Question

Perpendicular distance from the origin to the line joining the points $(a\cos \theta ,a\sin \theta ),(a\cos \varphi ,a\sin \varphi )$ is

• A. $2a\cos (\theta -\varphi )$
• B. $a\cos \left(\frac{\theta -\varphi }{2}\right)$
• C. $4a\cos \left(\frac{\theta -\varphi }{2}\right)$
• D. $a\cos \left(\frac{\theta +\varphi }{2}\right)$

Answer 1

The correct option is B $a\cos \left(\frac{\theta -\varphi }{2}\right)$

Slope of the line
$=\frac{asin\varphi -asin\theta }{acos\varphi -acos\theta }$
Hence the equation of the line will be
$\frac{y-asin\theta }{x-acos\theta }=\frac{asin\varphi -asin\theta }{acos\varphi -acos\theta }$
$\frac{y-asin\theta }{x-acos\theta }=\frac{sin\varphi -sin\theta }{cos\varphi -cos\theta }$
$(cos\varphi -cos\theta )y-(sin\varphi -sin\theta )x-asin\theta (cos\varphi -cos\theta )+acos\theta (sin\varphi -sin\theta )=0$
$(cos\varphi -cos\theta )y-(sin\varphi -sin\theta )x+a(cos\theta sin\varphi -sin\theta cos\varphi )+a(sin\theta cos\theta -cos\theta sin\theta )=0$
$(cos\varphi -cos\theta )y-(sin\varphi -sin\theta )x+a(cos\theta sin\varphi -sin\theta cos\varphi )=0$
$(cos\varphi -cos\theta )y-(sin\varphi -sin\theta )x+a(cos\theta sin\varphi -sin\theta cos\varphi )=0$
$(cos\varphi -cos\theta )y-(sin\varphi -sin\theta )x+asin(\varphi -\theta )=0$
Hence perpendicular distance from the origin will be
$d=\frac{|asin(\varphi -\theta )|}{\sqrt{(cos\varphi -cos\theta {)}^2+(sin\varphi -sin\theta {)}^2}}$
$=\frac{|asin(\varphi -\theta )|}{\sqrt{(co{s}^2\varphi +si{n}^2\varphi )+(co{s}^2\theta +si{n}^2\theta )-2(cos\theta .cos\varphi +sin\theta .sin\theta )}}$
$=\frac{|asin(\varphi -\theta )|}{\sqrt{2-2(cos\theta .cos\varphi +sin\theta .sin\theta )}}$
$=\frac{|asin(\varphi -\theta )|}{\sqrt{2(1-cos(\theta -\varphi \left)\right)}}$
$=\frac{|asin(\varphi -\theta )|}{\sqrt{4si{n}^2\left(\frac{\theta -\varphi }{2}\right)}}$
$=\frac{|asin(\varphi -\theta )|}{2sin\left(\frac{\theta -\varphi }{2}\right)}$
$=\frac{\left|2asin\left(\frac{\varphi -\theta }{2}\right)cos\left(\frac{\varphi -\theta }{2}\right)\right|}{2sin\left(\frac{\theta -\varphi }{2}\right)}$
$=acos\left(\frac{\varphi -\theta }{2}\right)$
$=a\cos \left(\frac{\theta -\varphi }{2}\right)$