1. **State the problem:** Given the equations of the sides of a triangle: $$x + 2y = 4,$$ $$2x - y = 3,$$ $$x - y - 2 = 0,$$ prove that the triangle is right angled and find its area. 2. **Find the vertices of the triangle:** Solve the pairs of equations to find the intersection points. - Intersection of $$x + 2y = 4$$ and $$2x - y = 3$$: From the first, $$x = 4 - 2y$$. Substitute into second: $$2(4 - 2y) - y = 3 \Rightarrow 8 - 4y - y = 3 \Rightarrow 8 - 5y = 3 \Rightarrow 5y = 5 \Rightarrow y = 1$$. Then $$x = 4 - 2(1) = 2$$. So vertex $$A = (2,1)$$. - Intersection of $$2x - y = 3$$ and $$x - y - 2 = 0$$: Rewrite third as $$x - y = 2 \Rightarrow y = x - 2$$. Substitute into second: $$2x - (x - 2) = 3 \Rightarrow 2x - x + 2 = 3 \Rightarrow x + 2 = 3 \Rightarrow x = 1$$. Then $$y = 1 - 2 = -1$$. So vertex $$B = (1,-1)$$. - Intersection of $$x + 2y = 4$$ and $$x - y - 2 = 0$$: From third, $$x = y + 2$$. Substitute into first: $$(y + 2) + 2y = 4 \Rightarrow 3y + 2 = 4 \Rightarrow 3y = 2 \Rightarrow y = \frac{2}{3}$$. Then $$x = \frac{2}{3} + 2 = \frac{8}{3}$$. So vertex $$C = \left(\frac{8}{3}, \frac{2}{3}\right)$$. 3. **Check if the triangle is right angled:** Use the distance formula to find lengths of sides: $$AB = \sqrt{(2 - 1)^2 + (1 - (-1))^2} = \sqrt{1^2 + 2^2} = \sqrt{5}$$. $$BC = \sqrt{\left(1 - \frac{8}{3}\right)^2 + \left(-1 - \frac{2}{3}\right)^2} = \sqrt{\left(-\frac{5}{3}\right)^2 + \left(-\frac{5}{3}\right)^2} = \sqrt{\frac{25}{9} + \frac{25}{9}} = \sqrt{\frac{50}{9}} = \frac{5\sqrt{2}}{3}$$. $$AC = \sqrt{\left(2 - \frac{8}{3}\right)^2 + \left(1 - \frac{2}{3}\right)^2} = \sqrt{\left(-\frac{2}{3}\right)^2 + \left(\frac{1}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$$. Check Pythagoras theorem: $$AB^2 = 5,$$ $$AC^2 + BC^2 = \left(\frac{\sqrt{5}}{3}\right)^2 + \left(\frac{5\sqrt{2}}{3}\right)^2 = \frac{5}{9} + \frac{50}{9} = \frac{55}{9}$$. Not equal. Try other combinations: $$BC^2 = \frac{50}{9},$$ $$AB^2 + AC^2 = 5 + \frac{5}{9} = \frac{45}{9} + \frac{5}{9} = \frac{50}{9}$$. So $$BC^2 = AB^2 + AC^2$$, confirming the triangle is right angled at vertex $$A$$. 4. **Calculate the area:** Use the formula for area of triangle with vertices: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$. Substitute: $$= \frac{1}{2} |2( -1 - \frac{2}{3}) + 1(\frac{2}{3} - 1) + \frac{8}{3}(1 - (-1))|$$ $$= \frac{1}{2} |2(-\frac{5}{3}) + 1(-\frac{1}{3}) + \frac{8}{3}(2)|$$ $$= \frac{1}{2} | -\frac{10}{3} - \frac{1}{3} + \frac{16}{3}| = \frac{1}{2} |\frac{5}{3}| = \frac{5}{6}$$. Recalculate carefully: $$2(-1 - \frac{2}{3}) = 2(-\frac{5}{3}) = -\frac{10}{3}$$ $$1(\frac{2}{3} - 1) = 1(-\frac{1}{3}) = -\frac{1}{3}$$ $$\frac{8}{3}(1 - (-1)) = \frac{8}{3} \times 2 = \frac{16}{3}$$ Sum inside absolute value: $$-\frac{10}{3} - \frac{1}{3} + \frac{16}{3} = \frac{5}{3}$$ Area: $$\frac{1}{2} \times \frac{5}{3} = \frac{5}{6}$$ This contradicts the given area 7.5. Check vertices again for calculation error: Vertices: $$A = (2,1), B = (1,-1), C = \left(\frac{8}{3}, \frac{2}{3}\right)$$ Use vector method for area: Vectors: $$\vec{AB} = (1 - 2, -1 - 1) = (-1, -2)$$ $$\vec{AC} = \left(\frac{8}{3} - 2, \frac{2}{3} - 1\right) = \left(\frac{2}{3}, -\frac{1}{3}\right)$$ Area: $$\frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} |(-1)(-\frac{1}{3}) - (-2)(\frac{2}{3})| = \frac{1}{2} |\frac{1}{3} + \frac{4}{3}| = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}$$ Still 0.8333, not 7.5. Re-examine vertices calculation: Intersection of $$x + 2y = 4$$ and $$x - y - 2 = 0$$: From third, $$x = y + 2$$. Substitute into first: $$(y + 2) + 2y = 4 \Rightarrow 3y + 2 = 4 \Rightarrow 3y = 2 \Rightarrow y = \frac{2}{3}$$ $$x = \frac{2}{3} + 2 = \frac{8}{3}$$ Correct. Intersection of $$2x - y = 3$$ and $$x - y - 2 = 0$$: From third, $$y = x - 2$$. Substitute into second: $$2x - (x - 2) = 3 \Rightarrow x + 2 = 3 \Rightarrow x = 1$$ $$y = 1 - 2 = -1$$ Correct. Intersection of $$x + 2y = 4$$ and $$2x - y = 3$$: From first, $$x = 4 - 2y$$. Substitute into second: $$2(4 - 2y) - y = 3 \Rightarrow 8 - 4y - y = 3 \Rightarrow 8 - 5y = 3 \Rightarrow 5y = 5 \Rightarrow y = 1$$ $$x = 4 - 2(1) = 2$$ Correct. Vertices are correct. Check area formula again: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ Substitute: $$= \frac{1}{2} |2(-1 - \frac{2}{3}) + 1(\frac{2}{3} - 1) + \frac{8}{3}(1 - (-1))|$$ Calculate each term: $$2(-1 - \frac{2}{3}) = 2(-\frac{5}{3}) = -\frac{10}{3}$$ $$1(\frac{2}{3} - 1) = 1(-\frac{1}{3}) = -\frac{1}{3}$$ $$\frac{8}{3}(1 - (-1)) = \frac{8}{3} \times 2 = \frac{16}{3}$$ Sum: $$-\frac{10}{3} - \frac{1}{3} + \frac{16}{3} = \frac{5}{3}$$ Area: $$\frac{1}{2} \times \frac{5}{3} = \frac{5}{6}$$ This is approximately 0.8333, not 7.5. Check if the problem meant area 7.5 or if a scale factor is missing. Alternatively, calculate area using base and height: Use side $$AB$$ as base: Length $$AB = \sqrt{5}$$. Find height from $$C$$ to line $$AB$$. Equation of $$AB$$: Points $$A(2,1)$$ and $$B(1,-1)$$. Slope: $$m = \frac{-1 - 1}{1 - 2} = \frac{-2}{-1} = 2$$. Equation: $$y - 1 = 2(x - 2) \Rightarrow y = 2x - 3$$. Distance from $$C\left(\frac{8}{3}, \frac{2}{3}\right)$$ to line $$y = 2x - 3$$: Rewrite line as: $$2x - y - 3 = 0$$. Distance formula: $$d = \frac{|2(\frac{8}{3}) - \frac{2}{3} - 3|}{\sqrt{2^2 + (-1)^2}} = \frac{|\frac{16}{3} - \frac{2}{3} - 3|}{\sqrt{5}} = \frac{|\frac{14}{3} - 3|}{\sqrt{5}} = \frac{|\frac{14}{3} - \frac{9}{3}|}{\sqrt{5}} = \frac{\frac{5}{3}}{\sqrt{5}} = \frac{5}{3\sqrt{5}} = \frac{\sqrt{5}}{3}$$. Area: $$\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \sqrt{5} \times \frac{\sqrt{5}}{3} = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}$$ Again 0.8333. **Conclusion:** The triangle is right angled (proved by Pythagoras) but the area is $$\frac{5}{6}$$ square units, not 7.5. **Final answer:** - The triangle formed by the given lines is right angled at vertex $$A(2,1)$$. - The area of the triangle is $$\frac{5}{6}$$ square units. If the problem states area 7.5, please verify the equations or units.