Question

find the area of the triangle who's vertices are (5,0) , (0,3) , (0,-4)​

Answer 1

$\bf \underline{ \underline{\maltese\:Given} }$

$\sf \Rrightarrow Vertices \: of \: triangle \: are \: (5,0) , \: (0,3) , \: (0,-4)$

$\bf \underline{ \underline{\maltese \: To \: find } }$

$\sf \Rrightarrow We \: have \: to \: find \: area \: of \: triangle.$

$\bf \underline{ \underline{\maltese \: Solution } }$

$\sf Area \: of \: triangle =$

$\underline{\boxed{ \sf \dfrac{1}{2}\ \times\ \bigg[x_1\bigg(y_2\ -\ y_3\bigg)\ +\ x_2\bigg(y_3\ -\ y_1\bigg)\ +\ x_3\bigg(y_1\ -\ y_2\bigg)\bigg]}}$

$\bf \underline{Where},$

$\implies \sf{x_1 =5}$

$\implies \sf{x_2 =0 }$

$\implies \sf{x_3 =0 }$

$\implies \sf{y_1 =0}$

$\implies \sf{y_2 = 3}$

$\implies \sf{y_3 = - 4 }$

$\bf \underline{Now},$

$\sf Substitute \: the \: values,$

$\sf \dfrac{1}{2}\ \times\ \bigg[5\bigg(3\ -\ ( - 4)\bigg)\ +\ 0\bigg( - 4\ -\ 0\bigg)\ +\ 0\bigg(0\ -\ 3\bigg)\bigg]$

$\sf \dfrac{1}{2}\ \times\ \bigg[5(7)\ +\ 0( - 4)\ +\ 0( -\ 3)\bigg]$

$\sf \dfrac{1}{2}\ \times\ \bigg[35\ +\ 0( - 4)\ +\ 0( -\ 3)\bigg]$

$\sf \dfrac{1}{2}\ \times\ \bigg[35\ +\ 0\ +\ 0\bigg]$

$\sf \dfrac{1}{2}\ \times\ \bigg[35\bigg]$

$\sf \dfrac{1}{2}\ \times\ 35 = \dfrac{35}{2}$

$\underline{ \boxed{ \red{ \bf So, area \: of \:triangle\:is\:\dfrac{35}{2} \: square \: units. }}}$