Find the coordinates of the points of intersection of the straight lines whose equations are
$x cos ϕ_{1} + y sin ϕ_{1} = a$ and $x cos ϕ_{2} + y sin ϕ_{2} = 1$ .

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Solution

$x cos ϕ_{1} + y sin ϕ_{1} = a . . . . . (i) x cos ϕ_{2} + y sin ϕ_{2} = 1 x cos ϕ_{2} = 1 - y sin ϕ_{2} x = \frac{1 - y sin ϕ_{2}}{cos ϕ_{2}}$

substituting $x$ in $(i)$

$(\frac{1 - y sin ϕ_{2}}{cos ϕ_{2}}) cos ϕ_{1} + y sin ϕ_{1} = a \frac{cos ϕ_{1} - y cos ϕ_{1} sin ϕ_{2} + y sin ϕ_{1} cos ϕ_{2}}{cos ϕ_{2}} = a (sin ϕ_{1} cos ϕ_{2} - cos ϕ_{1} sin ϕ_{2}) y + cos ϕ_{1} = a cos ϕ_{2} sin (ϕ_{1} - ϕ_{2}) y = a cos ϕ_{2} - cos ϕ_{1}$
$\Rightarrow y = \frac{a cos ϕ_{2} - cos ϕ_{1}}{sin (ϕ_{1} - ϕ_{2})}$
$x = \frac{1 - y sin ϕ_{2}}{cos ϕ_{2}}$
$x = \frac{1 - (\frac{a cos ϕ_{2} - cos ϕ_{1}}{sin (ϕ_{1} - ϕ_{2})}) sin ϕ_{2}}{cos ϕ_{2}}$
$x = \frac{sin (ϕ_{1} - ϕ_{2}) - (a cos ϕ_{2} - cos ϕ_{1}) sin ϕ_{2}}{sin (ϕ_{1} - ϕ_{2}) cos ϕ_{2}}$
$x = \frac{sin (ϕ_{1} - ϕ_{2}) - a cos ϕ_{2} sin ϕ_{2} + cos ϕ_{1} sin ϕ_{2}}{sin (ϕ_{1} - ϕ_{2}) cos ϕ_{2}}$

So the point of intersection is $(\frac{sin (ϕ_{1} - ϕ_{2}) - a cos ϕ_{2} sin ϕ_{2} + cos ϕ_{1} sin ϕ_{2}}{sin (ϕ_{1} - ϕ_{2}) cos ϕ_{2}}, \frac{a cos ϕ_{2} - cos ϕ_{1}}{sin (ϕ_{1} - ϕ_{2})})$

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