Question

Find equation of line Passing through points (a cos alpha,a sin alpha) and ( a cis bita,a sin beta)

Answer 1

The equation of line is,

$\large{\cot(\frac{ \alpha + \beta }{2})) x + y - asin \alpha - acos \alpha ( \cot( \frac{ \alpha + \beta }{2})) = 0 \: }$

Given points :

( a cosα, a sinα), ( a cosβ, a sinβ)

The equation of line passing through,

( x₁, y₁), (x₂, y₂) is,

y - y₁ = m ( x - x₁)

Here, m = (y₂ - y₁) ÷ (x₂ - x₁)

So for the given points, Let's find the slope m.

$m = \dfrac{ a \sin( \beta ) - a \sin( \alpha ) }{a \cos( \beta ) - a \cos( \alpha ) } \\ \\ m = \dfrac{ a (\sin( \beta ) - \sin( \alpha )) }{a( \cos( \beta ) - \cos( \alpha ) ) } \\ \\ m = \dfrac{ \sin( \beta ) - \sin( \alpha ) }{ \cos( \beta ) - \cos( \alpha ) } \\ \\ m \: = \frac{2 \cos( \frac{ \alpha + \beta }{2}) . \sin(\frac{ \alpha - \beta }{2}) }{ - 2 \sin( \frac{ \alpha + \beta }{2}) . \sin(\frac{ \alpha - \beta }{2})} \\ \\ m = - \cot (\frac{ \alpha + \beta }{2})$

Now, The equation of line joining the points is,

$y - a \sin( \alpha ) = - \cot(\frac{ \alpha + \beta }{2}))(x - a \cos( \alpha ) \\ \\ \cot(\frac{ \alpha + \beta }{2})) x + y - asin \alpha - acos \alpha ( \cot( \frac{ \alpha + \beta }{2})) = 0$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

The equation of any line passing through the points

$(x_1, y_1) \: \: and \: \: (x_2, y_2) \: \: is \:$

$\sf{\displaystyle \: \frac{y - y_1 }{x -x_1 } = \frac{y_2 - y_1}{x_2 - x_1}}$

TO DETERMINE

The equation of the line passing through the points

$\sf{(a \cos \alpha \: , \: a \sin \alpha ) \: \: and \: \: (a \cos \beta \: , \: a \sin \beta )}$

CALCULATION

The equation of the line passing through the points

$\sf{(a \cos \alpha \: , \: a \sin \alpha ) \: \: and \: \: (a \cos \beta \: , \: a \sin \beta )} \: \: is \:$

$\displaystyle \: \sf {\: \frac{y - a \sin \alpha}{x -a \cos \alpha } = \frac{a \sin \beta - a \sin \alpha}{a \cos \beta \: -a \cos \alpha }\:}$

$\implies \displaystyle \: \sf {\: \frac{y - a \sin \alpha}{x -a \cos \alpha } = \frac{ \sin \beta - \sin \alpha}{ \cos \beta \: - \cos \alpha }\:}$

$\implies \displaystyle \: \sf {\: \frac{y - a \sin \alpha}{x -a \cos \alpha } = \frac{2 \cos \: \frac{ \beta + \alpha }{2} \sin \: \frac{ \beta - \alpha }{2} }{2 \sin \: \frac{ \beta + \alpha }{2} \sin \: \frac{ \alpha - \beta }{2} } \: \:}$

$\implies \displaystyle \: \sf {\: \frac{y - a \sin \alpha}{x -a \cos \alpha } = - \frac{ \cos \: \frac{ \beta + \alpha }{2} }{ \sin \: \frac{ \beta + \alpha }{2} } \: \:}$

$\implies \displaystyle \: \sf {\: \frac{y - a \sin \alpha}{x -a \cos \alpha } = - \frac{ \cos \: \frac{ \beta + \alpha }{2} }{ \sin \: \frac{ \beta + \alpha }{2} } \: \:}$

$\displaystyle \sf{\implies \: y \sin \: \frac{ \beta + \alpha }{2}\: - a \sin \alpha \sin \: \frac{ \beta + \alpha }{2} = - x \: \cos \: \frac{ \beta + \alpha }{2} + a \cos \alpha \cos \: \frac{ \beta + \alpha }{2} \: }$

$\displaystyle \sf{\implies \: x \: \cos \: \frac{ \beta + \alpha }{2} + y \sin \: \frac{ \beta + \alpha }{2}\: = a \cos \alpha \cos \: \frac{ \beta + \alpha }{2} + a \sin \alpha \sin \: \frac{ \beta + \alpha }{2} \: }$

$\displaystyle \sf{\implies \: x \: \cos \: \frac{ \beta + \alpha }{2} + y \sin \: \frac{ \beta + \alpha }{2}\: = a ( \cos \alpha \cos \: \frac{ \beta + \alpha }{2} + \sin \alpha \sin \: \frac{ \beta + \alpha }{2}) \: }$

$\displaystyle \sf{\implies \: x \: \cos \: \frac{ \beta + \alpha }{2} + y \sin \: \frac{ \beta + \alpha }{2}\: = a \cos( \alpha - \frac{ \beta + \alpha }{2} )\: }$

$\displaystyle \sf{\implies \: x \: \cos \: \frac{ \beta + \alpha }{2} + y \sin \: \frac{ \beta + \alpha }{2}\: = a \cos( \frac{ \alpha - \beta }{2} )\: }$