Find the slope and the equation of the line passing through the points:

The slope of the equation can be calculated using

$\mathrm{m}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}} \Rightarrow \frac{-3-3}{-5-5}=\frac{-6}{-10}$

$\mathrm{~m}=\frac{3}{5}$

Now using two point form of the equation of a line

$\mathrm{y}-\mathrm{y}_{1}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\left(\mathrm{x}-\mathrm{x}_{1}\right)$ where $\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}=$ slope of line

$y-3=\frac{3}{5}(x-5) \Rightarrow 5(y-3)=3(x-5)$

$3 x-15-5 y+15=0$

$3 x-5 y=0$

So, required equation of line is 3x - 5y = 0.