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Find the area of the triangle whose vertices are:
(i) $(2, 3), (-1, 0), (2, -4)$
(ii) $(-5, -1), (3, -5), (5, 2)$
To do:
We have to find the area of the given triangles.
Solution:
(i) Let $A (2, 3), B (-1, 0)$ and $C (2, -4)$ be the vertices of a $\triangle ABC$.
We know that,
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle $ABC=\frac{1}{2}[2(0+4)+(-1)(-4-3)+2(3-0)]$
$=\frac{1}{2}[8+7+6]$
$=\frac{21}{2}$
$=10.5$ sq. units.
The area of the given triangle is $10.5$ sq. units.
(ii) Let $A (-5, -1), B (3, -5)$ and $C (5, 2)$ be the vertices of a $\triangle ABC$.
We know that,
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle $ABC=\frac{1}{2}[(-5)(-5-2)+3(2+1)+5(-1+5)]$
$=\frac{1}{2}[35+9+20]$
$=\frac{1}{2} \times 64$
$=32$ sq. units.
The area of the given triangle is $32$ sq. units.
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