Two sides of a parallelogram are along the lines 4x+5y=0 and 7x+2y=0. If the equation of one of the diagonals of the parallelogram is 11x+7y=9, then other diagonal passes through the point:
Solutions · 1
Official solution
Both the lines pass through origin.
point D is equal to intersection of 4x + 5y = 0 \& 11x + 7y = 9
So, coordinates of point \[D = \left( {{5 \over 3}, - {4 \over 3}} \right)\]
Also, point B is point of intersection of 7x + 2y = 0 and 11x + 7y = 9
So, coordinates of point \[B = \left( { - {2 \over 3},{7 \over 3}} \right)\]
diagonals of parallelogram intersect at middle let middle point of B, D
\[ \Rightarrow \left( {{{{5 \over 3} - {2 \over 3}} \over 2},{{{{ - 4} \over 3} + {7 \over 3}} \over 2}} \right) = \left( {{1 \over 2},{1 \over 2}} \right)\]
equation of diagonal AC
\[ \Rightarrow (y - 0) = {{{1 \over \alpha } - 0} \over {{1 \over \alpha } - 0}}(x - 0)\]
y=x
diagonal AC passes through (2, 2)