Question

If $$P_1,P_2,P_3$$ ​ be the perpendicular from the points $$(m^2,2m),(m{m}^{\prime },m+m^{\prime })$$and $$({m^{\prime }}^2,2m^{\prime })$$ respectively on the line $$x\cos a+y\sin a+{\displaystyle \frac{{\sin }^{2}a}{{\cos }^{2}a}}=0$$, then $$P_1,P_2,P_3$$​ are in
A. A.P.
B. G.P.
C. H.P.
D. None of these

Answer 1

Hint: In order to find the solution of the given question that is if $$P_1,P_2,P_3$$ ​ be the perpendicular from the points $$(m^2,2m),(m{m}^{\prime },m+m^{\prime })$$and $$({m^{\prime }}^2,2m^{\prime })$$ respectively on the line $$x\cos a+y\sin a+{\displaystyle \frac{{\sin }^{2}a}{{\cos }^{2}a}}=0$$, then $$P_1,P_2,P_3$$​ are in A.P./ G.P./H.P./ or none of these. Apply the formula of the perpendicular distance of a line $$ax+by+c$$ from the point $$(x,y)$$ is $$d=\left|{\displaystyle \frac{ax+by+c}{\sqrt{a^2+b^2}}}\right|$$. Now use this formula to find the perpendicular distances $$P_1$$​: given by line $$x\cos a+y\sin a+{\displaystyle \frac{{\sin }^{2}a}{{\cos }^{2}a}}=0$$ from point $$(m^2,2m)$$; $$P_2$$: given by line $$x\cos a+y\sin a+{\displaystyle \frac{{\sin }^{2}a}{{\cos }^{2}a}}=0$$ from point $$(m{m}^{\prime },m+m^{\prime })$$; $$P_3$$: given by line $$x\cos a+y\sin a+{\displaystyle \frac{{\sin }^{2}a}{{\cos }^{2}a}}=0$$ from point $$({m^{\prime }}^2,2m^{\prime })$$. After this check for the conditions if distance $$P_1,P_2,P_3$$ are in A.P. then the difference between any two consecutive terms will be constant. Then check for the conditions if distance $$P_1,P_2,P_3$$ are in G.P. then any element after the first is obtained by multiplying the preceding element by a constant. To check for the conditions if distance $$P_1,P_2,P_3$$ are in H.P. then the reciprocals of the terms of a sequence are in arithmetic progression.

Complete step by step solution:
According to the question, given line in the question is as follows:
$$x\cos a+y\sin a+{\displaystyle \frac{{\sin }^{2}a}{{\cos }^{2}a}}=0...\left(1\right)$$
We know that the perpendicular distance of a line $$ax+by+c$$ from the point $$(x,y)$$ is $$d=\left|{\displaystyle \frac{ax+by+c}{\sqrt{a^2+b^2}}}\right|$$.
Therefore, $$P_1$$​ is the perpendicular distance of given line $$\left(1\right)$$ from point $$(m^2,2m)$$ is as follows:

