1. **State the problem:** We are given three lines forming a triangle with equations: $$x + 2y = 4,$$ $$2x - y = 3,$$ $$x - y = 2.$$ 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 $x + 2y = 4$ and $2x - y = 3$: From $x + 2y = 4$, express $x = 4 - 2y$. Substitute into $2x - y = 3$: $$2(4 - 2y) - y = 3 \implies 8 - 4y - y = 3 \implies 8 - 5y = 3 \implies 5y = 5 \implies y = 1.$$ Then $x = 4 - 2(1) = 2$. So vertex $A = (2,1)$. - Intersection of $2x - y = 3$ and $x - y = 2$: From $x - y = 2$, express $y = x - 2$. Substitute into $2x - y = 3$: $$2x - (x - 2) = 3 \implies 2x - x + 2 = 3 \implies x + 2 = 3 \implies x = 1.$$ Then $y = 1 - 2 = -1$. So vertex $B = (1,-1)$. - Intersection of $x + 2y = 4$ and $x - y = 2$: From $x - y = 2$, express $x = y + 2$. Substitute into $x + 2y = 4$: $$(y + 2) + 2y = 4 \implies 3y + 2 = 4 \implies 3y = 2 \implies y = \frac{2}{3}.$$ Then $x = \frac{2}{3} + 2 = \frac{8}{3}$. So vertex $C = \left(\frac{8}{3}, \frac{2}{3}\right)$. 3. **Calculate the lengths of the sides using distance formula:** Distance between points $(x_1,y_1)$ and $(x_2,y_2)$ is: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$ - Side $AB$: $$AB = \sqrt{(1 - 2)^2 + (-1 - 1)^2} = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}.$$ - Side $BC$: $$BC = \sqrt{\left(\frac{8}{3} - 1\right)^2 + \left(\frac{2}{3} + 1\right)^2} = \sqrt{\left(\frac{5}{3}\right)^2 + \left(\frac{5}{3}\right)^2} = \sqrt{\frac{25}{9} + \frac{25}{9}} = \sqrt{\frac{50}{9}} = \frac{5\sqrt{2}}{3}.$$ - Side $CA$: $$CA = \sqrt{\left(\frac{8}{3} - 2\right)^2 + \left(\frac{2}{3} - 1\right)^2} = \sqrt{\left(\frac{8}{3} - \frac{6}{3}\right)^2 + \left(\frac{2}{3} - \frac{3}{3}\right)^2} = \sqrt{\left(\frac{2}{3}\right)^2 + \left(-\frac{1}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}.$$ 4. **Check for right angle using Pythagoras theorem:** Check if $AB^2 + CA^2 = BC^2$: $$AB^2 + CA^2 = (\sqrt{5})^2 + \left(\frac{\sqrt{5}}{3}\right)^2 = 5 + \frac{5}{9} = \frac{45}{9} + \frac{5}{9} = \frac{50}{9}.$$ $$BC^2 = \left(\frac{5\sqrt{2}}{3}\right)^2 = \frac{25 \times 2}{9} = \frac{50}{9}.$$ Since $AB^2 + CA^2 = BC^2$, the triangle is right-angled at vertex $A$. 5. **Calculate the area of the triangle:** Area of right triangle = $\frac{1}{2} \times$ (two perpendicular sides). Here, sides $AB$ and $CA$ are perpendicular. $$\text{Area} = \frac{1}{2} \times AB \times CA = \frac{1}{2} \times \sqrt{5} \times \frac{\sqrt{5}}{3} = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6} \times 1 = 2.5.$$ This seems inconsistent with the problem statement, so let's verify using the coordinate formula for area: Area = $$\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ Substitute $A=(2,1)$, $B=(1,-1)$, $C=\left(\frac{8}{3}, \frac{2}{3}\right)$: $$= \frac{1}{2} |2(-1 - \frac{2}{3}) + 1\left(\frac{2}{3} - 1\right) + \frac{8}{3}(1 + 1)|$$ $$= \frac{1}{2} |2\left(-\frac{5}{3}\right) + 1\left(-\frac{1}{3}\right) + \frac{8}{3} \times 2|$$ $$= \frac{1}{2} \left| -\frac{10}{3} - \frac{1}{3} + \frac{16}{3} \right| = \frac{1}{2} \left| \frac{5}{3} \right| = \frac{5}{6} = 0.8333.$$ This is not 7.5, so re-check the calculations: Recalculate $CA$ length: $$CA = \sqrt{\left(\frac{8}{3} - 2\right)^2 + \left(\frac{2}{3} - 1\right)^2} = \sqrt{\left(\frac{8}{3} - \frac{6}{3}\right)^2 + \left(\frac{2}{3} - \frac{3}{3}\right)^2} = \sqrt{\left(\frac{2}{3}\right)^2 + \left(-\frac{1}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}.$$ Recalculate $AB$ length: $$AB = \sqrt{(1 - 2)^2 + (-1 - 1)^2} = \sqrt{1 + 4} = \sqrt{5}.$$ Area using $\frac{1}{2} \times AB \times CA$: $$= \frac{1}{2} \times \sqrt{5} \times \frac{\sqrt{5}}{3} = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6} = 0.8333.$$ This is too small, so the error is in identifying the right angle or vertices. Check if right angle is at $B$ or $C$: - $AB^2 = 5$ - $BC^2 = \left(\frac{5\sqrt{2}}{3}\right)^2 = \frac{50}{9} \approx 5.555$ - $CA^2 = \left(\frac{\sqrt{5}}{3}\right)^2 = \frac{5}{9} \approx 0.555$ Check $AB^2 + BC^2 = CA^2$? $$5 + 5.555 = 10.555 \neq 0.555$$ Check $BC^2 + CA^2 = AB^2$? $$5.555 + 0.555 = 6.11 \neq 5$$ Check $AB^2 + CA^2 = BC^2$? $$5 + 0.555 = 5.555 = BC^2$$ So right angle is at $A$. Area should be $\frac{1}{2} \times AB \times CA = \frac{1}{2} \times \sqrt{5} \times \frac{\sqrt{5}}{3} = \frac{5}{6} = 0.8333$ which contradicts the problem statement. Re-express vertices as decimals: $A = (2,1)$ $B = (1,-1)$ $C = (2.6667, 0.6667)$ Calculate area using determinant formula: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ $$= \frac{1}{2} |2(-1 - 0.6667) + 1(0.6667 - 1) + 2.6667(1 + 1)|$$ $$= \frac{1}{2} |2(-1.6667) + 1(-0.3333) + 2.6667(2)|$$ $$= \frac{1}{2} |-3.3334 - 0.3333 + 5.3334| = \frac{1}{2} |1.6667| = 0.8333.$$ This confirms area is approximately 0.8333, not 7.5. **Conclusion:** The triangle is right-angled at $A$, but the area is approximately 0.8333 square units, not 7.5. If the problem states area is 7.5, possibly the equations or vertices need rechecking or scaling. --- **Final answers:** - The triangle is right-angled at vertex $A = (2,1)$. - The area of the triangle is approximately $\boxed{0.8333}$ square units.