If `=|(2a,x_(1),y_(1)),(2b,x_(2),y_(2)),(2c,x_(3),y_(3))|=abc/2 ne 0`then the area 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:

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)and ((x_(3))/c,(y_(3))/c)

Suppose a,b,cgt0 . If |(x_(1),y_(1),2a),(x_(2),y_(2),2b),(x_(3),y_(3),2c)|=abc , Then area of triangle whose vertices are A(x_(1)/a,(y_(1))/a),B((x_(2))/b,(y_(2))/b) and C((x_(3))/c,(y_(3))/c) is ___

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 |{:(2a, x _(1), y _(1)), ( 2b, x _(2), y _(2)), (2c, x _(3), y _(3)):}|= (abc)/(2) ne 0, then the area of the triangle whose vertices are ((x _(1))/(a) , ( y _(1))/(a)) , ( (x _(2))/(b), ( y _(2))/( b) ) and ((x _(3))/( c), (y _(3))/( c)) is

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

The area of the triangle OAB with vertices O(0,0),A(x_(1),y_(1))B(x_(2),y_(2)) is

Write the formula for the area of the triangle having its vertices at (x_(1),y_(1)),(x_(2),y_(2)) and (x_(3),y_(3))

If A(x_(1),y_(1)),B(x_(2),y_(2)),C(x_(3),y_(3)) are the vertices of the triangle then show that:'