$$P_1=\left|{\displaystyle \frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}}\right|$$
$$\Rightarrow P_1=\left|{\displaystyle \frac{m^2\cos a+2m\sin a+{\displaystyle \frac{{\sin }^{2}a}{{\cos }^{2}a}}}{\sqrt{{\cos }^{2}a+{\sin }^{2}a}}}\right|$$
$$\Rightarrow P_1=\left|{\displaystyle \frac{m^2{\cos }^{3}a+2m\sin a{\cos }^{2}a+{\sin }^{2}a}{\sqrt{{\cos }^{2}a+{\sin }^{2}a}\left({\cos }^{2}a\right)}}\right|$$
$$\Rightarrow P_1=\left|{\displaystyle \frac{m^2{\cos }^{3}a+2m\sin a{\cos }^{2}a+{\sin }^{2}a}{\sqrt{1}\left({\cos }^{2}a\right)}}\right|$$
$$\Rightarrow P_1=\left|{\displaystyle \frac{m^2{\cos }^{3}a+2m\sin a{\cos }^{2}a+{\sin }^{2}a}{{\cos }^{2}a}}\right|...\left(2\right)$$
We know that the perpendicular distance of a line $$ax+by+c$$ from the point $$(x,y)$$ is $$d=\left|{\displaystyle \frac{ax+by+c}{\sqrt{a^2+b^2}}}\right|$$.
Therefore,$$P_2$$​ is the perpendicular distance of given line $$\left(1\right)$$ from point $$(m{m}^{\prime },m+m^{\prime })$$
$$P_2=\left|{\displaystyle \frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}}\right|$$
$$\Rightarrow P_2=\left|{\displaystyle \frac{m{m}^{\prime }\cos a+m\sin a+m^{\prime }\sin a+{\displaystyle \frac{{\sin }^{2}a}{{\cos }^{2}a}}}{\sqrt{{\cos }^{2}a+{\sin }^{2}a}}}\right|$$
$$\Rightarrow P_2=\left|{\displaystyle \frac{m{m}^{\prime }{\cos }^{3}a+m\sin a{\cos }^{2}a+m^{\prime }\sin a{\cos }^{2}a+{\sin }^{2}a}{\sqrt{{\cos }^{2}a+{\sin }^{2}a}\left({\cos }^{2}a\right)}}\right|$$
$$\Rightarrow P_2=\left|{\displaystyle \frac{m{m}^{\prime }{\cos }^{3}a+m\sin a{\cos }^{2}a+m^{\prime }\sin a{\cos }^{2}a+{\sin }^{2}a}{\sqrt{1}\left({\cos }^{2}a\right)}}\right|$$
$$\Rightarrow P_2=\left|{\displaystyle \frac{m{m}^{\prime }{\cos }^{3}a+m\sin a{\cos }^{2}a+m^{\prime }\sin a{\cos }^{2}a+{\sin }^{2}a}{{\cos }^{2}a}}\right|...\left(3\right)$$
We know that the perpendicular distance of a line $$ax+by+c$$ from the point $$(x,y)$$ is $$d=\left|{\displaystyle \frac{ax+by+c}{\sqrt{a^2+b^2}}}\right|$$.
Therefore,$$P_3$$​ is the perpendicular distance of given line $$\left(1\right)$$ from point $$({m^{\prime }}^2,2m^{\prime })$$
$$P_3=\left|{\displaystyle \frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}}\right|$$
$$\Rightarrow P_3=\left|{\displaystyle \frac{{m^{\prime }}^2\cos a+2m^{\prime }\sin a+{\displaystyle \frac{{\sin }^{2}a}{{\cos }^{2}a}}}{\sqrt{{\cos }^{2}a+{\sin }^{2}a}}}\right|$$
$$\Rightarrow P_3=\left|{\displaystyle \frac{{m^{\prime }}^2{\cos }^{3}a+2m^{\prime }\sin a{\cos }^{2}a+{\sin }^{2}a}{\sqrt{{\cos }^{2}a+{\sin }^{2}a}\left({\cos }^{2}a\right)}}\right|$$
$$\Rightarrow P_3=\left|{\displaystyle \frac{{m^{\prime }}^2{\cos }^{3}a+2m^{\prime }\sin a{\cos }^{2}a+{\sin }^{2}a}{\sqrt{1}\left({\cos }^{2}a\right)}}\right|$$
$$\Rightarrow P_3=\left|{\displaystyle \frac{{m^{\prime }}^2{\cos }^{3}a+2m^{\prime }\sin a{\cos }^{2}a+{\sin }^{2}a}{{\cos }^{2}a}}\right|...\left(4\right)$$
Now, checking for the conditions if distance $$P_1,P_2,P_3$$ are in A.P. then the difference between any two consecutive terms will be constant. But this does not satisfy with $$P_1,P_2,P_3$$. Therefore, $$P_1,P_2,P_3$$ are not in A.P.
Now check for the conditions if distance $$P_1,P_2,P_3$$ are in G.P. then any element after the first is obtained by multiplying the preceding element by a constant.
We can clearly see from the equation $$\left(2\right)$$, $$\left(3\right)$$ and $$\left(4\right)$$, we get:
$${P_2}^2=P_1{P}_3$$
Hence, we can write $$P_1,P_2,P_3$$ in G.P. form.
Therefore, $$P_1,P_2,P_3$$ are in G.P.

So, the correct answer is “Option B”.

Note: Students get confused between the concepts of A.P. and G.P., its important to remember that a sequence is in A.P. if the difference between any two consecutive terms will be constant and a sequence is in G.P. then any element after the first is obtained by multiplying the preceding element by a constant.