1. **State the problem:** We are given three lines forming a triangle: - Line 1: $$x + 2y = 4$$ - Line 2: $$2x - y = 3$$ - Line 3: $$x - y - 2 = 0$$ We need to prove the triangle is right angled and find its area. 2. **Find the vertices of the triangle by solving pairs of equations:** - Intersection of Line 1 and Line 2: Solve $$x + 2y = 4$$ and $$2x - y = 3$$. From Line 1: $$x = 4 - 2y$$. Substitute into Line 2: $$2(4 - 2y) - y = 3$$ Simplify: $$8 - 4y - y = 3$$ $$8 - 5y = 3$$ $$-5y = 3 - 8 = -5$$ $$y = 1$$ Substitute back to find $$x = 4 - 2(1) = 2$$. So vertex A is $$(2,1)$$. - Intersection of Line 2 and Line 3: Solve $$2x - y = 3$$ and $$x - y - 2 = 0$$. From Line 3: $$y = x - 2$$. Substitute into Line 2: $$2x - (x - 2) = 3$$ Simplify: $$2x - x + 2 = 3$$ $$x + 2 = 3$$ $$x = 1$$ Then $$y = 1 - 2 = -1$$. So vertex B is $$(1,-1)$$. - Intersection of Line 1 and Line 3: Solve $$x + 2y = 4$$ and $$x - y - 2 = 0$$. From Line 3: $$x = y + 2$$. Substitute into Line 1: $$(y + 2) + 2y = 4$$ Simplify: $$3y + 2 = 4$$ $$3y = 2$$ $$y = \frac{2}{3}$$ Then $$x = \frac{2}{3} + 2 = \frac{8}{3}$$. So vertex C is $$\left(\frac{8}{3}, \frac{2}{3}\right)$$. 3. **Calculate the lengths of the sides using distance formula:** Distance between points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is $$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$. - Side AB: $$\sqrt{(1 - 2)^2 + (-1 - 1)^2} = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$ - Side BC: $$\sqrt{\left(\frac{8}{3} - 1\right)^2 + \left(\frac{2}{3} + 1\right)^2} = \sqrt{\left(\frac{5}{3}\right)^2 + \left(\frac{5}{3}\right)^2} = \sqrt{\frac{25}{9} + \frac{25}{9}} = \sqrt{\frac{50}{9}} = \frac{5\sqrt{2}}{3}$$ - Side CA: $$\sqrt{\left(\frac{8}{3} - 2\right)^2 + \left(\frac{2}{3} - 1\right)^2} = \sqrt{\left(\frac{2}{3}\right)^2 + \left(-\frac{1}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$$ 4. **Check for right angle using Pythagoras theorem:** The triangle is right angled if the square of the longest side equals the sum of squares of the other two sides. Longest side is BC: $$\left(\frac{5\sqrt{2}}{3}\right)^2 = \frac{25 \times 2}{9} = \frac{50}{9}$$ Sum of squares of AB and CA: $$(\sqrt{5})^2 + \left(\frac{\sqrt{5}}{3}\right)^2 = 5 + \frac{5}{9} = \frac{45}{9} + \frac{5}{9} = \frac{50}{9}$$ Since both are equal, the triangle is right angled at vertex A. 5. **Calculate the area of the triangle:** Area formula for triangle with vertices $$(x_1,y_1), (x_2,y_2), (x_3,y_3)$$ is $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ Substitute vertices A(2,1), B(1,-1), C(8/3, 2/3): $$\frac{1}{2} | 2(-1 - \frac{2}{3}) + 1(\frac{2}{3} - 1) + \frac{8}{3}(1 + 1) |$$ Simplify inside: $$2\left(-\frac{5}{3}\right) + 1\left(-\frac{1}{3}\right) + \frac{8}{3} \times 2$$ $$= -\frac{10}{3} - \frac{1}{3} + \frac{16}{3} = \frac{5}{3}$$ Area: $$\frac{1}{2} \times \frac{5}{3} = \frac{5}{6}$$ This seems inconsistent with the problem statement, so let's verify using base and height from right angle. Since right angle is at A, use sides AB and AC as legs: Length AB = $$\sqrt{5}$$, length AC = $$\frac{\sqrt{5}}{3}$$. Area = $$\frac{1}{2} \times \sqrt{5} \times \frac{\sqrt{5}}{3} = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}$$. This is $$7.5$$ not matching the problem's 7.5. Let's check the calculations again for vertices or distances. Recalculate vertex C: From Line 3: $$x - y = 2$$ From Line 1: $$x + 2y = 4$$ Add both equations: $$(x - y) + (x + 2y) = 2 + 4$$ $$2x + y = 6$$ From Line 3: $$y = x - 2$$ Substitute into $$2x + y = 6$$: $$2x + (x - 2) = 6$$ $$3x - 2 = 6$$ $$3x = 8 \Rightarrow x = \frac{8}{3}$$ Then $$y = \frac{8}{3} - 2 = \frac{8}{3} - \frac{6}{3} = \frac{2}{3}$$. So vertex C is correct. Let's calculate area using coordinate geometry 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: $$2(-1 - \frac{2}{3}) + 1(\frac{2}{3} - 1) + \frac{8}{3}(1 + 1)$$ Calculate each term: $$2 \times -\frac{5}{3} = -\frac{10}{3}$$ $$1 \times -\frac{1}{3} = -\frac{1}{3}$$ $$\frac{8}{3} \times 2 = \frac{16}{3}$$ Sum: $$-\frac{10}{3} - \frac{1}{3} + \frac{16}{3} = \frac{5}{3}$$ Area: $$\frac{1}{2} \times \frac{5}{3} = \frac{5}{6} = 0.8333$$ This is not 7.5, so let's try using base and height from the right angle vertex. Use AB as base: length $$\sqrt{5} \approx 2.236$$ Find height by distance from C to line AB. Equation of AB: Points A(2,1), 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 point C $$\left(\frac{8}{3}, \frac{2}{3}\right)$$ to line $$y = 2x - 3$$ is Use formula for distance from point to line $$Ax + By + C = 0$$: Rewrite line as $$2x - y - 3 = 0$$ Distance $$d = \frac{|2(\frac{8}{3}) - 1(\frac{2}{3}) - 3|}{\sqrt{2^2 + (-1)^2}} = \frac{|\frac{16}{3} - \frac{2}{3} - 3|}{\sqrt{5}} = \frac{|\frac{14}{3} - 3|}{\sqrt{5}} = \frac{|\frac{14}{3} - \frac{9}{3}|}{\sqrt{5}} = \frac{\frac{5}{3}}{\sqrt{5}} = \frac{5}{3\sqrt{5}}$$ Simplify: $$d = \frac{5}{3\sqrt{5}} = \frac{5\sqrt{5}}{3 \times 5} = \frac{\sqrt{5}}{3}$$ Area = $$\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \sqrt{5} \times \frac{\sqrt{5}}{3} = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}$$ This confirms area is $$\frac{5}{6}$$ square units, not 7.5. **Final conclusion:** The triangle formed by the given lines is right angled (right angle at vertex A) and its area is $$\frac{5}{6}$$ square units. If the problem states area 7.5, please verify the equations or vertices again.