1. **State the problem:** We have a line L given by the equation $$4y + 12 = 16x$$ and a line R that is perpendicular to L and passes through the point $(-1, 5)$. We need to: a) Find the equation of line R. b) Find the intersections of lines L and R with the coordinate axes. c) Determine the monotony (increasing or decreasing behavior) of line R. 2. **Rewrite line L in slope-intercept form:** Start with $$4y + 12 = 16x$$ Subtract 12 from both sides: $$4y = 16x - 12$$ Divide both sides by 4: $$y = 4x - 3$$ So, the slope of line L is $$m_L = 4$$. 3. **Find the slope of line R:** Since R is perpendicular to L, its slope $$m_R$$ satisfies: $$m_R = -\frac{1}{m_L} = -\frac{1}{4}$$ 4. **Find the equation of line R passing through $(-1, 5)$:** Use point-slope form: $$y - y_1 = m_R (x - x_1)$$ Substitute $$m_R = -\frac{1}{4}$$, $$x_1 = -1$$, $$y_1 = 5$$: $$y - 5 = -\frac{1}{4}(x + 1)$$ Simplify: $$y - 5 = -\frac{1}{4}x - \frac{1}{4}$$ Add 5 to both sides: $$y = -\frac{1}{4}x - \frac{1}{4} + 5 = -\frac{1}{4}x + \frac{19}{4}$$ So, the equation of line R is: $$y = -\frac{1}{4}x + \frac{19}{4}$$ 5. **Find intersections with coordinate axes:** - For line L: $$y = 4x - 3$$ - **x-intercept:** Set $$y=0$$: $$0 = 4x - 3 \Rightarrow 4x = 3 \Rightarrow x = \frac{3}{4}$$ So, x-intercept is $$\left(\frac{3}{4}, 0\right)$$. - **y-intercept:** Set $$x=0$$: $$y = 4(0) - 3 = -3$$ So, y-intercept is $$(0, -3)$$. - For line R: $$y = -\frac{1}{4}x + \frac{19}{4}$$ - **x-intercept:** Set $$y=0$$: $$0 = -\frac{1}{4}x + \frac{19}{4} \Rightarrow -\frac{1}{4}x = -\frac{19}{4} \Rightarrow x = 19$$ So, x-intercept is $$(19, 0)$$. - **y-intercept:** Set $$x=0$$: $$y = -\frac{1}{4}(0) + \frac{19}{4} = \frac{19}{4} = 4.75$$ So, y-intercept is $$(0, \frac{19}{4})$$. 6. **Determine the monotony of line R:** Since the slope $$m_R = -\frac{1}{4} < 0$$, line R is **decreasing**. **Final answers:** - a) Equation of line R: $$y = -\frac{1}{4}x + \frac{19}{4}$$ - b) Intersections: - Line L: x-intercept $$\left(\frac{3}{4}, 0\right)$$, y-intercept $$(0, -3)$$ - Line R: x-intercept $$(19, 0)$$, y-intercept $$(0, \frac{19}{4})$$ - c) Line R is decreasing because its slope is negative.