Question

Find the area of triangle whose vertices are (3,4) (5,6) (7,8)

Answer 1

Area =1/2*[x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2)
=0.5*[4*(3-1)+1*(1-7)+5*(7-3)]
=0.5*(4*2+1*(-6)+5*4)
=0.5*(8-6+20)
=0.5*22
=11

Answer 2

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

$\green{\tt{\therefore{Area\:of\:triangle=0\:sq\:units}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{: \implies Coordinate \: of \: A= (3,4) } \\ \\ \tt{: \implies Coordinate \: of \: B = (5,6) } \\ \\ \tt{: \implies Coordinate \: of \: C = (7,8) } \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies Area \: of \: triangle = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} | x_{1} ( y_{2} - y_{3}) + x_{2}( y_{3} - y_{1}) + x_{3}( y_{1} - y_{2} ) | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |3(6 - 8) + 5(8 - 4) + 7(4- 6)| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |3\times -2 + 5\times 4 + 7 \times -2 | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |-6+ 20 -14| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} \times 0} \\ \\ \green{\tt{: \implies Area \: of \: triangle =0 \: sq \: units}} \\ \\ \purple{\bold{Some \: formula \: related \: to \: coordinate \: geometery}} \\ \pink{\tt{ \circ \: Distance \: formula = \sqrt{ (x_{2} - x_{1})^{2} + ( y_{2} - y_{1} )^{2} } }} \\ \\ \pink{\tt{ \circ \: Section \: formula = x= \frac{m x_{2} + n x_{1} }{m + n} }}$