let algebraic sum of perpendicular distances from points (2,0) (0,2) (1,1) to a variable straight line be zero then line passes through a fixed point whose coordinates are

Dear Shubhankar Sinha,

Ans:- Let the eq of the line be ax+by+c=0

it's distances from the three points are

D1=(2a+0b+c)/K..................(1)

D2=((0a+2b+c)/K...................(2)

D3=(a+b+c)/K.......................(3)

Where K=√(a²+b²)

Adding 1, 2, 3 we get

D1+D2+D3=(3a+3b+3c)/K

                 =3(a+b+c)/K

As the sum is 0 hence a+b+c=0

So, the line passes through theFixed point (1,1)         [ ax+by+c=0  putting (1,1) point in x,y we get a1+b1+c=a+b+c=0]

 

Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We
are all IITians and here to help you in your IIT JEE preparation.

All the best Shubhankar Sinha !!!

 

Regards,

Askiitians Experts
Soumyajit Das IIT Kharagpur

Let the variable line be px+qy+1=0 then given condition P1+P2+P3=3p+3q+3=0Or p+q+1=0Condition of upper eq show that the variable line passes through the point (1,1)

let eq.will be ax+by+c=0 now alzebric sum of perpendicular distance will 0 so the. Distance is 3a+c/k+3b+c/k+2a+2b+c/k=0. And k=root of a^+b^. Now. Take lcm and after solve we get. 5a+5b+3c=0. Now divided by 3 to make 3is1. 5a/3+5b/3+c=0. So. Comparing on eq.ax+by+c. The point. Is x=5/3. And. y=5/3 so. (x,y)=(5/3,5/3)