A) 1
B) 8
C) -4
D) -2
Correct Answer: C
Solution :
Given lines are x = 1, y = 2 and \[{{a}^{2}}x+2y-20=0\]. Since, \[a=1,\,y=2\] and \[{{a}^{2}}x+2y-20=0\]are concurrent, i.e x = 1, y = 2 and \[{{a}^{2}}x+2y-20=0\] having a common solution. So, x = 1, y = 2 is a solution of given equation \[{{a}^{2}}.\,1+2\,.\,2-20=0\Rightarrow {{a}^{2}}-16=0\] \[\Rightarrow \,\,\,\,{{a}^{2}}=16\,\,\,\Rightarrow \,\,a=-4,\,4\]You need to login to perform this action.
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