Let the angles made with the positive x-axis by two straight lines drawn from the point P(2,3) and meeting the line x+y=6 at a distance 32 from the point P be θ1 and θ2
Solutions · 1
Official solution
Let any point Q(t,6−t) on line x+y=6 such that.
distance PQ=32⇒PQ2=32.
using distance formula for points P(2,3) and Q(t,6−t).
(2−t)2+(3−6+t)2=32(2−t)2+(t−3)2=324+t2−4t+t2+9−6t=322t2−10t+13=326t2−30t+39−2=06t2−30t+37=0
let t1&t2 are roots of this equation.
t1+t2=630=5t1t2=637.
So, there are two point Q1(t1,6−t1)&Q2(t2,6−t2).
It is given that angle made with positive x -axis by straight lines from P(2,3) to Q1&Q2 is θ1&θ2 respectively.
This means tanθ1= slope of line PQ1.
tanθ1=2−t13−(6−t1)⇒θ1=2−t1t1−3
similarly, tanθ2= slope of line PQ2tanθ2=2−t2t2−3⇒θ2=tan−1(2−t2t2−3)
adding equation (1) \& (2)
θ1+θ2=tan−1(2−t1t1−3)+tan−1(2−t2t2−3)=tan−1((∣1−(2−t1t1−3)(2−t2t2−3)2−t1t1−3+2−t2t2−3))∣,tan−1x+tan−1y=tan−1(1−xyx+y)=tan−14−2(t1+t2)+t1t2−(9−3(t1+t2)+t1t2)2t1−6−t1t2+3t1+2t2−6−t1t2+3t1θ1+θ2=tan−1(t1+t2)−55(t1+t2)−2t1t2−12
substitute value of t1+t2=5&t1t2=637θ1+θ2=tan−15−55×5−2×637−12θ1+θ2=tan−1(∞)=2π