11z lies at the intersection of the lines joining the origin with (18,24) and the line joining (25,0) with (-7,24)
\hspace{-16}$Find Complex no. $\mathbf{z}$ which satisfy\\\\ $\mathbf{\mid z \mid+\mid z-25 \mid+\mid z-18-24i \mid+\mid z+7-24i \mid=70}$
Ans:: z=9+12i
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341And since the given points form a paralellogram, its the midpoint of the diagonals
341We will establish that 70 is the least sum of distances from the four points.
Take A(origin) and B(18,24). We know that the least sum of distance from these two points is obtained by any point on the line segment AB.
Similarly for C(25,0) and D(-7,24) the least sum of distance is for a point lying on CD.
Hence if there is a point of intersection internal to both AB amd CD, this point is the unique point with min sum of distance which here is 70.
