1. **State the Problem:** Given the line equation $$6y - 8x - 10 = 0$$, find: I. The slope and intercept of line L₁. II. The equation of the line perpendicular to L₁ passing through (4,2). 2. For triangle OAB with vertices O(0,0), A(0,6), B(6,0): I. Sketch the triangle. II. Find the equations of the three medians. III. Show that point (2,2) lies on all three medians. --- ### Step 1: Slope and Intercept of Line L₁ 1. Start with $$6y - 8x - 10 = 0$$ 2. Rewrite in slope-intercept form $$y = mx + c$$: $$6y = 8x + 10$$ $$y = \frac{8}{6}x + \frac{10}{6} = \frac{4}{3}x + \frac{5}{3}$$ 3. So, the slope is $$m = \frac{4}{3}$$ and the y-intercept is $$c = \frac{5}{3}$$. --- ### Step 2: Equation of Line Perpendicular to L₁ through (4,2) 1. The slope of the perpendicular line is the negative reciprocal of $$\frac{4}{3}$$: $$m_\perp = -\frac{3}{4}$$ 2. Use point-slope form for line through (4,2): $$y - 2 = -\frac{3}{4}(x - 4)$$ 3. Simplify: $$y - 2 = -\frac{3}{4}x + 3$$ $$y = -\frac{3}{4}x + 5$$ This is the perpendicular line equation. --- ### Step 3: Sketch Triangle OAB 1. Triangle vertices are O(0,0), A(0,6), and B(6,0). 2. Plot these points on a coordinate plane and connect them to see the right triangle. --- ### Step 4: Equations of the Three Medians 1. Median from O to midpoint of AB: - Midpoint of AB: $$\left(\frac{0+6}{2}, \frac{6+0}{2}\right) = (3,3)$$ - Equation of median OA midpoint: Slope from O(0,0) to (3,3): $$m = \frac{3-0}{3-0} = 1$$ Equation: $$y = x$$ 2. Median from A to midpoint of OB: - Midpoint of OB: $$\left(\frac{0+6}{2}, \frac{0+0}{2}\right) = (3,0)$$ - Slope from A(0,6) to (3,0): $$m = \frac{0-6}{3-0} = -2$$ Equation with point A: $$y - 6 = -2(x - 0)$$ $$y = -2x + 6$$ 3. Median from B to midpoint of OA: - Midpoint of OA: $$\left(\frac{0+0}{2}, \frac{0+6}{2}\right) = (0,3)$$ - Slope from B(6,0) to (0,3): $$m = \frac{3-0}{0-6} = \frac{3}{-6} = -\frac{1}{2}$$ Equation with point B: $$y - 0 = -\frac{1}{2}(x - 6)$$ $$y = -\frac{1}{2}x + 3$$ --- ### Step 5: Verify (2,2) Lies on All Three Medians 1. Check median 1: $$y=x$$ At (2,2), $$2 = 2$$ true. 2. Check median 2: $$y = -2x + 6$$ At (2,2): $$2 = -2(2) + 6 = -4 + 6 = 2$$ true. 3. Check median 3: $$y = -\frac{1}{2}x + 3$$ At (2,2): $$2 = -\frac{1}{2}(2) + 3 = -1 + 3 = 2$$ true. Thus, the point (2,2) lies on all three medians, showing they are concurrent. --- ### Final Answers: - Slope of L₁: $$\frac{4}{3}$$ - Intercept of L₁: $$\frac{5}{3}$$ - Perpendicular line through (4,2): $$y = -\frac{3}{4}x + 5$$ - Equations of medians: 1. $$y = x$$ 2. $$y = -2x + 6$$ 3. $$y = -\frac{1}{2}x + 3$$ - Point (2,2) lies on all medians (concurrent).