Question

Find area of triangle whose vertices are (3,5), (5,8) and (1,2).​

Answer 1

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

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

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

$\green{\underline \bold{Given:}} \\ \tt: \implies Coordinate \: of \: A= (3,5) \\ \\ \tt: \implies Coordinate \: of \: B = (5,8) \\ \\ \tt: \implies Coordinate \: of \: C = (1,2) \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Area \: of \: triangle = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies Area \: of \: triangle = \frac{1}{2} |( x_{1}( y_{2} - y_{3}) + x_{2} ( y_{3} - y_{1} ) + x_{3}( y_{1} - y_{2}))| \\ \\ \tt: \implies Area \: of \: triangle = \frac{1}{2}|(3(8 - 2) + 5(2 - 5) + 1(5 - 8))| \\ \\ \tt: \implies Area \: of \: triangle = \frac{1}{2}|(3 \times 6 + 5 \times ( - 3) + 1 \times ( - 3) )|\\ \\ \tt: \implies Area \: of \: triangle = \frac{1}{2}|(18 - 15 - 3)| \\ \\ \tt: \implies Area \: of \: triangle = \frac{1}{2}|(18 - 18)| \\ \\ \tt: \implies Area \: of \: triangle = \frac{1}{2} \times 0 \\ \\ \green{\tt: \implies Area \: of \: triangle = 0 \: units}$

Answer 2

Answer:-

∆area = 1/2 [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)

∆area = 1/2 [3(8 - 2) + 5(2 - 5) + 1(5b- 8)]

∆area = 1/2 (18-15-3)

∆area = 1/2 × 0

∆area = 0

Hence, the area of a triangle is 0 units.

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