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Question 38 Find the Area of a Triangle whose verticesare (-3, 1), (1, -3) and (2,3)7 Sq.Units9 Sq.Units11 Sq.Units14 Sq.Units |
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Answer» Step-by-step explanation: Given:-Given POINTS are (-3, 1), (1, -3) and (2,3) To find:-Find the Area of a TRIANGLE whose VERTICES are (-3, 1), (1, -3) and (2,3) ? Solution:-Given points are (-3, 1), (1, -3) and (2,3) Let (x1, y1)=(-3,1)=>x1=-3 and y1 = 1 Let (x2, y2)=(1,-3)=>x2=1 and y2 = -3 Let (x3, y3)=(2,3)=>x3=2 and y3=3 We know that The Area of a triangle formed by whose vertices are( x1 ,y1 )and (x2, y2 )and (x3, y3) is ∆=(1/2) | x1(y2-y3) +x2(y3-y1) + x3(y1-y2) | sq.units On SUBSTITUTING the values in the above formula =>∆=(1/2) | (-3)(-3-3)+(1)(3-1)+(2)(1-(-3)) | sq.units =>∆=(1/2) | (-3)(-6)+(1)(2)+(2)(4) | sq.units =>∆=(1/2) | 18+2+8 | sq.units =>∆=(1/2) | 28 | sq.units =>∆=28/2 sq.units =>∆=14 sq.units Answer:-The area of the given triangle formed by the given vertices is 14 sq.units Used formulae:-1) The Area of a triangle formed by whose vertices are( x1 ,y1 )and (x2, y2 )and (x3, y3) is ∆=(1/2) | x1(y2-y3) +x2(y3-y1) + x3(y1-y2) | sq.units |
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