Question

find area of the triangle whose vertices are A (10,-6) ,B(2,5) and c (-1,3)​

Answer 1

24.5 sq unit

by putting the area of triangle formulae

Answer 2

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

$\green{\tt{\therefore{Area\:of\:triangle=24.5\:sq\:units}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{: \implies Coordinate \: of \: A= (10,-6) } \\ \\ \tt{: \implies Coordinate \: of \: B = (2,5) } \\ \\ \tt{: \implies Coordinate \: of \: C = (-1,3) } \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies Area \: of \: triangle = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} | x_{1} ( y_{2} - y_{3}) + x_{2}( y_{3} - y_{1}) + x_{3}( y_{1} - y_{2} ) | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |10(5- 3) + 2(3-( - 6))- 1(-6 - 5)| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |10 \times 2 + 2\times 9 -1 \times -11| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |20+ 18 +11| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} \times 49} \\ \\ \green{\tt{: \implies Area \: of \: triangle =24.5 \: sq \: units}} \\ \\ \purple{\bold{Some \: formula \: related \: to \: coordinate \: geometery}} \\ \pink{\tt{ \circ \: Distance \: formula = \sqrt{ (x_{2} - x_{1})^{2} + ( y_{2} - y_{1} )^{2} } }} \\ \\ \pink{\tt{ \circ \: Section \: formula = x= \frac{m x_{2} + n x_{1} }{m + n} }}$