Statement 1: If the sum of algebraic distances from point `A(1,1),B(2,3),C(0,2)` is zero on the line `a x+b y+c=0,` then `a+3b+c=0` Statement 2: The centroid of the triangle is (1, 2)

Statement 1: If the sum of algebraic distances from point A(1,1),B(2,3),C(0,2) is zero on the line ax+by+c=0, then a+3b+c=0 statement 2: The centroid of the triangle is (1,2)

Statement I If sum of algebraic distances from points A(1,2),B(2,3),C(6,1) is zero on the line ax+by+c = 0 then 2a+3b + c = 0 , Statement II The centroid of the triangle is (3,2)

Statement I If sum of algebraic distances from points A(1,2),B(2,3),C(6,1) is zero on the line ax+by+c = 0 then 2a+3b + c = 0 , Statement II The centroid of the triangle is (3,2)

Statement I If sum of algebraic distances from points A(1,2),B(2,3),C(6,1) is zero on the line ax+by+c = 0 then 2a+3b + c = 0 , Statement II The centroid of the triangle is (3,2)

The algebraic sum of the perpendicular distances from the points A(-2,0),B(0,2) and C(1,1) to a variable line be zero,then all such lines

The algebraic sum of the perpendicular distances from A(x_1, y_1) , B(x_2, y_2) and C(x_3, y_3) to a variable line is zero. Then the line passes through (A) the orthocentre of triangleABC (B) centroid of triangleABC (C) incentre of triangleABC (D) circumcentre of triangleABC

L is a variable line such that the algebraic sum of the distances of the points (1,1) (2,0) and (0,2) from the line is equal to zero. The line L will always pass through a. (1,1) b. (2,1) c. (1,2) d. none of these

L is a variable line such that the algebraic sum of the distances of the points (1,1) (2,0) and (0,2) from the line is equal to zero. The line L will always pass through a. (1,1) b. (2,1) c. (1,2) d. none of these

The algebraic sum of distances of the line ax + by + 2 = 0 from (1,2), (2,1) and (3,5) is zero and the lines bx - ay + 4 = 0 and 3x + 4y + 5=0 cut the coordinate axes at concyclic points. Then (a) a+b=-2/7 (b) area of triangle formed by the line ax+by+2=0 with coordinate axes is 14/5 (c) line ax+by+3=0 always passes through the point (-1,1) (d) max {a,b}=5/7

The algebraic sum of distances of the line ax + by + 2 = 0 from (1,2), (2,1) and (3,5) is zero and the lines bx - ay + 4 = 0 and 3x + 4y + 5=0 cut the coordinate axes at concyclic points. Then (a) a+b=-2/7 (b) area of triangle formed by the line ax+by+2=0 with coordinate axes is 14/5 (c) line ax+by+3=0 always passes through the point (-1,1) (d) max {a,b}=5/7