Question

find the area of triangle whose vertices are (-1, 3), (2, 7), (0, 0)​

Answer 1

Given vertices of a triangle :-

• (-1,3)
• (2,7)
• (0,0)

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We know that, area of triangle with given vertices is given by

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$\star{\boxed{\sf{\orange{Area = \dfrac{1}{2} \bigg| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \bigg|}}}}$

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Here,

• $\sf{x_1 = -1\: , \: y_1 = 3}$
• $\sf{x_2 = 2\: , \: y_2 = 7}$
• $\sf{x_3 = 0 \: , \: y_3 = 0}$

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Putting the values

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$\tt:\implies{Area = \dfrac{1}{2} \bigg| -1(7 - 0) + 2(0 - 3) + 0(3 - 7) \bigg|}$

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$\tt:\implies{Area = \dfrac{1}{2} \bigg| -1(7) + 2(-3) + 0(-4) \bigg|}$

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$\tt:\implies{Area = \dfrac{1}{2} \bigg| -7 - 6 + 0 \bigg|}$

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$\tt:\implies{Area = \dfrac{1}{2} \bigg| -13 \bigg|}$

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$\tt:\implies{Area = \dfrac{1}{2} \times 13}$

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$\bf:\implies{Area = \dfrac{13}{2}}$

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Hence,

• Area of the given triangle is 13/2 square units