Question

How do you find the intersection of the diagonals of a shape/the intersection of lines in a shape?

Answer 1

Firs we have to gave some coordinates to the intersection point and the sahpe like this:

With the coordinades B and C we can place an equation for the line BC

$\begin{gathered} y_2-y_1=m(x_2-x_1) \\ m=\frac{2-5}{6-(-6)} \\ m=\frac{-3}{12}=-\frac{1}{4} \end{gathered}$

Now we can find the intersection with the y axis with the general formula for a line, in any of the coordinates (I use the coordinate of C):

$\begin{gathered} y=mx+c \\ 2=-\frac{1}{4}(6)+c \end{gathered}$

and then we solve for c

$\begin{gathered} 2=-\frac{3}{2}+c \\ c=2+\frac{3}{2} \\ c=\frac{7}{2} \\ c=3.5 \end{gathered}$

Now we repeat the prosses for the segment CD, so the slope will be:

$\begin{gathered} m=\frac{2-(-9)}{6-2} \\ m=\frac{11}{4} \end{gathered}$

Now we can writte the equation of the line like:

$y_2-y_1=m(x_2-x_1)$

We can replace the coordinate (x1, y1) for (x1, 0) where x1 is the intersection, and replace the coordinate (x2, y2) for any or the coordinates we know, so I will use the coordinate C)

$2-0=\frac{11}{4}(6-x_1)$

and we solve for x1

$\begin{gathered} \frac{4}{11}2=6-x_1 \\ x_1=6-\frac{8}{11} \\ x_1=5.28 \end{gathered}$

Now we repeat the prosedure for the line EB, so the slope is going to be

$\begin{gathered} m=\frac{5-(-6)}{-6-(-10)} \\ m=\frac{11}{4} \end{gathered}$