1. **State the problem:** We have a line $l$ passing through the point $(4,2)$. The line intersects the x-axis at $(x_1,0)$ and the y-axis at $(0,y_1)$. The triangle formed by this line and the coordinate axes has an area of 25 square units. 2. **Write the equation of the line in terms of slope $m$:** Given the point-slope form: $$y - 2 = m(x - 4)$$ Expanding: $$y - 2 = mx - 4m$$ Rearranging to standard form $ax + by + c = 0$: $$mx - y - 4m + 2 = 0$$ 3. **Find the intercepts in terms of $m$:** - To find the x-intercept, set $y=0$: $$mx - 0 - 4m + 2 = 0 \implies mx = 4m - 2 \implies x = \frac{4m - 2}{m}$$ - To find the y-intercept, set $x=0$: $$m(0) - y - 4m + 2 = 0 \implies -y = 4m - 2 \implies y = 2 - 4m$$ So the intercepts are: $$x_1 = \frac{4m - 2}{m}, \quad y_1 = 2 - 4m$$ 4. **Use the area formula for the triangle formed by the axes and the line:** The area $A$ of the triangle formed by the intercepts is: $$A = \frac{1}{2} |x_1 y_1| = 25$$ Substitute $x_1$ and $y_1$: $$\frac{1}{2} \left| \frac{4m - 2}{m} (2 - 4m) \right| = 25$$ Multiply both sides by 2: $$\left| \frac{4m - 2}{m} (2 - 4m) \right| = 50$$ 5. **Simplify the expression inside the absolute value:** $$\frac{4m - 2}{m} (2 - 4m) = \frac{(4m - 2)(2 - 4m)}{m}$$ Expand numerator: $$(4m)(2) = 8m, \quad (4m)(-4m) = -16m^2, \quad (-2)(2) = -4, \quad (-2)(-4m) = 8m$$ Sum: $$8m - 16m^2 - 4 + 8m = -16m^2 + 16m - 4$$ So expression is: $$\frac{-16m^2 + 16m - 4}{m}$$ 6. **Rewrite the equation:** $$\left| \frac{-16m^2 + 16m - 4}{m} \right| = 50$$ Multiply both sides by $|m|$: $$| -16m^2 + 16m - 4 | = 50 |m|$$ 7. **Solve the quadratic inside the absolute value:** Let: $$Q(m) = -16m^2 + 16m - 4$$ Divide by -4 for simplicity: $$4m^2 - 4m + 1 = 0$$ Discriminant: $$\Delta = (-4)^2 - 4 \times 4 \times 1 = 16 - 16 = 0$$ So $Q(m)$ has a double root at: $$m = \frac{4}{2 \times 4} = \frac{4}{8} = \frac{1}{2}$$ 8. **Analyze the absolute value equation:** Since $Q(m) = -16m^2 + 16m - 4$, rewrite original equation: $$|Q(m)| = 50 |m|$$ 9. **Rewrite $Q(m)$ as:** $$Q(m) = -16(m^2 - m + \frac{1}{4}) = -16(m - \frac{1}{2})^2$$ So: $$|Q(m)| = 16 (m - \frac{1}{2})^2$$ 10. **Substitute back:** $$16 (m - \frac{1}{2})^2 = 50 |m|$$ Divide both sides by 2: $$8 (m - \frac{1}{2})^2 = 25 |m|$$ 11. **Solve for $m$ considering $m > 0$ and $m < 0$ separately:** - For $m > 0$: $$8 (m - \frac{1}{2})^2 = 25 m$$ Expand: $$8 (m^2 - m + \frac{1}{4}) = 25 m$$ $$8 m^2 - 8 m + 2 = 25 m$$ Bring all terms to one side: $$8 m^2 - 8 m + 2 - 25 m = 0 \implies 8 m^2 - 33 m + 2 = 0$$ - For $m < 0$: $$8 (m - \frac{1}{2})^2 = -25 m$$ Expand left side as above: $$8 m^2 - 8 m + 2 = -25 m$$ Bring all terms to one side: $$8 m^2 - 8 m + 2 + 25 m = 0 \implies 8 m^2 + 17 m + 2 = 0$$ 12. **Solve the quadratic equations:** - For $m > 0$: $$8 m^2 - 33 m + 2 = 0$$ Discriminant: $$\Delta = (-33)^2 - 4 \times 8 \times 2 = 1089 - 64 = 1025$$ Roots: $$m = \frac{33 \pm \sqrt{1025}}{16}$$ Only positive roots are valid here. - For $m < 0$: $$8 m^2 + 17 m + 2 = 0$$ Discriminant: $$\Delta = 17^2 - 4 \times 8 \times 2 = 289 - 64 = 225$$ Roots: $$m = \frac{-17 \pm 15}{16}$$ Possible roots: $$m = \frac{-17 + 15}{16} = \frac{-2}{16} = -\frac{1}{8}$$ $$m = \frac{-17 - 15}{16} = \frac{-32}{16} = -2$$ Both negative roots are valid here. 13. **Summary of possible slopes:** $$m \approx \frac{33 + 32.0156}{16} = 4.13, \quad m \approx \frac{33 - 32.0156}{16} = 0.0615$$ and $$m = -\frac{1}{8} = -0.125, \quad m = -2$$ 14. **Find intercepts for each $m$ using:** $$x_1 = \frac{4m - 2}{m}, \quad y_1 = 2 - 4m$$ This completes the solution for the first question. **Final answers:** - Equation of line: $$mx - y - 4m + 2 = 0$$ - Intercepts: $$x_1 = \frac{4m - 2}{m}, \quad y_1 = 2 - 4m$$ - Possible slopes $m$ satisfy: $$16 (m - \frac{1}{2})^2 = 50 |m|$$ with approximate solutions: $$m \approx 4.13, 0.0615, -0.125, -2$$