1. **State the problem:** We are given the equations of the sides of a triangle: $$x + 2y = 4,$$ $$2x - y = 3,$$ $$x - y - 2 = 0.$$ We need to prove that the triangle formed by these lines 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 $$2x - y = 3$$: From the first equation, $$x = 4 - 2y$$. Substitute into the 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 the third equation as $$x - y = 2 \Rightarrow y = x - 2$$. Substitute into the 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 the third equation, $$x = y + 2$$. Substitute into the 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:** Calculate the slopes of the sides. - 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 $$CA$$: $$m_{CA} = \frac{\frac{2}{3} - 1}{\frac{8}{3} - 2} = \frac{-\frac{1}{3}}{\frac{8}{3} - \frac{6}{3}} = \frac{-\frac{1}{3}}{\frac{2}{3}} = -\frac{1}{3} \times \frac{3}{2} = -\frac{1}{2}.$$ Check if any two sides are perpendicular by verifying if the product of their slopes is $$-1$$. - $$m_{AB} \times m_{BC} = 2 \times 1 = 2 \neq -1$$ - $$m_{BC} \times m_{CA} = 1 \times (-\frac{1}{2}) = -\frac{1}{2} \neq -1$$ - $$m_{CA} \times m_{AB} = -\frac{1}{2} \times 2 = -1$$ Since $$m_{CA} \times m_{AB} = -1$$, sides $$CA$$ and $$AB$$ are perpendicular, so the triangle is right-angled at vertex $$A$$. 4. **Calculate the area of the triangle:** Use the formula for the area of a triangle given coordinates: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ Substitute $$A(2,1), B(1,-1), C\left(\frac{8}{3}, \frac{2}{3}\right)$$: $$= \frac{1}{2} |2(-1 - \frac{2}{3}) + 1(\frac{2}{3} - 1) + \frac{8}{3}(1 - (-1))|$$ Calculate inside: $$-1 - \frac{2}{3} = -\frac{3}{3} - \frac{2}{3} = -\frac{5}{3}$$ $$\frac{2}{3} - 1 = \frac{2}{3} - \frac{3}{3} = -\frac{1}{3}$$ $$1 - (-1) = 2$$ So, $$= \frac{1}{2} |2 \times (-\frac{5}{3}) + 1 \times (-\frac{1}{3}) + \frac{8}{3} \times 2|$$ $$= \frac{1}{2} | -\frac{10}{3} - \frac{1}{3} + \frac{16}{3} | = \frac{1}{2} | \frac{5}{3} | = \frac{5}{6}.$$ This seems incorrect; re-check calculations carefully. Recalculate the sum inside the absolute value: $$2(-\frac{5}{3}) = -\frac{10}{3}$$ $$1(-\frac{1}{3}) = -\frac{1}{3}$$ $$\frac{8}{3} \times 2 = \frac{16}{3}$$ Sum: $$-\frac{10}{3} - \frac{1}{3} + \frac{16}{3} = \frac{-10 - 1 + 16}{3} = \frac{5}{3}$$ Area: $$= \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}$$ This is not 7.5, so let's try another method: use the base and height from the right angle. Since the right angle is at $$A$$, the sides $$AB$$ and $$AC$$ are perpendicular and can be considered base and height. Calculate lengths: $$AB = \sqrt{(1 - 2)^2 + (-1 - 1)^2} = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}.$$ $$AC = \sqrt{\left(\frac{8}{3} - 2\right)^2 + \left(\frac{2}{3} - 1\right)^2} = \sqrt{\left(\frac{8}{3} - \frac{6}{3}\right)^2 + \left(\frac{2}{3} - \frac{3}{3}\right)^2} = \sqrt{\left(\frac{2}{3}\right)^2 + \left(-\frac{1}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}.$$ Area: $$= \frac{1}{2} \times AB \times AC = \frac{1}{2} \times \sqrt{5} \times \frac{\sqrt{5}}{3} = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}.$$ This is again $$\frac{5}{6}$$, which is approximately 0.8333, not 7.5. Re-examine the vertices for calculation errors. Check vertex $$C$$ coordinates: From $$x - y - 2 = 0$$, $$x = y + 2$$. Substitute into $$x + 2y = 4$$: $$(y + 2) + 2y = 4 \Rightarrow 3y + 2 = 4 \Rightarrow 3y = 2 \Rightarrow y = \frac{2}{3}.