`|(x_1,y_1,1),(x_2,y_2,1),(x_3,y_3,1)|=|(a_1,b_1,1),(a_2,b_2,1),(a_3,b_3,1)|` then the two triangles with vertices `(x_(1), y_(1)), (x_(2), y_(2)), (x_(3), y_(3)) and (a_(1), b_(1)), (a_(2), b_(2)), (a_(3), b_(3))` are

If |{:(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1):}|=|{:(a_(1),b_(1),1),(a_(2),b_(2),1),(a_(3),b_(3),1):}| , then the two triangles with vertices (x_(1),y_(1)) , (x_(2),y_(2)) , (x_(3),y_(3)) and (a_(1),b_(1)) , (a_(2),b_(2)) , (a_(3),b_(3)) must be congruent.

The centroid of the triangle with vertices at A(x_(1),y_(1)),B(x_(2),y_(2)),c(x_(3),y_(3)) is

If |x_1y_1 1x_2y_2 1x_3y_3 1|=|a_1b_1 1a_2b_2 1a_3b_3 1| then the two triangles with vertices (x_1, y_1),(x_2,y_2),(x_3,y_3) and (a_1,b_1),(a_2,b_2),(a_3,b_3) are equal to area (b) similar congruent (d) none of these

The following question consist of two stateements, one labelled as the 'Assertion (A)' and the other as 'Reason (R)'. You are to examine these two statement carafully and select the answer. Assertion (A) : If two triangles with vertices (x_(1), y_(1)), (x_(2), y_(2)), (x_(3), y_(3)) and (a_(1), b_(1)), (a_(2), b_(2)), (a_(3), b_(3)) satisfy the relation Reason (R) : For the given triangles satisfying the above relation impolies that the triangles have equal area.

STATEMENT-1: If three points (x_(1),y_(1)),(x_(2),y_(2)),(x_(3),y_(3)) are collinear, then |{:(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1):}|=0 STATEMENT-2: If |{:(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1):}|=0 then the points (x_(1),y_(1)),(x_(2),y_(2)),(x_(3),y_(3)) will be collinear. STATEMENT-3: If lines a_(1)x+b_(1)y+c_(1)=0,a_(2)=0and a_(3)x+b_(3)y+c_(3)=0 are concurrent then |{:(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3)):}|=0

if |[2a,x_1,y_1],[2b,x_2,y_2],[2c,x_3,y_3]|=(abc)/2!=0 then the area of the triangle whose vertices are (x_1/a,y_1/a),(x_2/b,y_2/b),(x_3/c,y_3/c) is (A) (1/4)abc (B) (1/8)abc (C) 1/4 (D) 1/8

If the coordinates of the vertices of an equilateral triangle with sides of length a are (x_(1),y_(1)),(x_(2),y_(2)) and (x_(3),y_(3)) then |(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1)|^2=(3a^(4))/4