1. **State the problem:** We have two lines, $D_1$ and $D_2$. We know $D_1$ cuts the y-axis at $(0, -5)$ and is parallel to $D_2$ given by $x - 2y - 1 = 0$. We also know $D_1$ cuts the x-axis at $(-2, 0)$ and is perpendicular to another $D_2$ given by $3x - 2y + 1 = 0$. We need to find the equations of these lines. 2. **Find the equation of $D_1$ parallel to $D_2: x - 2y - 1 = 0$** - The slope of $D_2$ is found by rewriting it in slope-intercept form: $$x - 2y - 1 = 0 \implies -2y = -x + 1 \implies y = \frac{1}{2}x - \frac{1}{2}$$ - So slope $m = \frac{1}{2}$. - Since $D_1$ is parallel to $D_2$, $D_1$ has the same slope $m = \frac{1}{2}$. - $D_1$ cuts the y-axis at $(0, -5)$, so the y-intercept $b = -5$. - Equation of $D_1$ is: $$y = \frac{1}{2}x - 5$$ 3. **Find the equation of $D_1$ perpendicular to $D_2: 3x - 2y + 1 = 0$** - Find slope of $D_2$: $$3x - 2y + 1 = 0 \implies -2y = -3x - 1 \implies y = \frac{3}{2}x + \frac{1}{2}$$ - Slope of $D_2$ is $m_2 = \frac{3}{2}$. - Slope of $D_1$ perpendicular to $D_2$ is the negative reciprocal: $$m_1 = -\frac{1}{m_2} = -\frac{1}{\frac{3}{2}} = -\frac{2}{3}$$ - $D_1$ cuts the x-axis at $(-2, 0)$, so point $(-2, 0)$ lies on $D_1$. - Use point-slope form: $$y - y_1 = m(x - x_1)$$ $$y - 0 = -\frac{2}{3}(x - (-2)) = -\frac{2}{3}(x + 2)$$ - Simplify: $$y = -\frac{2}{3}x - \frac{4}{3}$$ **Final answers:** - Equation of $D_1$ parallel to $D_2 \equiv x - 2y - 1 = 0$ is: $$y = \frac{1}{2}x - 5$$ - Equation of $D_1$ perpendicular to $D_2 \equiv 3x - 2y + 1 = 0$ is: $$y = -\frac{2}{3}x - \frac{4}{3}$$