1. **Problem statement:** Find the equations of two lines passing through the point $(-2,4)$ that form triangles with the coordinate axes having an area of 9 square units. 2. **Formula for area of triangle formed by axes and line:** If a line intercepts the x-axis at $(a,0)$ and the y-axis at $(0,b)$, the area of the triangle formed with the axes is given by $$\text{Area} = \frac{1}{2} |a \times b|.$$ We want this area to be 9, so $$\frac{1}{2} |a b| = 9 \implies |a b| = 18.$$ 3. **Equation of the line in intercept form:** $$\frac{x}{a} + \frac{y}{b} = 1.$$ This line passes through $(-2,4)$, so substitute: $$\frac{-2}{a} + \frac{4}{b} = 1.$$ 4. **Express $b$ in terms of $a$ using the area condition:** $$|a b| = 18 \implies b = \frac{18}{a} \text{ or } b = -\frac{18}{a}.$$ 5. **Substitute $b$ into the point condition:** For $b = \frac{18}{a}$: $$\frac{-2}{a} + \frac{4}{18/a} = 1 \implies \frac{-2}{a} + \frac{4a}{18} = 1.$$ Multiply both sides by $18a$: $$-36 + 4a^2 = 18a.$$ Rearranged: $$4a^2 - 18a - 36 = 0.$$ Divide by 2: $$2a^2 - 9a - 18 = 0.$$ 6. **Solve quadratic for $a$:** $$a = \frac{9 \pm \sqrt{(-9)^2 - 4 \times 2 \times (-18)}}{2 \times 2} = \frac{9 \pm \sqrt{81 + 144}}{4} = \frac{9 \pm \sqrt{225}}{4} = \frac{9 \pm 15}{4}.$$ So, $$a_1 = \frac{24}{4} = 6, \quad a_2 = \frac{-6}{4} = -1.5.$$ 7. **Find corresponding $b$ values:** For $a=6$, $$b = \frac{18}{6} = 3.$$ For $a=-1.5$, $$b = \frac{18}{-1.5} = -12.$$ 8. **Write the two line equations:** For $(a,b) = (6,3)$: $$\frac{x}{6} + \frac{y}{3} = 1 \implies y = 3 - \frac{1}{2}x.$$ For $(a,b) = (-1.5,-12)$: $$\frac{x}{-1.5} + \frac{y}{-12} = 1 \implies -\frac{2}{3}x - \frac{1}{12}y = 1 \implies y = -12 - 8x.$$ **Final answers:** $$y = 3 - \frac{1}{2}x \quad \text{and} \quad y = -12 - 8x.$$