dear student let z=x+iy then z+iz= x+iy+ix+i2y=(x-y)+i(x+y)iz=ix+i2y=-y+ixso points are (x,y),(x-y,x+y) and (-y,x)area of triangle formed=xy1x-yx+y1-yx1=x(x+y-y)-(x-y)(y-x)-y(y-x-y)=xy+x2+y2-2xy+xy=x2+y2given area =50x2+y2=50x2+y2=50so z=x2+y2=50=52 regards

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dear student

l e t z = x + i y t h e n z + i z = x + i y + i x + i^{2} y = (x - y) + i (x + y) i z = i x + i^{2} y = - y + i x s o p o i n t s a r e (x, y), (x - y, x + y) a n d (- y, x) a r e a o f t r i a n g l e f o r m e d = |\begin{matrix} x & y & 1 \\ x - y & x + y & 1 \\ - y & x & 1 \end{matrix}| = x (x + y - y) - (x - y) (y - x) - y (y - x - y) = x y + x^{2} + y^{2} - 2 x y + x y = x^{2} + y^{2} g i v e n a r e a = 50 x^{2} + y^{2} = 50 \sqrt{x^{2} + y^{2}} = \sqrt{50} s o |z| = \sqrt{x^{2} + y^{2}} = \sqrt{50} = 5 \sqrt{2}

regards

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