Question

find the are of the triangle whose vertices are (2,3),(-10,),(2,-4) is​

Answer 1

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

$\sf Vertices \: of \: the \: triangle \: are :$

$\sf\implies(2, 3)$

$\sf\implies (- 1, 0)$

$\sf\implies(2, - 4)$

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

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

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

$\bf Area \: of \: triangle =$

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

$\bf \underline{Where},$

$\implies \sf{x_1 = 2}$

$\implies \sf{x_2 = - 1}$

$\implies \sf{x_3= 2}$

$\implies \sf{y_1 = 3}$

$\implies \sf{y_2 = 0}$

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

$\bf \underline{Now},$

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

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

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

$\sf \dfrac{1}{2}\ \times\ \bigg[ 8\ -\ (-7) +\ 6\bigg]$

$\sf \dfrac{1}{2}\ \times\ \bigg[ 8 +7 +\ 6\bigg]$

$\sf \dfrac{1}{2}\ \times\ \bigg[ 21\bigg]$

$\sf \dfrac{1}{2}\ \times21 =\dfrac{21}{2}$

$\underline{ \boxed{ \red{ \bf So, area \: of \: triangle \: is \:\dfrac{21}{2} \: square \: units. }}}$