Geo2.40

Find the point of intersection of diagonals of the quadrilateral with vertices $(1,2),(3,4),(2,1),(-1,-2)\,$

Let the given points be $A,B,C,D\,$ be the vertices of the quadrilateral.

Then the diagonals of this are the lines $AC,BD\,$

The equation of AC is $y-2=\frac{1-2}{2-1}(x-1),y-2=-x+1,x+y-3=0\,$

The equation of the other diagonal BD is $y-4=\frac{-2-4}{-1-3}(x-3)\,$

$4y-16=6x-18,6x-4y-2=0,3x-2y-1=0\,$

Solving AC and BD

$2(x+y-3)+(3x-2y-1)=0,5x-6-1=0,x=\frac{7}{5}\,$

Now the value o fy is $\frac{7}{5}+y-3=0,y=\frac{8}{5}\,$

Therefore,the point of intersection of the diagonals are $\left(\frac{7}{5},\frac{8}{5}\right)\,$