac{1}{2} ) {}^{2} } \\ \\ \sf : \implies \: \sin(60 \degree) = \sqrt{1 - \frac{1}{4} } \\ \\ \sf : \implies \: \sin(60 \degree) = \frac{ \sqrt{3} }{2}\end{gathered}

:⟹sin(θ)=

1−cos

2

(θ)

​

:⟹sin(60°)=

1−(

2

1

​

)

2

​

:⟹sin(60°)=

1−

4

1

​

​

:⟹sin(60°)=

2

3

​

​

​

\begin{gathered}\sf : \implies \: put \: the \: value \: of \: \sin( \theta) \: \: and \: \: \cos( \theta) \: \: and \: \: p \: \: \: in \: \: \: equation \\ \\ \sf : \implies \: x( \cos60 \degree ) + y( \sin 60 \degree ) = 4 \\ \\ \sf : \implies \: \frac{x}{2} + \frac{ \sqrt{3}y }{2} = 4 \\ \\ \sf : \implies \: x \: + \sqrt{3} y = 8\end{gathered}

:⟹putthevalueofsin(θ)andcos(θ)andpinequation

:⟹x(cos60°)+y(sin60°)=4

:⟹

2

x

​

+

2

3

​

y

​

=4

:⟹x+

3

​

y=8

​

\sf : \implies \: \green{{ \underline{equation \: of \: the \: line \: = x + \sqrt{3} y = 8}}}:⟹

equationoftheline=x+

3

​

y=8

​

Step-by-step explanation: