Question

THE AREA OF TRIANGLE OAB WHERE O IS THE ORIGIN AND A = (A,0),B=(0.B) IS 1) [AB]SQ UNITS 2) [AB]/3 SQ UNITS 3) 1/2[AB] SQ UNITS 4) 1/3[A+B]SQ UNITS?​

Answer 1

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$\boxed {3)\: \: \frac{1}{2} ab \: sq \: units}$

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${\huge{\underline{\sf {\pink{Explanation :}}}}}$

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Vertices of the triangle = (0 , 0) (a , 0) (0 , b)

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$\sf \: area = \frac{1}{2} |{x}_{1}({y}_{2} - {y}_{3}) + {x}_{2}({y}_{3} - {y}_{1} ) + {x}_{3}({y}_{1} - {y}_{2})|$

↬$\sf \: area \: = \frac{1}{2} |0(0 - b) + a(b - 0) + 0(0 - 0)|$

↬$\sf \: area = \frac{1}{2} |0 + ab + 0|$

↬$\sf \: area \: = \frac{1}{2} ab \: \: \: sq. \: units$

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${\huge{\underline{\sf {\pink{More \: Information :}}}}}$

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• $\sf \: distance \: formula = \sqrt{({x}_{2} - {x}_{1})^{2} +( {y}_{2} - {y}_{1})^{2}}$

• $\sf \: mid \: point \: formula \: = (\frac{{x}_{1} + {x}_{2}}{2} )( \frac{{y}_{1} + {y}_{2}}{2} )$