A line with slope of -3 intersects the positive x-axis at A and the positive y-axis at B. A second line intersects the x-axis at (8,0) and the y-axis at D. The lines intersect at (4,4) . What is the area of the shaded quadrilateral OBEC?
We can split the triangle into 2 portions as shown below.
First let's find the coordinates of Point A and Point B, which are both on the same line of slope -3 going through point 4, 4.
\(y = -3x + b \\ 4 = -12 + b \\ b = 16 \\ y = -3x+16\)
Because Point B is the y-intercept of this line, Point B is at (0, 16). We can plug in y = 0 to find Point A.
\(y = -3x+16 \\ 0 = -3x + 16 \\ -3x = -16 \\ x = \frac{16}{3}\)
So Point A is at (16/3, 0). First we can find the area of the red.
\(16 \cdot \frac{16}{3} \over 2 \\ \frac{216}{3} \cdot \frac{1}{2} \\ \frac{216}{6} = 36 \)
The red section has an area of 36. Moving onto blue, 8 - 5 1/3 = 2 2/3 = 8/3. Height is equal to 4.
\(4 \cdot \frac{8}{3} \over 2 \\ \frac{32}{3} \cdot \frac{1}{2} \\ \frac{32}{6} = 5 \frac{1}{3} \)
36 + 5 1/3 = 41 1/3