Question

Find the area of a triangle whose verticis are A ( 1 , -1) B ( -4, 6 ) C (-3, 5)​

Answer 1

$\bigstar\leadsto\boxed{\large\bf\red{Area=24\: square\:units }}$

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$\large\bf\underline{Correction\:in\: question:}$

• ➠ Find the area of a triangle whose verticis are A ( 1 , -1) B ( -4, 6 ) C (-3, -5)

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$\large\bf\underline{Given:}$

• ➨ A = ( 1 , -1 )
• ➨ B = ( -4 , 6 )
• ➨ C = ( -3 , -5 )

$\large\bf\underline{Area\:of\: triangle\:ABC:}$

$\underline{\boxed{\bf\pink{= \frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]}}}$

$\large\bf\underline{Here,}$

• $\tt{x_{1}=1}$
• $\tt{x_{2}=-4}$
• $\tt{x_{3}=-3}$
• $\tt{y_{1}=-1}$
• $\tt{y_{2}=6}$
• $\tt{y_{3}=-5}$

$\large\bf\underline{Putting\:values:}$

$\bf{⟼\frac{1}{2}\times [1(6-(-5))+(-4)(-5)-(-1))+(-3)(-1-6)]}$

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$\bf{⟼\frac{1}{2}\times [1(6-(-5))+(-4)((-5)+1)+(-3)(-7)]}$

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$\bf{⟼\frac{1}{2}\times [1(6+5)+(-4)(-5+1)+(-3)(-7)]}$

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$\bf{⟼\frac{1}{2}\times [1(11)+(-4)(-4)+(-3)(-7)]}$

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$\bf{⟼\frac{1}{2}\times[11+16+21]}$

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$\bf{⟼\frac{1}{2}\times 48}$

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$\bf{⟼24}$

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$\underline{\boxed{\bf{Hence,\:area=24 \:square\:units }}}$

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