1. **State the problem:** We are given three lines forming a triangle with equations $x+2y=4$, $2x-y=3$, and $x-y-2=0$. We need to prove the triangle is right angled and find its area. 2. **Find the vertices of the triangle:** The vertices are the intersection points of the lines. - Intersection of $x+2y=4$ and $2x-y=3$: Solve the system: $$\begin{cases} x+2y=4 \\ 2x - y=3 \end{cases}$$ Multiply the first equation by 1 and the second by 2 to eliminate $y$: $$x + 2y = 4$$ $$4x - 2y = 6$$ Add: $$5x = 10 \Rightarrow x=2$$ 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 $x - y - 2=0$: Rewrite the second as $x - y = 2$. Subtract the two equations: $$2x - y = 3$$ $$x - y = 2$$ Subtract second from 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 $x - y - 2 = 0$: From $x - y - 2=0$, we get $x = y + 2$. Substitute into $x + 2y = 4$: $$(y + 2) + 2y = 4 \Rightarrow 3y + 2 = 4 \Rightarrow 3y = 2 \Rightarrow y = \frac{2}{3}$$ 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:** Calculate the lengths of sides $AB$, $BC$, and $CA$: - $AB = \sqrt{(2-1)^2 + (1 - (-1))^2} = \sqrt{1^2 + 2^2} = \sqrt{5}$ - $BC = \sqrt{\left(\frac{8}{3} - 1\right)^2 + \left(\frac{2}{3} - (-1)\right)^2} = \sqrt{\left(\frac{5}{3}\right)^2 + \left(\frac{5}{3}\right)^2} = \sqrt{2 \times \left(\frac{25}{9}\right)} = \sqrt{\frac{50}{9}} = \frac{5\sqrt{2}}{3}$ - $CA = \sqrt{\left(\frac{8}{3} - 2\right)^2 + \left(\frac{2}{3} - 1\right)^2} = \sqrt{\left(\frac{2}{3}\right)^2 + \left(-\frac{1}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$ Check the Pythagorean theorem: - $AB^2 = 5$ - $BC^2 = \left(\frac{5\sqrt{2}}{3}\right)^2 = \frac{25 \times 2}{9} = \frac{50}{9}$ - $CA^2 = \left(\frac{\sqrt{5}}{3}\right)^2 = \frac{5}{9}$ Sum of $CA^2 + AB^2 = \frac{5}{9} + 5 = \frac{5}{9} + \frac{45}{9} = \frac{50}{9} = BC^2$ Since $CA^2 + AB^2 = BC^2$, the triangle is right angled at vertex $A$. 4. **Calculate the area of the triangle:** Area of a right triangle = $\frac{1}{2} \times$ (two perpendicular sides). The perpendicular sides are $AB = \sqrt{5}$ and $CA = \frac{\sqrt{5}}{3}$. Area = $\frac{1}{2} \times \sqrt{5} \times \frac{\sqrt{5}}{3} = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6} \approx 0.8333$ (This is incorrect, so let's use the coordinate formula for area). Alternatively, use the formula for area given vertices $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$: $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ Substitute: $$= \frac{1}{2} |2(-1 - \frac{2}{3}) + 1(\frac{2}{3} - 1) + \frac{8}{3}(1 - (-1))|$$ Calculate inside: $$= \frac{1}{2} |2(-\frac{5}{3}) + 1(-\frac{1}{3}) + \frac{8}{3} \times 2| = \frac{1}{2} |-\frac{10}{3} - \frac{1}{3} + \frac{16}{3}| = \frac{1}{2} |\frac{5}{3}| = \frac{5}{6}$$ This is $\frac{5}{6}$, which is approximately 0.8333, not 7.5. Check calculations again for vertices or area. Recalculate vertices carefully: - $A = (2,1)$ - $B = (1,-1)$ - $C = \left(\frac{8}{3}, \frac{2}{3}\right)$ Calculate area again: $$\text{Area} = \frac{1}{2} |2(-1 - \frac{2}{3}) + 1(\frac{2}{3} - 1) + \frac{8}{3}(1 - (-1))|$$ 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 $\frac{5}{6}$, not 7.5. Since the problem states area is 7.5, check if vertices or equations are correct or if the area is requested in different units. Alternatively, calculate area using base and height: Use line $x - y - 2=0$ as base. Length of base $BC$: $$BC = \sqrt{\left(\frac{8}{3} - 1\right)^2 + \left(\frac{2}{3} + 1\right)^2} = \sqrt{\left(\frac{5}{3}\right)^2 + \left(\frac{5}{3}\right)^2} = \frac{5\sqrt{2}}{3}$$ Distance from point $A(2,1)$ to line $x - y - 2=0$: $$d = \frac{|2 - 1 - 2|}{\sqrt{1^2 + (-1)^2}} = \frac{|-1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$$ Area = $\frac{1}{2} \times$ base $\times$ height = $$\frac{1}{2} \times \frac{5\sqrt{2}}{3} \times \frac{1}{\sqrt{2}} = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}$$ Again $\frac{5}{6}$. Therefore, the area is $\frac{5}{6}$ square units, not 7.5. **Final answers:** - The triangle is right angled at vertex $A$. - The area of the triangle is $\frac{5}{6}$ square units. If the problem insists area is 7.5, please verify the equations or units.