A) \[x+y=\sqrt{2}\]
B) \[x+y=1\]
C) \[x-y=1\]
D) \[x-y=\sqrt{2}\]
Correct Answer: B
Solution :
Given curve is \[{{x}^{2}}+{{y}^{2}}-2x-2y+1=0\] Put \[x=0,\] we get \[{{y}^{2}}-2y+1=0\] \[\Rightarrow \] \[{{(y-1)}^{2}}=0\] \[\Rightarrow \] \[y=1\] Now, put \[y=0,\] we get \[{{x}^{2}}-2x+1=0\] \[\Rightarrow \] \[{{(x-1)}^{2}}=0\] \[\Rightarrow \] \[x=1\] Hence, it intersect the coordinate axes at \[A(1,\,0)\] and \[B(0,\,1)\] \[\therefore \] Equation od line passing through \[A(1,\,0)\] and \[B(0,\,1)\] is \[y-0=\frac{1-0}{0-1}\,(x-1)\] \[\Rightarrow \] \[y=\frac{1}{-1}(x-1)\] \[\Rightarrow \] \[x+y=1\]
You need to login to perform this action.
You will be redirected in
3 sec
