A(6,4), B(8,4), C(10,6) and D(2,8) are the vertices of quadrilateral ABCD. Then the point of intersection of its diagonals is
(507,327)
We know that the slope(m) of the line formed by joining the points (x1,y1) and (x2,y2) is given by,
m=y2−y1x2−x1
Now, the slope of the diagonal AC
=6−410−6
=12
And the slope of the diagonal BD is
=8−42−8
=−23
Let P(x,y) be the point of intersection of these diagonals.
Then P(x,y) lies on both AC and BD.
Therefore y−6x−10=12 and y−8x−2=−23
⟹2y−12=x−10 and 3y−24=−2x+4
⟹2y−x=2 and 3y+2x=28
Solving the above two equations, we get, x=507 and y=327
Hence the point of intersection P of the diagonals AC and BD is (507,327)