1. **State the problem:** We need to find the equation of a line perpendicular to line A, which passes through points (2, -5) and (10, -1), and this perpendicular line must pass through the point (4, 3). 2. **Find the slope of line A:** The slope formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$. Calculate slope of A: $$m_A = \frac{-1 - (-5)}{10 - 2} = \frac{-1 + 5}{8} = \frac{4}{8} = \frac{1}{2}$$ 3. **Find the slope of the perpendicular line:** The slope of a line perpendicular to another is the negative reciprocal. $$m_{\perp} = -\frac{1}{m_A} = -\frac{1}{\frac{1}{2}} = -2$$ 4. **Use point-slope form to find the equation of the perpendicular line:** Point-slope form: $$y - y_1 = m(x - x_1)$$ Using point (4, 3) and slope $m_{\perp} = -2$: $$y - 3 = -2(x - 4)$$ 5. **Simplify the equation:** $$y - 3 = -2x + 8$$ Add 3 to both sides: $$y = -2x + 8 + 3$$ $$y = -2x + 11$$ 6. **Final answer:** The equation of the line perpendicular to A passing through (4, 3) is $$y = -2x + 11$$