Question

The area of a triangle whose vertices (points) are (0, 0), (3, 0) and (0, 4) ​

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

The area of a triangle whose vertices (points) are (0, 0), (3, 0) and (0, 4)

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{\pink{Given}}}}}$

• Vertices are A(0,0) , B(3,4) , C(0,4)

Let,

$\mapsto\sf{\:x_{1}\:=\:0,\:x_{2}\:=\:3,\:x_{3}\:=\:0}$

And,

$\mapsto\sf{\:y_{1}\:=\:0,\:y_{2}\:=\:4,\:y_{3}\:=\:4}$

$\Large{\underline{\mathfrak{\bf{\pink{Find}}}}}$

• Area of triangle .

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

we know,

$\small\sf{\blue{\:Area_{\triangle\:ABC}\:=\:\dfrac{1}{2}\big[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\big]}}$

Keep all values,

$\mapsto\sf{\:Area_{\triangle\:ABC}\:=\:\dfrac{1}{2}\left(0(4-4)+3(4-0)+0(0-4)\right)} \\ \\ \mapsto\sf{\:Area_{\triangle\:ABC}\:=\:\dfrac{1}{\cancel{2}}.(\cancel{12})} \\ \\ \mapsto\sf{\:Area_{\triangle\:ABC}\:=\:6\:square\:units}$

$\Large{\underline{\mathfrak{\bf{\pink{Hence}}}}}$

$\mapsto\sf{\:Area_{\triangle\:ABC}\:will\:be\:=\:6\:square\:unit}$