If z_1=2+3i,z_2=4-5i,and z_3 are three points | vedclass

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(A) $z_3 = \frac{z_1+z_2}{2} = \frac{(2+3i)+(4-5i)}{2} = \frac{6-2i}{2} = 3-i$. Given $5z_1+xz_2+yz_3=0$. Substituting the values: $5(2+3i) + x(4-5i)

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$-5$

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(A) $z_3 = \frac{z_1+z_2}{2} = \frac{(2+3i)+(4-5i)}{2} = \frac{6-2i}{2} = 3-i$. Given $5z_1+xz_2+yz_3=0$. Substituting the values: $5(2+3i) + x(4-5i) + y(3-i) = 0$. $(10+15i) + (4x-5xi) + (3y-yi) = 0$. Grouping real and imaginary parts: $(10+4x+3y) + i(15-5x-y) = 0$. Equating real and imaginary parts to zero: $4x+3y = -10$ $(i)$ $5x+y = 15$ (ii) From (ii),$y = 15-5x$. Substituting into $(i)$: $4x + 3(15-5x) = -10$. $4x + 45 - 15x = -10$. $-11x = -55 \Rightarrow x = 5$. $y = 15 - 5(5) = 15-25 = -10$. Therefore,$x+y = 5-10 = -5$.

If $z_1=2+3i$,$z_2=4-5i$,and $z_3$ are three points in the Argand plane such that $5z_1+xz_2+yz_3=0$ $(x, y \in R)$ and $z_3$ is the midpoint of the segment joining the points $z_1$ and $z_2$,then $x+y=$

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