What is the area bounded by 4 lines?

Question

Let's say I have 4 points.

• $A(65.3,101.48)$
• $B(61.3,102.52)$
• $C(73.13,102.48)$
• $D(72.96, 101.42)$

There are four lines that connect these points:

• Line 1 is $\overline{AB}$
• Line 2 is $\overline{BC}$
• Line 3 is $\overline{CD}$
• Line 4 is $\overline{DA}$

This would be clearer if you graph the points then connect the A to B to C to D to A. I need to get the area bounded by the 4 lines or inside the 4 points connected.

I thought of using integration and getting the area under the curves but I'm having difficulty with it. Do you know of other ways I can use to get the area?

If this helps, I got the equations of the lines:

• $\overline{AB}: y=-\frac{13}{50}x + \frac{59229}{500}$
• $\overline{BC}: y=-\frac{4}{1183}x + \frac{3038159}{29575}$
• $\overline{CD}: y=\frac{106}{17}x - \frac{300481}{850}$
• $\overline{DA}: y=-\frac{3}{383}x + \frac{1953137}{19150}$

I'd really appreciate the help! I need this for my research paper and I got the graph from my experiment. Thank you in advance!

Edit: From the answers so far, it would seem like geometry/vectors could be used but if someone could give an answer using integration, that would be great because I have to replace 2 of those lines later on with curves.

Answer 1

How is it more difficult than using geometry or vectors? Based on the question, it seems that the four points are joined by straight lines, making this a simple quadrilateral. If so, you could just use simpler methods like splitting the quadrilateral into triangles and calculating their areas? Or the other method suggested above. I get an answer of around 10.1365.

I could only see you needing to use integration if the lines connecting the four points are curves.

Answer 2

This would be easiest if you used a little bit about vectors, instead of integration.

It can be done with integration, but it may be better to just use what you know from geometry. For example, you could divide it into four triangles and calculate their respective areas using $\frac{1}{2}bh$.

On the other hand, given four vectors: $$(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4).$$

form the two vectors: $[x_2-x_1, y_2-y_1]$ and $[x_3-x_1,y_3-y_1]$ and take their determinant multiplied by $\frac{1}{2}$. This will give you the area of the first triangle**. Form the analogous vectors for the other triangle and do the same. Adding will yield the full area.

**The determinant of two vectors $[a,b],[c,d]$ is just $ad-bc$.

Answer 3

If you have a simple polygonal closed line with vertices in the points $P_k=(x_k,y_k)$, $k=0,\dots,N$, $P_N=P_0$, ordered counter-clockwise, its area is simply: $$ \frac 1 2 \sum_{k=1}^N (x_{k-1}-x_k)(y_{k-1}+y_k). $$

Just notice that each addend is the area of the trapezoid below the corresponding segment.