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

How do you find the intersection of the diagonals of a shape/the intersection of lines-example-1
User Eran Abir
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1 Answer

26 votes
26 votes

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=(2-5)/(6-(-6)) \\ m=(-3)/(12)=-(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=-(1)/(4)(6)+c \end{gathered}

and then we solve for c


\begin{gathered} 2=-(3)/(2)+c \\ c=2+(3)/(2) \\ c=(7)/(2) \\ c=3.5 \end{gathered}

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


\begin{gathered} m=(2-(-9))/(6-2) \\ m=(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=(11)/(4)(6-x_1)

and we solve for x1


\begin{gathered} (4)/(11)2=6-x_1 \\ x_1=6-(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=(5-(-6))/(-6-(-10)) \\ m=(11)/(4) \end{gathered}

How do you find the intersection of the diagonals of a shape/the intersection of lines-example-1
User Frawel
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3.3k points