Question

Find the area of triangle with verticle(1,1),(2,3) and (4,5)​

Answer 1

Explanation:

check the poto for the answer

Answer 2

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

$\sf Vertices = (1,1) \: , \: (2,3) \: and \: (4,5)$

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

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

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

$\sf Let \: the \: vertices \: be :$

$\bf \implies A(1,1) \: , \: B(2,3) \: and \:C (4,5)$

$\bf \underline{Now},$

$\underline{\boxed{\bf Area \ of \ triangle = \dfrac{1}{2} \times | \ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \ |}}$

$\bf \underline{Where},$

$\implies \: \sf x_1 = 1$

$\implies \: \sf x_2 = 2$

$\implies \: \sf x_3 = 4$

$\implies \: \sf y_1 = 1$

$\implies \: \sf y_2 = 3$

$\implies \: \sf y_3 = 5$

$\sf Now, substituting \: the \: values,$

$\sf\implies \dfrac{1}{2} \times | \ 1(3-5)+2(5-1)+4(1-3) \ |$

$\sf\implies \dfrac{1}{2} \times | \ 1(-2)+2(4)+4( - 2) \ |$

$\sf\implies \dfrac{1}{2} \times | -2+8+ (- 8) \ |$

$\sf\implies \dfrac{1}{2} \times | - 2 \ |$

$\sf\implies { \dfrac{1}{\cancel{2}} \times\cancel{2} } = 1$

$\sf \implies 1 \: units^{2}$

$\underline{ \boxed{ \bf \red{Area \: of \: the \: triangle = 1 \: units^{2} }}}$