1
Question
Find the area of the triangle whose vertices are (3,2), (−2,−3) and (2,3).
Find the area of the triangle whose vertices are (3,2), (−2,−3) and (2,3).
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Solution
The correct option is D 5 sq.unit
Area of a triangle with vertices (x1,y1); (x2,y2) and (x3,y3)=∣∣∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)2∣∣∣
Hence, substituting the points (x1,y1)=(3,2) ; (x2,y2)=(−2,−3) and (x3,y3)=(2,3) in the area formula, we get
Area = ∣∣∣3(−3−3)+(−2)(3−2)+2(2−(−3))2∣∣∣=5 sq. unit
Area of a triangle with vertices (x1,y1); (x2,y2) and (x3,y3)
=∣∣∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)2∣∣∣
Hence, substituting the points (x1,y1)=(3,2) ; (x2,y2)=(−2,−3) and (x3,y3)=(2,3)
in the area formula, we get
Area = ∣∣∣3(−3−3)+(−2)(3−2)+2(2−(−3))2∣∣∣=5 sq. unit
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