A(−3,−3),B(2,−2),C(2,3) and D(−4,4) are the vertices of quadrilateral ABCD. Then the point of intersection of its diagonals is
The correct option is C. (−311,311)
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
= 3−(−3)2−(−3)
=65
And the slope of the diagonal BD is
=4−(−2)−4−2
=−1
Let P(x,y) be the point of intersection of these diagonals.
Then P(x,y) lies on both AC and BD.
Therefore y−3x−2=65 and y−(−2)x−2=−1
⟹5y−15=6x−12 and y+2=−x+2
⟹6x−5y+3=0 and x+y=0
Solving the above two equations, we get, x=−311 and y=311
Hence the point of intersection P of the diagonals AC and BD is (−311,311).