question_answer

The fourth vertex D of a parallelogram ABCD whose three vertices are A (- 2, 3), 5(6, 7) and C(8, 3) is

A) (0, 1)

B) (0, -1)

C) (-1, 0)

D) (1, 0)

Correct Answer: B

Solution :

Let the fourth vertex of parallelogram, \[D\equiv \left( {{x}_{4}},\,{{y}_{4}} \right)\] and L, M be the middle points of AC and BD, respectively.

Then, \[L\equiv \left( \frac{-2+8}{2},\,\frac{3+3}{2} \right)\equiv \left( 3,\,3 \right)\]

[since, mid - point of a line segment having points \[\left( {{x}_{1}},\,{{y}_{1}} \right)\]and \[\left. =\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\,\frac{{{y}_{1}}+{{y}_{2}}}{2} \right) \right]\]

And     \[M\equiv \left( \frac{6+{{x}_{4}}}{2},\,\frac{7+{{y}_{4}}}{2} \right)\]

Since, ABCD is a parallelogram, therefore diagonals AC and BD will bisect each other. Hence, L and At are the same points.

\[\therefore \,\,\,3=\frac{6+{{x}_{4}}}{2}\] and \[3=\frac{7+{{y}_{4}}}{2}\]

\[\Rightarrow \,\,\,6=6+{{x}_{4}}\] and \[6=7+{{y}_{4}}\]

\[\Rightarrow \,\,\,{{x}_{4}}=0\] and      \[{{y}_{4}}=6-7\]

\[\therefore \,\,\,{{x}_{4}}=0\] and \[{{y}_{4}}=-1\]

Hence, the fourth vertex of parallelogram is \[D\equiv \left( {{x}_{4}},\,{{y}_{4}} \right)\equiv \left( 0,\,-1 \right)\]