Question

Prove that the length of perpendiculars from P (m^2,2m), Q (mn, m+n) and R(n^2,m 2n)
to the line x cos^2 + y sin cos + sin^2 = 0 are in G.P.​

Answer 1

Solution:

Given: Equation of line is x cos²θ + ysinθcosθ +sin²θ = 0 and points are P (m²,2m) , Q (mn,m+n) and R (n²,2n)

w.k.t., perpendicular distance of a point (x₁,y₁) from line ax+by+c = 0 is $\tt{ | \dfrac{ax_{1}+by_{1}+c}{\sqrt{a^{2}+b^{2}}} | }$

So, perpendicular distance of point P (m²,2m) from line is

$\tt{ d_{P} = | \dfrac{m^{2}cos^{2}\theta+2msin\theta cos\theta+sin^{2}\theta}{\sqrt{(cos^{2}\theta)^{2}+(sin\theta cos\theta)^{2}}} | }$

Similarly, perpendicular distance of point Q (mn,m+n) from line is

$\tt{ d_{Q} = | \dfrac{m\,ncos^{2}\theta+(m+n)sin\theta cos\theta+sin^{2}\theta}{\sqrt{(cos^{2}\theta)^{2}+(sin\theta cos\theta)^{2}}} | }$

And, perpendicular distance of point R (n²,2n) from line is

$\tt{ d_{R} = | \dfrac{n^{2}cos^{2}\theta+2nsin\theta cos\theta+sin^{2}\theta}{\sqrt{(cos^{2}\theta)^{2}+(sin\theta cos\theta)^{2}}} | }$

w.k.t., if a, b and c are in G.P., then b² = ac.

Now, $\tt{ d_{P} \times d_{R} = | \ \dfrac{(m^{2}cos^{2}\theta+2msin\theta cos\theta+sin^{2}\theta)}{\sqrt{(cos^{2}\theta)^{2}+(sin\theta cos\theta)^{2}}} \times \dfrac{(n^{2}cos^{2}\theta+2nsin\theta cos\theta+sin^{2}\theta)}{\sqrt{(cos^{2}\theta)^{2}+(sin\theta cos\theta)^{2}}} \ | }$

On simplification, we get

$\tt{ d_{P} \times d_{R} = \ \dfrac{[\ mncos^{2}\theta \ +\ (m+n)sin\theta cos\theta \ +\ sin^{2}\theta \ ]^{2}}{(cos^{2}\theta)^{2}+(sin\theta cos\theta)^{2}} \ }$          ----------[1]

Also, $\tt{ (d_{Q})^{2}} = \ \dfrac{[\ mncos^{2}\theta \ +\ (m+n)sin\theta cos\theta \ +\ sin^{2}\theta \ ]^{2}}{(cos^{2}\theta)^{2}+(sin\theta cos\theta)^{2}} \ }$          ----------[2]

From equations [1] and [2],

$\tt{ (d_{Q})^{2} = d_{P} \times d_{R} }$

Hence, $\textt{ d_{P} \ ,\ d_{Q} \ and\ d_{R}\ }$ are in G.P.

Hence proved