1. **State the problem:** Find the equations of two lines passing through the point $(-4, 2)$ whose perpendicular distance from the origin is 2. 2. **Recall the distance formula from a point to a line:** $$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$ where $ax + by + c = 0$ is the line equation and $(x_1, y_1)$ is the point. 3. **General line through $(-4, 2)$:** Using point-slope form: $$y - y_1 = m(x - x_1)$$ $$y - 2 = m(x + 4)$$ Rearranged to standard form: $$y - 2 = m x + 4m$$ $$y - m x = 2 + 4m$$ or $$-m x + y - (2 + 4m) = 0$$ Here, $a = -m$, $b = 1$, $c = -(2 + 4m)$. 4. **Apply distance formula for $d=2$ from origin $(0,0)$:** $$2 = \frac{|a \cdot 0 + b \cdot 0 + c|}{\sqrt{a^2 + b^2}} = \frac{|c|}{\sqrt{a^2 + b^2}} = \frac{|-(2 + 4m)|}{\sqrt{(-m)^2 + 1^2}} = \frac{|2 + 4m|}{\sqrt{m^2 + 1}}$$ 5. **Set up equation:** $$2 = \frac{|2 + 4m|}{\sqrt{m^2 + 1}}$$ Multiply both sides by $\sqrt{m^2 + 1}$: $$2 \sqrt{m^2 + 1} = |2 + 4m|$$ Square both sides: $$4(m^2 + 1) = (2 + 4m)^2$$ Expand right side: $$4m^2 + 4 = 4 + 16m + 16m^2$$ Simplify: $$4m^2 + 4 = 4 + 16m + 16m^2$$ Subtract $4 + 4m^2$ from both sides: $$0 = 16m + 12m^2$$ Factor: $$0 = 4m(3m + 4)$$ 6. **Solve for $m$:** $$4m = 0 \Rightarrow m = 0$$ $$3m + 4 = 0 \Rightarrow m = -\frac{4}{3}$$ 7. **Find line equations for each $m$:** - For $m=0$: $$y - 2 = 0(x + 4) \Rightarrow y = 2$$ Standard form: $$0 \cdot x + 1 \cdot y - 2 = 0$$ - For $m = -\frac{4}{3}$: $$y - 2 = -\frac{4}{3}(x + 4)$$ $$y - 2 = -\frac{4}{3}x - \frac{16}{3}$$ $$y = -\frac{4}{3}x - \frac{16}{3} + 2 = -\frac{4}{3}x - \frac{16}{3} + \frac{6}{3} = -\frac{4}{3}x - \frac{10}{3}$$ Multiply both sides by 3 to clear denominators: $$3y = -4x - 10$$ Rearranged: $$4x + 3y + 10 = 0$$ **Final answers:** $$y = 2$$ $$4x + 3y + 10 = 0$$