A) 3, 7
B) 5, 5
C) 6, 4
D) 8, 9
Correct Answer: B
Solution :
Let the equation of side AB be \[y=x+a\] Then, \[A\equiv (1-a,1),\]\[B\equiv (2,2+a).\]Equation of side AD is \[y-1=-\,\,(x-(1-a)).\]Hence, \[D\equiv \,\,(-2,4-a).\] Let \[C\equiv \,\,(h,k).\]Then, \[h+1-a=2-2\] \[\Rightarrow \]\[h=a-1\]and \[k+1=2+a+4-a\Rightarrow k=5\] Thus locus of C is y = 5.You need to login to perform this action.
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