1. **State the problem:** We need to find the equation of any line passing through the point $(-4,2)$. Then, find the equations of two lines through $(-4,2)$ whose perpendicular distance from the origin is 2. 2. **General form of a line:** The equation of a line can be written as $$Ax + By + C = 0$$ where $A$, $B$, and $C$ are constants. 3. **Condition for passing through $(-4,2)$:** Substitute $x=-4$ and $y=2$ into the line equation: $$A(-4) + B(2) + C = 0 \Rightarrow -4A + 2B + C = 0$$ 4. **Perpendicular distance from origin:** The distance $d$ from the origin $(0,0)$ to the line $Ax + By + C = 0$ is given by: $$d = \frac{|C|}{\sqrt{A^2 + B^2}}$$ We are given $d=2$, so: $$\frac{|C|}{\sqrt{A^2 + B^2}} = 2 \Rightarrow |C| = 2\sqrt{A^2 + B^2}$$ 5. **Express $C$ from step 3:** $$C = 4A - 2B$$ 6. **Substitute $C$ into distance formula:** $$|4A - 2B| = 2\sqrt{A^2 + B^2}$$ 7. **Square both sides:** $$ (4A - 2B)^2 = 4(A^2 + B^2) $$ Expand left side: $$16A^2 - 16AB + 4B^2 = 4A^2 + 4B^2$$ 8. **Simplify:** $$16A^2 - 16AB + 4B^2 - 4A^2 - 4B^2 = 0 \Rightarrow 12A^2 - 16AB = 0$$ 9. **Factor:** $$4A(3A - 4B) = 0$$ So either: - $4A=0 \Rightarrow A=0$ - or $3A - 4B=0 \Rightarrow 3A=4B \Rightarrow B=\frac{3}{4}A$ 10. **Case 1: $A=0$** From step 5, $C = 4(0) - 2B = -2B$ Distance condition: $$|C| = 2\sqrt{A^2 + B^2} = 2|B|$$ But $|C| = |-2B| = 2|B|$, so condition holds. Equation of line: $$0 \cdot x + B y + C = 0 \Rightarrow B y - 2B = 0 \Rightarrow y = 2$$ 11. **Case 2: $B=\frac{3}{4}A$** Substitute into $C$: $$C = 4A - 2B = 4A - 2 \times \frac{3}{4}A = 4A - \frac{3}{2}A = \frac{5}{2}A$$ Distance condition: $$|C| = 2\sqrt{A^2 + B^2} = 2\sqrt{A^2 + \left(\frac{3}{4}A\right)^2} = 2|A|\sqrt{1 + \frac{9}{16}} = 2|A|\sqrt{\frac{25}{16}} = 2|A| \times \frac{5}{4} = \frac{5}{2}|A|$$ Since $|C| = \frac{5}{2}|A|$, condition holds. Equation of line: $$A x + B y + C = 0 \Rightarrow A x + \frac{3}{4}A y + \frac{5}{2}A = 0$$ Divide both sides by $A$ (assuming $A \neq 0$): $$x + \frac{3}{4} y + \frac{5}{2} = 0$$ Multiply through by 4 to clear fractions: $$4x + 3y + 10 = 0$$ 12. **Final answers:** - Line 1: $$y = 2$$ - Line 2: $$4x + 3y + 10 = 0$$ These are the two lines through $(-4,2)$ whose perpendicular distance from the origin is 2.