$$ Then $$x = \frac{2}{3} + 2 = \frac{8}{3}$$. Coordinates are correct. Check the length calculations again: $$AC = \sqrt{\left(\frac{8}{3} - 2\right)^2 + \left(\frac{2}{3} - 1\right)^2} = \sqrt{\left(\frac{8}{3} - \frac{6}{3}\right)^2 + \left(\frac{2}{3} - \frac{3}{3}\right)^2} = \sqrt{\left(\frac{2}{3}\right)^2 + \left(-\frac{1}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}.$$ $$AB = \sqrt{(1 - 2)^2 + (-1 - 1)^2} = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}.$$ Area: $$= \frac{1}{2} \times \sqrt{5} \times \frac{\sqrt{5}}{3} = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}.$$ This is consistent but contradicts the problem statement. Try using the distance formula for all sides to check if the triangle is correct. - $$BC = \sqrt{\left(\frac{8}{3} - 1\right)^2 + \left(\frac{2}{3} - (-1)\right)^2} = \sqrt{\left(\frac{8}{3} - \frac{3}{3}\right)^2 + \left(\frac{2}{3} + 1\right)^2} = \sqrt{\left(\frac{5}{3}\right)^2 + \left(\frac{5}{3}\right)^2} = \sqrt{\frac{25}{9} + \frac{25}{9}} = \sqrt{\frac{50}{9}} = \frac{5\sqrt{2}}{3}.$$ Check if the triangle is right angled by Pythagoras: $$AB^2 + AC^2 = 5 + \frac{5}{9} = \frac{45}{9} + \frac{5}{9} = \frac{50}{9}$$ $$BC^2 = \left(\frac{5\sqrt{2}}{3}\right)^2 = \frac{25 \times 2}{9} = \frac{50}{9}$$ Since $$AB^2 + AC^2 = BC^2$$, the triangle is right angled at $$A$$. Now calculate the area using base $$AB$$ and height $$AC$$: $$\text{Area} = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times \sqrt{5} \times \frac{\sqrt{5}}{3} = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}.$$ This is approximately 0.8333, not 7.5. Check if the problem expects area in different units or if the lines are correct. Alternatively, calculate area using the formula for area of triangle formed by three lines: Area = $$\frac{|c_1(a_2b_3 - a_3b_2) + c_2(a_3b_1 - a_1b_3) + c_3(a_1b_2 - a_2b_1)|}{2|a_1b_2 - a_2b_1|}$$ Where lines are $$a_ix + b_iy + c_i = 0$$. Rewrite lines: 1) $$x + 2y - 4 = 0$$, so $$a_1=1, b_1=2, c_1=-4$$ 2) $$2x - y - 3 = 0$$, so $$a_2=2, b_2=-1, c_2=-3$$ 3) $$x - y - 2 = 0$$, so $$a_3=1, b_3=-1, c_3=-2$$ Calculate numerator: $$c_1(a_2b_3 - a_3b_2) = -4(2 \times (-1) - 1 \times (-1)) = -4(-2 + 1) = -4(-1) = 4$$ $$c_2(a_3b_1 - a_1b_3) = -3(1 \times 2 - 1 \times (-1)) = -3(2 + 1) = -3(3) = -9$$ $$c_3(a_1b_2 - a_2b_1) = -2(1 \times (-1) - 2 \times 2) = -2(-1 - 4) = -2(-5) = 10$$ Sum numerator: $$4 - 9 + 10 = 5$$ Calculate denominator: $$2|a_1b_2 - a_2b_1| = 2|1 \times (-1) - 2 \times 2| = 2|-1 - 4| = 2| -5| = 10$$ Area: $$= \frac{5}{10} = 0.5$$ This is even smaller. Since the problem states area is 7.5, possibly the lines or the problem statement has a typo or different scale. **Summary:** - The triangle is right angled at vertex $$A$$. - The area calculated from vertices is $$\frac{5}{6}$$ square units. - The problem's stated area 7.5 does not match calculations with given lines. **Final answer:** The triangle formed by the lines is right angled at $$A(2,1)$$. The area of the triangle is $$\frac{5}{6}$$ square units (approximately 0.8333), not 7.5. If the problem expects 7.5, please verify the line equations or units.