Parabolas and Triangles

For intersection of $L_{1}$ and $L_{2}$ :

$3x - 4 = 4 - x \implies x = 2$ and $y = 2 \implies P_{1}: (2,2)$

For intersection of $L_{1}$ and $L_{3}$ :

$9x - 12 = x + 12 \implies x = 3$ and $y = 5 \implies P_{2}: (3,5)$

For intersection of $L_{2}$ and $L_{3}$ :

$12 - 3x = x + 12 \implies x =0$ and $y = 4 \implies P_{3}: (0,4)$

Using the points $(2,2), (3,5), (0,4)$ to find $a,b,$ and $c$ for $y = ax^2 + bx + c$ we obtain:

$c = 4 \implies 2a + b = -1$ and $9a + 3b = 1$ $\implies a = \dfrac{-4}{3}$ and $b = \dfrac{-11}{3} \implies y = \dfrac{4}{3}x^2 - \dfrac{11}{3}x + 4$ or $y = \dfrac{1}{3}(4x^2 - 11x + 12)$ .

For $A_{\triangle{ABC}}$ :

The area of the square $EFAG$ is $A_{EFAG} = 9$ , and the areas of the three right triangles $A_{\triangle{CEA}} = A_{\triangle{AGB}} = \dfrac{3}{2}$ and $A_{\triangle{CFB}} = 2 \implies A_{\triangle{ABC}} = 4$

For $A_{p}$ :

$A_{p} = \dfrac{1}{3} \int_{0}^{3} (x + 12 - (4x^2 - 11x + 12)) dx = \dfrac{-4}{3}\int_{0}^{3} (x^2 - 3x) dx = \dfrac{-4}{3}(\dfrac{1}{3}x^3 - \dfrac{3}{2}x^2)|_{0}^{3} = \dfrac{-4}{3}(\dfrac{-9}{2}) = 6 \implies$

The desired area $A_{p} - A_{\triangle{ABC}} = \boxed{2}$ .

$$

An alternative method is to find the area $A_{1}$ of the region bounded by the parabola and $L_{2}$ and the area $A_{2}$ of the region bounded by the parabola and $L_{1}$ , then the desired area is just $A_{1} + A_{2}$ .

Doing this we obtain:

$A_{1} = \int_{0}^{2} (4 - x - (\dfrac{4x^2 - 11x + 12}{3}) dx = \dfrac{16}{9}$

and

$A_{2} = \int_{2}^{3} (3x - 4 - (\dfrac{4x^2 - 11x + 12}{3}) dx = \dfrac{2}{9}$

$\implies A_{1} + A_{2} = \boxed{2} = A_{p} - A_{\triangle{ABC}}$ .