1. **State the problem:** Find the area of triangle $\triangle ABC$ where lines $l_1: y = -2x + 6$ and $l_2: y = x + 3$ intersect at point $A$. Line $l_1$ meets the x-axis at $C$, and line $l_2$ meets the y-axis at $B$ and the x-axis at $D$. We need the area of $\triangle ABC$. 2. **Find coordinates of points:** - Point $A$ is the intersection of $l_1$ and $l_2$. Set $-2x + 6 = x + 3$: $$-2x + 6 = x + 3$$ $$-2x - x = 3 - 6$$ $$-3x = -3$$ $$x = \frac{-3}{-3} = 1$$ Substitute $x=1$ into $l_2$: $y = 1 + 3 = 4$. So, $A = (1,4)$. - Point $C$ is where $l_1$ meets the x-axis ($y=0$): $$0 = -2x + 6$$ $$2x = 6$$ $$x = 3$$ So, $C = (3,0)$. - Point $B$ is where $l_2$ meets the y-axis ($x=0$): $$y = 0 + 3 = 3$$ So, $B = (0,3)$. 3. **Calculate area of $\triangle ABC$:** Use the formula for area of triangle given coordinates: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ Let $A=(x_1,y_1)=(1,4)$, $B=(x_2,y_2)=(0,3)$, $C=(x_3,y_3)=(3,0)$: $$\text{Area} = \frac{1}{2} |1(3 - 0) + 0(0 - 4) + 3(4 - 3)|$$ $$= \frac{1}{2} |1 \times 3 + 0 + 3 \times 1| = \frac{1}{2} |3 + 3| = \frac{1}{2} \times 6 = 3$$ 4. **Answer:** The area of $\triangle ABC$ is $3$. Therefore, the correct choice is **D 3**.