1. **Problem statement:** We have a line $l$ passing through the point $(4, 2)$ that 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. **(a) Write the equation of the line in terms of slope $m$:** The slope-intercept form of the line is $y = mx + c$. Since the line passes through $(4, 2)$, substitute to find $c$: $$2 = m \times 4 + c \implies c = 2 - 4m$$ So the line equation is: $$y = mx + 2 - 4m$$ Rearranged to standard form $ax + by + c = 0$: $$mx - y + 2 - 4m = 0$$ 3. **(b) Find the intercepts in terms of $m$:** - x-intercept $(x_1, 0)$: set $y=0$ in the line equation: $$0 = mx_1 + 2 - 4m \implies mx_1 = 4m - 2 \implies x_1 = \frac{4m - 2}{m}$$ Show cancellation: $$x_1 = \frac{\cancel{m}(4 - \frac{2}{m})}{\cancel{m}} = 4 - \frac{2}{m}$$ - y-intercept $(0, y_1)$: set $x=0$: $$y_1 = m \times 0 + 2 - 4m = 2 - 4m$$ 4. **(c) Use the area of the triangle to find possible values of $m$:** Area of triangle formed by intercepts is: $$\text{Area} = \frac{1}{2} |x_1| |y_1| = 25$$ Substitute $x_1$ and $y_1$: $$\frac{1}{2} \left|4 - \frac{2}{m}\right| |2 - 4m| = 25$$ Multiply both sides by 2: $$\left|4 - \frac{2}{m}\right| |2 - 4m| = 50$$ Rewrite $4 - \frac{2}{m}$ as $\frac{4m - 2}{m}$: $$\left|\frac{4m - 2}{m}\right| |2 - 4m| = 50$$ Multiply both sides by $|m|$: $$|4m - 2| |2 - 4m| = 50 |m|$$ Note $|2 - 4m| = |4m - 2|$, so: $$|4m - 2|^2 = 50 |m|$$ Square the absolute value: $$(4m - 2)^2 = 50 |m|$$ Expand: $$16m^2 - 16m + 4 = 50 |m|$$ Let $t = |m|$, then: $$16m^2 - 16m + 4 = 50 t$$ Since $m^2 = t^2$, consider cases: **Case 1: $m > 0$ so $t = m$** $$16m^2 - 16m + 4 = 50 m$$ Rearranged: $$16m^2 - 66m + 4 = 0$$ Use quadratic formula: $$m = \frac{66 \pm \sqrt{66^2 - 4 \times 16 \times 4}}{2 \times 16} = \frac{66 \pm \sqrt{4356 - 256}}{32} = \frac{66 \pm \sqrt{4100}}{32}$$ $$\sqrt{4100} = 10 \sqrt{41}$$ So: $$m = \frac{66 \pm 10 \sqrt{41}}{32}$$ **Case 2: $m < 0$ so $t = -m$** $$16m^2 - 16m + 4 = -50 m$$ Rearranged: $$16m^2 + 34m + 4 = 0$$ Quadratic formula: $$m = \frac{-34 \pm \sqrt{34^2 - 4 \times 16 \times 4}}{2 \times 16} = \frac{-34 \pm \sqrt{1156 - 256}}{32} = \frac{-34 \pm \sqrt{900}}{32}$$ $$\sqrt{900} = 30$$ So: $$m = \frac{-34 \pm 30}{32}$$ Possible values: $$m = \frac{-34 + 30}{32} = \frac{-4}{32} = -\frac{1}{8}$$ $$m = \frac{-34 - 30}{32} = \frac{-64}{32} = -2$$ From case 1, approximate values: $$m \approx \frac{66 + 64.03}{32} = \frac{130.03}{32} \approx 4.06$$ $$m \approx \frac{66 - 64.03}{32} = \frac{1.97}{32} \approx 0.06$$ Since the problem asks for two possible values, the relevant solutions are: $$m = -2 \quad \text{and} \quad m = -\frac{1}{8}$$ 5. **(d) Find the acute angle between the two lines with slopes $m_1 = -2$ and $m_2 = -\frac{1}{8}$:** Formula for angle $\theta$ between two lines with slopes $m_1$ and $m_2$: $$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$ Calculate numerator: $$m_1 - m_2 = -2 - \left(-\frac{1}{8}\right) = -2 + \frac{1}{8} = -\frac{16}{8} + \frac{1}{8} = -\frac{15}{8}$$ Calculate denominator: $$1 + m_1 m_2 = 1 + (-2) \times \left(-\frac{1}{8}\right) = 1 + \frac{2}{8} = 1 + \frac{1}{4} = \frac{5}{4}$$ So: $$\tan \theta = \left| \frac{-\frac{15}{8}}{\frac{5}{4}} \right| = \left| -\frac{15}{8} \times \frac{4}{5} \right| = \frac{15 \times 4}{8 \times 5} = \frac{60}{40} = 1.5$$ Find $\theta$: $$\theta = \arctan(1.5) \approx 56.31^\circ$$ The acute angle between the lines is approximately $56^\circ$. **Final answers:** - (a) Equation: $$mx - y + 2 - 4m = 0$$ - (b) Intercepts: $$x_1 = 4 - \frac{2}{m}, \quad y_1 = 2 - 4m$$ - (c) Possible slopes: $$m = -2 \quad \text{or} \quad m = -\frac{1}{8}$$ - (d) Acute angle between lines: $$56^\circ$$