1. **Problem statement:** The line $l$ passes through the point $(4, 2)$ and forms a triangle with the coordinate axes with area 25 square units. The line cuts the x-axis at $(x_1, 0)$ and the y-axis at $(0, y_1)$. We need to: (a) Write the equation of the line in terms of slope $m$ in the form $ax + by + c = 0$. (b) Find the intercepts $(x_1, 0)$ and $(0, y_1)$ in terms of $m$. (c) Use the area condition to find possible values of $m$. (d) Find the acute angle between the two lines. 2. **(a) Equation of the line in terms of $m$:** The slope-intercept form is $y = mx + c$. Since the line passes through $(4, 2)$, substitute: $$2 = m \times 4 + c \implies c = 2 - 4m$$ So the line is: $$y = mx + 2 - 4m$$ Rearranged to standard form: $$mx - y + (2 - 4m) = 0$$ This is the required form $ax + by + c = 0$ with $a = m$, $b = -1$, $c = 2 - 4m$. 3. **(b) Find intercepts in terms of $m$:** - x-intercept $(x_1, 0)$: set $y=0$ in the line equation: $$0 = m x_1 + 2 - 4m \implies m x_1 = 4m - 2 \implies x_1 = \frac{4m - 2}{m}$$ 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 area to find $m$:** Area of triangle formed by intercepts and axes 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 $4 - \frac{2}{m} = \frac{4m - 2}{m}$: $$\left|\frac{4m - 2}{m}\right| |2 - 4m| = 50$$ Multiply numerator and denominator: $$\frac{|4m - 2|}{|m|} |2 - 4m| = 50$$ Multiply both sides by $|m|$: $$|4m - 2| |2 - 4m| = 50 |m|$$ Note $2 - 4m = -2(2m - 1)$ and $4m - 2 = 2(2m - 1)$, so: $$|4m - 2| |2 - 4m| = |2(2m - 1)| | -2(2m - 1)| = 4 |2m - 1|^2$$ So: $$4 |2m - 1|^2 = 50 |m|$$ Divide both sides by 2: $$2 |2m - 1|^2 = 25 |m|$$ Rewrite: $$2 (2m - 1)^2 = 25 |m|$$ Square is always positive, so: $$2 (2m - 1)^2 = 25 |m|$$ Let $m > 0$ or $m < 0$ and solve separately. For $m > 0$: $$2 (2m - 1)^2 = 25 m$$ Expand: $$2 (4m^2 - 4m + 1) = 25 m$$ $$8 m^2 - 8 m + 2 = 25 m$$ Bring all terms to one side: $$8 m^2 - 8 m + 2 - 25 m = 0 \implies 8 m^2 - 33 m + 2 = 0$$ Solve quadratic: $$m = \frac{33 \pm \sqrt{33^2 - 4 \times 8 \times 2}}{2 \times 8} = \frac{33 \pm \sqrt{1089 - 64}}{16} = \frac{33 \pm \sqrt{1025}}{16}$$ Approximate: $$m \approx \frac{33 \pm 32.0156}{16}$$ Two roots: $$m_1 \approx \frac{33 + 32.0156}{16} = 4.06$$ $$m_2 \approx \frac{33 - 32.0156}{16} = 0.0615$$ For $m < 0$, $|m| = -m$: $$2 (2m - 1)^2 = 25 (-m)$$ $$2 (4m^2 - 4m + 1) = -25 m$$ $$8 m^2 - 8 m + 2 = -25 m$$ $$8 m^2 - 8 m + 2 + 25 m = 0 \implies 8 m^2 + 17 m + 2 = 0$$ Solve quadratic: $$m = \frac{-17 \pm \sqrt{17^2 - 4 \times 8 \times 2}}{16} = \frac{-17 \pm \sqrt{289 - 64}}{16} = \frac{-17 \pm \sqrt{225}}{16}$$ $$m = \frac{-17 \pm 15}{16}$$ Two roots: $$m_3 = \frac{-17 + 15}{16} = \frac{-2}{16} = -0.125$$ $$m_4 = \frac{-17 - 15}{16} = \frac{-32}{16} = -2$$ So possible values of $m$ are approximately: $$m \approx 4.06, 0.0615, -0.125, -2$$ 5. **(d) Find the acute angle between the two lines:** The problem states two possible values of $m$ for the line. The two lines form an acute angle. We consider the two positive slopes $m_1 = 4.06$ and $m_2 = 0.0615$ (or the two negative slopes, but acute angle is positive). 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: $$m_1 - m_2 = 4.06 - 0.0615 = 3.9985$$ Calculate denominator: $$1 + m_1 m_2 = 1 + 4.06 \times 0.0615 = 1 + 0.249 = 1.249$$ Calculate tangent: $$\tan \theta = \frac{3.9985}{1.249} = 3.201$$ Find angle: $$\theta = \arctan(3.201) \approx 72.3^\circ$$ The acute angle is approximately $72^\circ$. **Final answers:** (a) Equation: $$m x - y + 2 - 4 m = 0$$ (b) Intercepts: $$x_1 = 4 - \frac{2}{m}, \quad y_1 = 2 - 4 m$$ (c) Possible $m$ values: approximately $$4.06, 0.0615, -0.125, -2$$ (d) Acute angle between two lines (taking positive slopes): approximately $$72^\circ$$