Geo2.40

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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)\,


Main Page:Geometry:Straight Lines-I