1. **Problem:** Write an equation for the line that is perpendicular to the given line and passes through $(-1,3)$. The given line is $-y=5-10x$. 2. **Put the given line into slope-intercept form.** Start with: $$-y=5-10x$$ Multiply both sides by $-1$: $$y=-5+10x$$ So the slope of the given line is $10$. 3. **Find the slope of the perpendicular line.** Perpendicular slopes are negative reciprocals, so: $$m=-\frac{1}{10}$$ 4. **Use point-slope form.** The formula is: $$y-y_1=m(x-x_1)$$ Substitute $m=-\frac{1}{10}$ and $(-1,3)$: $$y-3=-\frac{1}{10}(x-(-1))$$ $$y-3=-\frac{1}{10}(x+1)$$ 5. **Simplify to get slope-intercept form.** Distribute: $$y-3=-\frac{1}{10}x-\frac{1}{10}$$ Add $3$ to both sides: $$y=-\frac{1}{10}x+\frac{29}{10}$$ 6. **Final answer:** $$y=-\frac{1}{10}x+\frac{29}{10}$$