In fig., the area of a triangle ABC$( in\ sq.\ units)$ is:
$( A) \ 15$
$( B) \ 10$
$( C) \ 7.5$
$( D) \ 2.5$
"\n
Given: In figure three vertices A, B and C.
What to do: To find the area of a triangle ABC$( in\ sq.\ units)$.
Solution:
And we know that area of a triangle with vertices $( x_{1} ,\ y_{1})$, $(x_{2},\ y_{2})$ and $(x_{3},\ y_{3})$
$\frac{1}{2}[ x_{1}( y_{2} -y_{3}) +x_{2}( y_{3} -y_{1}) +x_{3}( y_{1} -y_{2})]$
Here as shown in the fig. vertices $A( 1,\ 3) ,\ B( -1,\ 0) \ and\ C( 4,\ 0)$}
$\therefore$ Area of $\vartriangle ABC=\frac{1}{2}[ 1( 0-0) -1( 0-3) +4( 3-0)]$
$=\frac{1}{2}( 0+3+12)$
$=\frac{15}{2}$
$=7.5$ sq.unit
$\therefore$ Option $( C)$ is correct.
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