Question

find the area of the triangle whose vertices are (0,0),(5,3) and (1,9)​

Answer 1

the area of the triangle (0,0) (5,3) (1,9) is 21

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Area\:of\:triangle=21\:sq\:units}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ \tt{: \implies Coordinate \: of \: A= (0,0) } \\ \\ \tt{: \implies Coordinate \: of \: B = (5,3) } \\ \\ \tt{: \implies Coordinate \: of \: C = (1,9) } \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies Area \: of \: triangle = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} | x_{1} ( y_{2} - y_{3}) + x_{2}( y_{3} - y_{1}) + x_{3}( y_{1} - y_{2} ) | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |0(3- 9) + 5(9 - 0) + 1(0 - 3)| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |0 \times -6 + 5\times 9 + 1 \times -3 | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |0 + 45-3| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} \times 42} \\ \\ \green{\tt{: \implies Area \: of \: triangle =21 \: sq \: units}} \\ \\ \purple{\bold{Some \: formula \: related \: to \: coordinate \: geometery}} \\ \pink{\tt{ \circ \: Distance \: formula = \sqrt{ (x_{2} - x_{1})^{2} + ( y_{2} - y_{1} )^{2} } }} \\ \\ \pink{\tt{ \circ \: Section \: formula = x= \frac{m x_{2} + n x_{1} }{m + n} }}$