1. **State the problem:** We have two perpendicular lines T and K intersecting at point (3,2). Line T passes through (3,2) and (2,5). We need to find which of the given points lies on line K. 2. **Find the slope of line T:** The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. $$m_T = \frac{5 - 2}{2 - 3} = \frac{3}{-1} = -3$$ 3. **Find the slope of line K:** Since lines T and K are perpendicular, their slopes satisfy $m_T \times m_K = -1$. $$m_K = -\frac{1}{m_T} = -\frac{1}{-3} = \frac{1}{3}$$ 4. **Write the equation of line K:** Using point-slope form with point (3,2) and slope $\frac{1}{3}$: $$y - 2 = \frac{1}{3}(x - 3)$$ Simplify: $$y = \frac{1}{3}x - 1 + 2 = \frac{1}{3}x + 1$$ 5. **Check which point lies on line K:** Substitute each point into $y = \frac{1}{3}x + 1$. - For (5,3): $3 \stackrel{?}{=} \frac{1}{3} \times 5 + 1 = \frac{5}{3} + 1 = \frac{8}{3} \approx 2.67$ (No) - For (-2,1): $1 \stackrel{?}{=} \frac{1}{3} \times (-2) + 1 = -\frac{2}{3} + 1 = \frac{1}{3} \approx 0.33$ (No) - For (1,2): $2 \stackrel{?}{=} \frac{1}{3} \times 1 + 1 = \frac{1}{3} + 1 = \frac{4}{3} \approx 1.33$ (No) - For (6,3): $3 \stackrel{?}{=} \frac{1}{3} \times 6 + 1 = 2 + 1 = 3$ (Yes) **Answer:** The point (6,3) lies on line K.