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Question

The are of ABC with vertices A(a, 0), O(0, 0) and B(0, b) in square units is

(a) ab                               (b)  12ab                         (c) 12a2b2                        (d)   12b2


Solution

A (a , 0), O (0 , 0) and B (0 , b)  are the vertices of ΔABC. Then

(x1=a,y1=0),(x2=0,y2=0),(x3=0,y3=b)

Area of triangle ABC

12[x1(y2y3)+x2(y3y1)+x3(y1y2)]

12[a(0b)+0(b0)+0(00)]

=  12[a(b)]

=     |12[a(b)]|

=   12ab

answer :- (b) 12ab 

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