Question

find the area of triangle, the coordinates of whose vertices are (3,2), (5,7) and (-5,-7)

Answer 1

Let,

(3,2) = (x1,y1)

(5,7) = (x2,y2)

(-5,-7) = (x3,y3)

Now,

Area of triangle = 1/2 [x1 (y2-y3) + x2 (y3-y1) + x3 (y1-y2)]

= 1/2 [3 (7-(-7)) + 5 ((-5)-2) + (-5) (2-7)

= 1/2 [3 (14) + 5 (-7) - 5 (-5)]

= 1/2 [ 42 - 35 + 25]

= 1/2 [32]

= 32/2

= 16 sq. units

I hope this helped you!

Answer 2

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

$\green{\tt{\therefore{Area\:of\:triangle=11\:sq\:units}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{: \implies Coordinate \: of \: A= (3,2) } \\ \\ \tt{: \implies Coordinate \: of \: B = (5,7) } \\ \\ \tt{: \implies Coordinate \: of \: C = (-5,-7) } \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies Area \: of \: triangle = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} | x_{1} ( y_{2} - y_{3}) + x_{2}( y_{3} - y_{1}) + x_{3}( y_{1} - y_{2} ) | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |3(7-(- 7)) + 5(-7 - 2) - 5(2 - 7)| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |3 \times 14 + 5\times -9 -5 \times -5 | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |42 - 45 +25| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} \times 22} \\ \\ \green{\tt{: \implies Area \: of \: triangle =11 \: sq \: units}} \\ \\ \purple{\bold{Some \: formula \: related \: to \: coordinate \: geometery}} \\ \pink{\tt{ \circ \: Distance \: formula = \sqrt{ (x_{2} - x_{1})^{2} + ( y_{2} - y_{1} )^{2} } }} \\ \\ \pink{\tt{ \circ \: Section \: formula = x= \frac{m x_{2} + n x_{1} }{m + n} }}$