#### Answer

$y=-\displaystyle \frac{1}{2}x+12$

#### Work Step by Step

We solve the given equation for y (write it in slope-intercept form)
$6x-3y=5$
$-3y=-6x+5$
$y=\displaystyle \frac{-6}{-3}x+\frac{5}{-3}$
$y=2x-\displaystyle \frac{5}{3}$
The slope of the line is $m_{1}=2$, and any line perpendicular to it has slope $m_{2}$
such that
$m_{2}=-\displaystyle \frac{1}{m_{1}}$
$m_{2}=-\displaystyle \frac{1}{2}$
Now, find the perpendicular line through $(4,10)$.
The point-slope equation of a line containing the point $(x_{1},y_{1})$, with slope $m$ is
$y=y_{1}+m(x-x_{1})$
Given $(x_{1},y_{1})=(4,10)$ and $m=-\displaystyle \frac{1}{2}$
$y=10+(-\displaystyle \frac{1}{2})(x-4)$
$y=10-\displaystyle \frac{1}{2}x+2$
$y=-\displaystyle \frac{1}{2}x+12$