A) \[(0,1)\]
B) \[(0,-1)\]
C) \[(-1,0)\]
D) \[(1,0)\]
Correct Answer: B
Solution :
| [b] Let the fourth vertex of parallelogram be \[D({{x}_{4}},{{y}_{4}})\] and LM the mid-points of AC and BD, respectively. |
| \[\therefore \,\,\,\,\,\,\,\,L=\left( \frac{-2+8}{2},\frac{3+3}{2} \right)=\,(3,3)\] |
| and \[M=\left( \frac{{{x}_{4}}+6}{2},\frac{{{y}_{4}}+7}{2} \right)\] |
| Since, ABCD is a parallelogram, therefore diagonals AC and BD will bisect each other. |
| Hence, L and M are the same points. |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\frac{{{x}_{4}}+6}{2}=3\] and \[\frac{{{y}_{4}}+7}{2}=3\] |
| \[\Rightarrow \,\,\,\,\,\,\,{{x}_{4}}+6=6\] and \[{{y}_{4}}+7=6\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}_{4}}=0\] and \[{{y}_{4}}=6-7\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}_{4}}=0\] and \[{{y}_{4}}=-1\] |
| Hence, the fourth vertex of parallelogram is \[D(0,-1)\]. |
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