Question

Given point A(6,30) and point B(24,6), equation of line AB is 4x+3y=114. Point (0,λ)is a point on y-axis such that 0< λ <38. For all positions of P, angle APB is maximum when point P is.

Answer 1

Answer with explanation:

It is given that , in ΔAPB , we have to find point P on y-axis,such that,  0< λ <38,angle APB is maximum.

Coordinates of A(6,30), B (24,6) and P (0,λ ).

We will use Cosine formula to solve this problem.

⇒2×AP×BP×Cos P=AP²+BP²-AB²

Using Distance formula to, find AP, BP and AB

$AP=\sqrt{(6-0)^2+(30-\lambda )^2}\\\\=\sqrt{36+900+(\lambda )^2-60\lambda}\\\\=\sqrt{(\lambda )^2-60\lambda+936}\\\\BP=\sqrt{(24-0)^2+(6-\lambda)^2}\\\\=\sqrt{576+36+(\lambda)^2-12\lambda}\\\\=\sqrt{612+(\lambda)^2-12\lambda}\\\\AB=\sqrt{(24-6)^2+(6-30)^2}\\\\AB=\sqrt{324+576}\\\\AB=\sqrt{900}\\\\AB=30$

$\Rightarrow 2 \times \sqrt{(\lambda )^2-60\lambda+936} \times\sqrt{612+(\lambda)^2-12\lambda}\times \cos P=(\lambda )^2-60\lambda+936+612+(\lambda)^2-12\lambda-900\\\\\Rightarrow 2 \times \sqrt{(\lambda )^2-60\lambda+936} \times\sqrt{612+(\lambda)^2-12\lambda}\times \cos P=2(\lambda )^2-72\lambda+648\\\\\rightarrow \cos P=\frac{(\lambda )^2-36\lambda+324}{ \sqrt{(\lambda )^2-60\lambda+936} \times\sqrt{612+(\lambda)^2-12\lambda}}\\\\\rightarrow  \cos P =0$

$\rightarrow\frac{(\lambda )^2-36\lambda+324}{ \sqrt{(\lambda )^2-60\lambda+936} \times\sqrt{612+(\lambda)^2-12\lambda}}=0\\\\(\lambda )^2-36\lambda+324=0\\\\\rightarrow[(\lambda ) -18]^2=0\\\\\rightarrow(\lambda )=18$

Position of point P= (0,18)

Answer 2

Answer:

Step-by-step explanation:

Method 1(the basic approach )Dekho koi bi general point lo y axis pe. Uske baad uska angle likh lo tan k Form me using slope formlua and then dono slope kae angle ka tan lelo by formula m1-m2/1+m1-m2. And then u will find that for lambda between (0,38) tan is always positive therefore maximum value tab aaegi jab denominator zero ho jaaega, and that value will be lambda =18.

Method2 (toppers method to solve in 5 seconds): if u will make a rough Diagram then see that by symmetry maximum angle tabhi aaega when y coordinate will be the AM of other y coordinates ( sort of reflection from plane mirror) then u will get the answer lambda =30+6/2=18.....