If $z_1=2+3 i, z_2=4-5 i$ and $z_3$ are are three points in the Argand plane such that $5…
If $z_1=2+3 i, z_2=4-5 i$ and $z_3$ are are three points in the Argand plane such that $5 \mathrm{z}_1+\mathrm{xz}_2+\mathrm{yz}_3=0$ $(x, y \in \mathbb{R})$ and $z_3$ is the midpoint of the segment joining the points $\mathrm{z}_1$ and $\mathrm{z}_2$ then $\mathrm{x}+\mathrm{y}=$
$-5$
$0$
$4$
$-1$
Solution
$z_3=\frac{z_1+z_2}{2}=3-i$
Now, $5 z_1+x z_2+\mathrm{yz}_3=0$
$
\begin{aligned}
& \Rightarrow(10+4 x+3 y)+i(15-5 x-y)=0 \\
& \Rightarrow 4 x+3 y=-10 \ldots \text { (i) and } 15-5 x-y=0 \ldots
\end{aligned}
$
From equations (i) and (ii), we get $x=5, y=-10$
$
\therefore x+y=5-10=-5
$