1. **State the problem:** We are given the equations of the sides of a triangle: $x+2y=4$, $2x-y=3$, and $x-y-2=0$. We need to prove that the triangle is right angled and find its area, which is claimed to be 7.5 square units. 2. **Find the vertices of the triangle:** The vertices are the intersection points of the pairs of lines. - Intersection of $x+2y=4$ and $2x-y=3$: $$\begin{cases} x+2y=4 \\ 2x - y=3 \end{cases}$$ Multiply the first equation by 1 and the second by 2 to eliminate $y$: $$\begin{cases} x+2y=4 \\ 4x - 2y=6 \end{cases}$$ Add the two equations: $$x + 2y + 4x - 2y = 4 + 6 \Rightarrow 5x = 10 \Rightarrow x = 2$$ Substitute $x=2$ into $x+2y=4$: $$2 + 2y = 4 \Rightarrow 2y = 2 \Rightarrow y = 1$$ So, vertex $A = (2,1)$. - Intersection of $2x - y = 3$ and $x - y - 2 = 0$ (or $x - y = 2$): $$\begin{cases} 2x - y = 3 \\ x - y = 2 \end{cases}$$ Subtract the second from the first: $$(2x - y) - (x - y) = 3 - 2 \Rightarrow x = 1$$ Substitute $x=1$ into $x - y = 2$: $$1 - y = 2 \Rightarrow y = -1$$ So, vertex $B = (1,-1)$. - Intersection of $x + 2y = 4$ and $x - y = 2$: $$\begin{cases} x + 2y = 4 \\ x - y = 2 \end{cases}$$ Subtract the second from the first: $$(x + 2y) - (x - y) = 4 - 2 \Rightarrow 3y = 2 \Rightarrow y = \frac{2}{3}$$ Substitute $y=\frac{2}{3}$ into $x - y = 2$: $$x - \frac{2}{3} = 2 \Rightarrow x = 2 + \frac{2}{3} = \frac{8}{3}$$ So, vertex $C = \left(\frac{8}{3}, \frac{2}{3}\right)$. 3. **Check if the triangle is right angled:** Calculate the slopes of the sides and check if any two are perpendicular (product of slopes = -1). - Slope of $AB$: $$m_{AB} = \frac{y_B - y_A}{x_B - x_A} = \frac{-1 - 1}{1 - 2} = \frac{-2}{-1} = 2$$ - Slope of $BC$: $$m_{BC} = \frac{\frac{2}{3} - (-1)}{\frac{8}{3} - 1} = \frac{\frac{2}{3} + 1}{\frac{8}{3} - \frac{3}{3}} = \frac{\frac{5}{3}}{\frac{5}{3}} = 1$$ - Slope of $CA$: $$m_{CA} = \frac{\frac{2}{3} - 1}{\frac{8}{3} - 2} = \frac{\frac{2}{3} - \frac{3}{3}}{\frac{8}{3} - \frac{6}{3}} = \frac{-\frac{1}{3}}{\frac{2}{3}} = -\frac{1}{2}$$ Check products: $$m_{AB} \times m_{CA} = 2 \times \left(-\frac{1}{2}\right) = -1$$ Since the product is $-1$, the sides $AB$ and $CA$ are perpendicular, so the triangle is right angled at vertex $A$. 4. **Calculate the area of the triangle:** Use the formula for area given coordinates: $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ Substitute $$A=(2,1), B=(1,-1), C=\left(\frac{8}{3}, \frac{2}{3}\right)$$: $$\frac{1}{2} \left| 2(-1 - \frac{2}{3}) + 1(\frac{2}{3} - 1) + \frac{8}{3}(1 + 1) \right|$$ Simplify inside the absolute value: $$2\left(-\frac{5}{3}\right) + 1\left(-\frac{1}{3}\right) + \frac{8}{3} \times 2 = -\frac{10}{3} - \frac{1}{3} + \frac{16}{3} = \frac{5}{3}$$ Area: $$\frac{1}{2} \times \frac{5}{3} = \frac{5}{6}$$ Wait, this is not 7.5. Let's check calculation carefully. Recalculate the expression inside absolute value: $$2(-1 - \frac{2}{3}) = 2 \times \left(-\frac{5}{3}\right) = -\frac{10}{3}$$ $$1(\frac{2}{3} - 1) = 1 \times \left(-\frac{1}{3}\right) = -\frac{1}{3}$$ $$\frac{8}{3}(1 + 1) = \frac{8}{3} \times 2 = \frac{16}{3}$$ 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 not 7.5. Let's try another method: use base and height from right angle. Since angle at $A$ is right angled, use lengths $AB$ and $AC$ as base and height. Calculate length $AB$: $$AB = \sqrt{(1-2)^2 + (-1-1)^2} = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$ Calculate length $AC$: $$AC = \sqrt{\left(\frac{8}{3} - 2\right)^2 + \left(\frac{2}{3} - 1\right)^2} = \sqrt{\left(\frac{8}{3} - \frac{6}{3}\right)^2 + \left(\frac{2}{3} - \frac{3}{3}\right)^2} = \sqrt{\left(\frac{2}{3}\right)^2 + \left(-\frac{1}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$$ Area: $$\frac{1}{2} \times AB \times AC = \frac{1}{2} \times \sqrt{5} \times \frac{\sqrt{5}}{3} = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}$$ Still not 7.5. The problem states area is 7.5, so let's check if vertices are correct or if the problem expects a different interpretation. Check if the vertices are correct by substituting back: - For $C$, $x - y - 2 = 0$: $$\frac{8}{3} - \frac{2}{3} - 2 = \frac{6}{3} - 2 = 2 - 2 = 0$$ Correct. - For $B$, $2x - y = 3$: $$2(1) - (-1) = 2 + 1 = 3$$ Correct. - For $A$, $x + 2y = 4$: $$2 + 2(1) = 2 + 2 = 4$$ Correct. All vertices are correct. Check if the problem meant area 7.5 or 5/6. The area calculated is $$\frac{5}{6} \approx 0.8333$$, which is much less than 7.5. Try calculating area using the determinant formula with decimals: $$\frac{1}{2} |2(-1 - 0.6667) + 1(0.6667 - 1) + 2.6667(1 + 1)|$$ Calculate inside: $$2(-1.6667) + 1(-0.3333) + 2.6667(2) = -3.3334 - 0.3333 + 5.3334 = 1.6667$$ Area: $$\frac{1}{2} \times 1.6667 = 0.8333$$ Still 0.8333. **Conclusion:** The triangle is right angled at $A$ but the area is $$\frac{5}{6}$$ square units, not 7.5. If the problem insists area is 7.5, there might be a typo or different scaling. **Final answer:** The triangle formed by the lines is right angled at vertex $A(2,1)$. The area of the triangle is $$\frac{5}{6}$$ square units.