Area bounded by lines in First Quadrant

Question

We are given 2 equations

$3x+2y=14$,--------(1)

$2x-3y=5$ ----------(2)

Below is the graph plot for it.

We need to find the area bounded by these two lines in the first quadrant. Red Line is line 1 and Blue line is line 2.

It seems like we only need to find area bounded by the line (1) in the first quadrant

And so I did $A=\int_0^{4.667}(\frac{14-3x}{2})dx=16.33$ but my answer was wrong.

Can someone please let me know where I am wrong? Correct Answer was 15.25

The above problem is not a homework problem, but a question asked in GATE.

Answer 1

From the above graph the point of intersection of the lines $3x+2y=14$ and $2x-3y=5$ is $x=4$ and $y=1$

Now consider the area of the small triangle with points $\left(\dfrac52,0\right),\left(\dfrac{14}{3},0\right),\left(4,3\right)$ is $$\dfrac12\times\left(\dfrac{13}{6}\right)\times1=\dfrac{13}{2}$$ Now consider the area of the bigger triangle with points $\left(0,0\right),\left(\dfrac{14}{3}\right),(0,7)$ is $$\dfrac{1}{2}\times\dfrac{14}{3}\times7=\dfrac{49}{3}$$

Now, the needed area is $\dfrac{49}{3}-\dfrac{13}{12}\approx15.25$

Answer 2

You can do this one with the shoelace algorithm, if you know the vertices of any polygon there area is:

$\frac 12 \left|\begin{matrix}0&0\\2.5&0\\4&1\\0&7\end{matrix}\right|$

Shoelaces... Multiply in diagonal lines and subtract the multiplications of the lines in the direction in the opposite direction. When you get to the bottom row, loop around to the top row.

$\frac 12 (0\cdot 0 + 2.5\cdot 1 + 4\cdot 7 + 0\cdot 0 - 2.5\cdot 0 - 4\cdot 0 - 0\cdot 1 - 0\cdot 7) = \frac 12(2.5+ 28) = 15.25$

If you travel counter clockwise around the vertices you will get a positive number. If you go the wrong way you will need to flip the sign.