JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 19)
Two sides of a parallelogram are along the lines, x + y = 3 & x – y + 3 = 0. If its diagonals intersect at (2, 4), then one of its vertex is :
(2, 1)
(2, 6)
(3, 5)
(3, 6)
Explanation
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Solving
$$\matrix{ {x + y = 3} \cr {x - y = - 3} \cr } \,\, > \,\,A\left( {0,3} \right)$$
and $${{{x_1} + 0} \over 2} = 2;\,\,{x_i} = 4$$
similarly y1 = 5
C $$ \Rightarrow $$ (4, 5)
Now equation of BC is x $$-$$ y = $$-$$ 1
and equation of CD is x + y = 9
Solving x + y = 9 and x $$-$$ y = $$-$$ 3
Point D is (3, 6)
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