1. **Problem statement:** We have a line $l$ passing through the point $(1,1)$ that intersects the x-axis at $(x,0)$ and the y-axis at $(0,y)$. The triangle formed by this line and the coordinate axes has an area of 25 square units. 2. **Part (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 $(1,1)$, substitute to find $c$: $$1 = m \times 1 + c \implies c = 1 - m$$ So the line equation is: $$y = mx + 1 - m$$ Rearranged to standard form $ax + by + c = 0$: $$mx - y + 1 - m = 0$$ 3. **Part (b): Find the x- and y-intercepts in terms of $m$** - x-intercept: set $y=0$: $$0 = mx + 1 - m \implies mx = m - 1 \implies x = \frac{m - 1}{m}$$ - y-intercept: set $x=0$: $$y = m \times 0 + 1 - m = 1 - m$$ So the intercepts are: $$\left(\frac{m - 1}{m}, 0\right) \quad \text{and} \quad (0, 1 - m)$$ 4. **Part (c): Use the area of the triangle to find possible values of $m$** The area of the triangle formed by the intercepts is: $$\text{Area} = \frac{1}{2} \times \left| x \right| \times \left| y \right| = 25$$ Substitute the intercepts: $$\frac{1}{2} \times \left| \frac{m - 1}{m} \right| \times \left| 1 - m \right| = 25$$ Simplify: $$\left| \frac{m - 1}{m} \right| \times \left| 1 - m \right| = 50$$ Note $|m-1| = |1-m|$, so: $$\left| \frac{m - 1}{m} \right| \times |m - 1| = 50$$ $$\frac{|m - 1|^2}{|m|} = 50$$ Multiply both sides by $|m|$: $$|m - 1|^2 = 50 |m|$$ Let $m$ be positive or negative; square both sides: $$(m - 1)^2 = 50 |m|$$ Consider two cases: **Case 1: $m > 0$** $$(m - 1)^2 = 50 m$$ Expand: $$m^2 - 2m + 1 = 50 m$$ Bring all terms to one side: $$m^2 - 52 m + 1 = 0$$ Use quadratic formula: $$m = \frac{52 \pm \sqrt{52^2 - 4 \times 1 \times 1}}{2} = \frac{52 \pm \sqrt{2704 - 4}}{2} = \frac{52 \pm \sqrt{2700}}{2}$$ $$\sqrt{2700} = 10 \sqrt{27} = 10 \times 3 \sqrt{3} = 30 \sqrt{3}$$ So: $$m = \frac{52 \pm 30 \sqrt{3}}{2} = 26 \pm 15 \sqrt{3}$$ **Case 2: $m < 0$** Then $|m| = -m$, so: $$(m - 1)^2 = 50 (-m)$$ $$m^2 - 2m + 1 = -50 m$$ $$m^2 + 48 m + 1 = 0$$ Quadratic formula: $$m = \frac{-48 \pm \sqrt{48^2 - 4}}{2} = \frac{-48 \pm \sqrt{2304 - 4}}{2} = \frac{-48 \pm \sqrt{2300}}{2}$$ $$\sqrt{2300} = 10 \sqrt{23}$$ So: $$m = \frac{-48 \pm 10 \sqrt{23}}{2} = -24 \pm 5 \sqrt{23}$$ 5. **Part (d): Find the acute angle between the two lines** The two possible slopes are: $$m_1 = 26 + 15 \sqrt{3}$$ $$m_2 = 26 - 15 \sqrt{3}$$ The formula for the angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is: $$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$ Calculate numerator: $$m_1 - m_2 = (26 + 15 \sqrt{3}) - (26 - 15 \sqrt{3}) = 30 \sqrt{3}$$ Calculate denominator: $$1 + m_1 m_2$$ First find $m_1 m_2$: $$(26 + 15 \sqrt{3})(26 - 15 \sqrt{3}) = 26^2 - (15 \sqrt{3})^2 = 676 - 225 \times 3 = 676 - 675 = 1$$ So denominator: $$1 + 1 = 2$$ Therefore: $$\tan \theta = \frac{30 \sqrt{3}}{2} = 15 \sqrt{3}$$ Calculate $\theta$: $$\theta = \arctan(15 \sqrt{3})$$ Since $15 \sqrt{3}$ is large, $\theta$ is close to 90 degrees, but we want the acute angle between the two lines, so: $$\theta \approx 89.8^\circ$$ **Final answers:** - (a) Equation: $$mx - y + 1 - m = 0$$ - (b) Intercepts: $$\left(\frac{m - 1}{m}, 0\right), (0, 1 - m)$$ - (c) Possible $m$ values: $$m = 26 \pm 15 \sqrt{3}$$ $$m = -24 \pm 5 \sqrt{23}$$ - (d) Acute angle between lines: $$\approx 90^\circ$$