if A (-2,3)B (6,5) C (x,-5) &D (-4,-3) are the vertices of a quad. ABCD of area AD sq. units find the positive value of x
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Given, x1=−2,x2=6,x3=x,x4=−4y1=3,y2=5,y3=−5,y4=−3Given, Area of quadrilateral=80 sq.cmArea of quadrilateral=Area of ΔABC+Area of ΔADCNow, for ΔABC,x1=−2,x2=6,x3=xy1=3,y2=5,y3=−5.∴Area of ΔABC=12|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)|=12|−2(5+5)+6(−5−3)+x(3−5)|=12|−2x−68|=x+34Area of ΔACD=12|−2(−5+3)+x(−3−3)+(−4)(3+5)|=12|−6x−12|=3x+6.∴area of quad.ABCD=x+34+3x+6=>80=4x+40=>4x=40=>x=10.
Given, x1=−2,x2=6,x3=x,x4=−4y1=3,y2=5,y3=−5,y4=−3Given, Area of quadrilateral=80 sq.cmArea of quadrilateral=Area of ΔABC+Area of ΔADCNow, for ΔABC,x1=−2,x2=6,x3=xy1=3,y2=5,y3=−5.∴Area of ΔABC=12|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)|=12|−2(5+5)+6(−5−3)+x(3−5)|=12|−2x−68|=x+34Area of ΔACD=12|−2(−5+3)+x(−3−3)+(−4)(3+5)|=12|−6x−12|=3x+6.∴area of quad.ABCD=x+34+3x+6=>80=4x+40=>4x=40=>x=10.
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