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Exercise C · Q2

Q.Find the area of the triangle with vertices (−2,−3)(-2,-3), (−1,−8)(-1,-8) and (3,2)(3,2).

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✓ Free question

Using the determinant/coordinate formula, the area is 1515 sq units.

Area of a triangle with vertices (x1,y1),(x2,y2),(x3,y3)(x_1,y_1),(x_2,y_2),(x_3,y_3):

Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣.\text{Area}=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.

  1. Take (x1,y1)=(−2,−3), (x2,y2)=(−1,−8), (x3,y3)=(3,2)(x_1,y_1)=(-2,-3),\ (x_2,y_2)=(-1,-8),\ (x_3,y_3)=(3,2).
  2. Substitute:

Area=12∣(−2)(−8−2)+(−1)(2−(−3))+3(−3−(−8))∣\text{Area}=\frac12\left|(-2)(-8-2)+(-1)(2-(-3))+3(-3-(-8))\right|

  1. Simplify each term: (−2)(−10)=20(-2)(-10)=20, (−1)(5)=−5(-1)(5)=-5, 3(5)=153(5)=15.
  2. Area=12∣20−5+15∣=12∣30∣=15\text{Area}=\dfrac12\left|20-5+15\right|=\dfrac12\left|30\right|=15.
✓Final answer

Area =15=15 square units.

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