if A(-2,3) B(6,5 ) C (x,-5) D(-4,-3) are the vertices of a quadrilateral ABCD of area 80sq unit then find positive value of x


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(y2y3)+x2(y3y1)+x3(y1y2)|=12|2(5+5)+6(53)+x(35)|=12|2x68|=x+34Area of ΔACD=12|2(5+3)+x(33)+(4)(3+5)|=12|6x12|=3x+6.area of quad.ABCD=x+34+3x+6=>80=4x+40=>4x=40=>x=10.

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