Question

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

Answer 1

Answer:

$\Large{\underline{\underline{\bf{QuEsTiOn:-}}}}$

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

$\mathfrak{\huge{\pink{\underline{\underline{Answer :}}}}}$

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,

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

Equation of line passing points (x₁,y₁) and (x₂,y₂) is

y-y₁ =y₂-y₁/x₂-x₁ (x-x₁)

or, y-asinα = asinβ-asinα/acosβ-acosα (x-acosα)

or, , (y-asinα)(cosβ-cosα) = (sinβ-sinα)(x-acosα)

This is required equation.