Question

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

(a) ab (b) $\frac{1}{2}ab$ (c) $\frac{1}{2}{a}^2{b}^2$ (d) $\frac{1}{2}{b}^2$

Answer 1

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

$(x_1=a,y_1=0),(x_2=0,y_2=0),(x_3=0,y_3=b)$

Area of triangle ABC

= $\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

= $\frac{1}{2}\left[a\right(0-b)+0(b-0)+0(0-0\left)\right]$

= $\frac{1}{2}\left[a\right(-b\left)\right]$

= $|\frac{1}{2}\left[a\right(-b\left)\right]|$

= $\frac{1}{2}ab$

answer :- (b) $\frac{1}{2}ab$