Question

find the area of the triangle whose vertices are (0,0),(1,2),(4,3)

Answer 1

Answer:

Let A(0,0)=(x1,y1), B(1,2)=(x2,y2) and C(4,3)=(x3,y3)

ar(ABC)=1÷2{x1(y2-y3)+x2(y3-y1)+x3(y1-y2)}

ar(ABC)=1÷2{0(2-3)+1(3-0)+4(0-2)}

ar(ABC)=1÷2{0(-1)+1(3)+4(-2)}

ar(ABC)=1÷2(0+3-8)

ar(ABC)=1÷2(-5)

ar(ABC)=-2.5 sq. units

Step-by-step explanation:

Answer 2

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

$\green{\tt{\therefore{Area\:of\:triangle=2.5\:sq\:units}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{: \implies Coordinate \: of \: A= (0,0) } \\ \\ \tt{: \implies Coordinate \: of \: B = (1,2) } \\ \\ \tt{: \implies Coordinate \: of \: C = (4,3) } \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies Area \: of \: triangle = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} | x_{1} ( y_{2} - y_{3}) + x_{2}( y_{3} - y_{1}) + x_{3}( y_{1} - y_{2} ) | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |0(2 - 3) + 1(3 - 0) + 4(0- 2)| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |0\times -1 + 1\times 3 + 4 \times -2 | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |-5| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} \times 5} \\ \\ \green{\tt{: \implies Area \: of \: triangle =2.5 \: sq \: units}} \\ \\ \purple{\bold{Some \: formula \: related \: to \: coordinate \: geometery}} \\ \pink{\tt{ \circ \: Distance \: formula = \sqrt{ (x_{2} - x_{1})^{2} + ( y_{2} - y_{1} )^{2} } }} \\ \\ \pink{\tt{ \circ \: Section \: formula = x= \frac{m x_{2} + n x_{1} }{m + n} }}$