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# 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 Askiitians Expert Soumyajit IIT-Kharagpur
28 Points
11 years ago

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]

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Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We

All the best Shubhankar Sinha !!!

Regards,

Soumyajit Das IIT Kharagpur

6 years ago
let the algebraic sum of the perpendicular distance from (3,0),(0,3) and (2,2) to a variable line is zero, then the line passes through a fixed point whose co-ordinates are?
4 years ago
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)
3 years ago
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)