1. **Stating the problem:** We need to find the shape of the region defined by the system of inequalities: $$4x + y \geq 8$$ $$3x + 4y \leq 24$$ $$x + 6y \geq 12$$ 2. **Understanding the problem:** Each inequality represents a half-plane bounded by a line. The solution region is the intersection of these half-planes. 3. **Find the boundary lines by converting inequalities to equalities:** - Line 1: $$4x + y = 8$$ - Line 2: $$3x + 4y = 24$$ - Line 3: $$x + 6y = 12$$ 4. **Find the intersection points of these lines to determine vertices of the polygon:** - Intersection of Line 1 and Line 2: $$\begin{cases}4x + y = 8 \\ 3x + 4y = 24\end{cases}$$ Multiply first equation by 4: $$16x + 4y = 32$$ Subtract second equation: $$16x + 4y - (3x + 4y) = 32 - 24$$ $$13x = 8$$ $$x = \frac{8}{13}$$ Substitute back: $$4 \times \frac{8}{13} + y = 8$$ $$\frac{32}{13} + y = 8$$ $$y = 8 - \frac{32}{13} = \frac{104}{13} - \frac{32}{13} = \frac{72}{13}$$ - Intersection of Line 2 and Line 3: $$\begin{cases}3x + 4y = 24 \\ x + 6y = 12\end{cases}$$ Multiply second equation by 3: $$3x + 18y = 36$$ Subtract first equation: $$3x + 18y - (3x + 4y) = 36 - 24$$ $$14y = 12$$ $$y = \frac{12}{14} = \frac{6}{7}$$ Substitute back: $$x + 6 \times \frac{6}{7} = 12$$ $$x + \frac{36}{7} = 12$$ $$x = 12 - \frac{36}{7} = \frac{84}{7} - \frac{36}{7} = \frac{48}{7}$$ - Intersection of Line 1 and Line 3: $$\begin{cases}4x + y = 8 \\ x + 6y = 12\end{cases}$$ Multiply second equation by 4: $$4x + 24y = 48$$ Subtract first equation: $$4x + 24y - (4x + y) = 48 - 8$$ $$23y = 40$$ $$y = \frac{40}{23}$$ Substitute back: $$x + 6 \times \frac{40}{23} = 12$$ $$x + \frac{240}{23} = 12$$ $$x = 12 - \frac{240}{23} = \frac{276}{23} - \frac{240}{23} = \frac{36}{23}$$ 5. **Vertices of the polygon are:** $$\left(\frac{8}{13}, \frac{72}{13}\right), \left(\frac{48}{7}, \frac{6}{7}\right), \left(\frac{36}{23}, \frac{40}{23}\right)$$ 6. **Conclusion:** Since there are three vertices, the region is a triangle. 7. **Check the type of triangle:** Calculate slopes of sides: - Between $$\left(\frac{8}{13}, \frac{72}{13}\right)$$ and $$\left(\frac{48}{7}, \frac{6}{7}\right)$$: $$m_1 = \frac{\frac{6}{7} - \frac{72}{13}}{\frac{48}{7} - \frac{8}{13}} = \frac{\frac{78}{91} - \frac{504}{91}}{\frac{336}{91} - \frac{56}{91}} = \frac{-\frac{426}{91}}{\frac{280}{91}} = -\frac{426}{280} = -\frac{213}{140}$$ - Between $$\left(\frac{48}{7}, \frac{6}{7}\right)$$ and $$\left(\frac{36}{23}, \frac{40}{23}\right)$$: $$m_2 = \frac{\frac{40}{23} - \frac{6}{7}}{\frac{36}{23} - \frac{48}{7}} = \frac{\frac{280}{161} - \frac{138}{161}}{\frac{252}{161} - \frac{1104}{161}} = \frac{\frac{142}{161}}{-\frac{852}{161}} = -\frac{142}{852} = -\frac{71}{426}$$ - Between $$\left(\frac{36}{23}, \frac{40}{23}\right)$$ and $$\left(\frac{8}{13}, \frac{72}{13}\right)$$: $$m_3 = \frac{\frac{72}{13} - \frac{40}{23}}{\frac{8}{13} - \frac{36}{23}} = \frac{\frac{1272}{299} - \frac{520}{299}}{\frac{184}{299} - \frac{468}{299}} = \frac{\frac{752}{299}}{-\frac{284}{299}} = -\frac{752}{284} = -\frac{188}{71}$$ Check if any product of slopes equals -1 (perpendicular): $$m_1 \times m_2 = -\frac{213}{140} \times -\frac{71}{426} = \frac{213 \times 71}{140 \times 426} \neq 1$$ $$m_2 \times m_3 = -\frac{71}{426} \times -\frac{188}{71} = \frac{188}{426} \neq 1$$ $$m_1 \times m_3 = -\frac{213}{140} \times -\frac{188}{71} = \frac{213 \times 188}{140 \times 71} \neq 1$$ No right angle, so the triangle is not right-angled. **Final answer:** The region is a triangle but not a right triangle. **Answer choice:** B. Segitiga sembarang