A) \[6,4\]
B) \[-\,\,9,3\]
C) \[9,3\]
D) \[7,9\]
Correct Answer: B
Solution :
According to question, PQ = 10 \[\sqrt{{{(10-2)}^{2}}+{{(y+3)}^{2}}}=10\] [\[\because \]Distance\[=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]] \[\sqrt{{{(8)}^{2}}+{{(y+3)}^{2}}}=10\] \[\sqrt{64+{{y}^{2}}+9+6y}=10\] \[\sqrt{{{y}^{2}}+6y+73}=10\] On squaring both sides, we get \[{{y}^{2}}+6y+73=100\] \[\Rightarrow \] \[{{y}^{2}}+6y-27=0\] \[\Rightarrow \] \[{{y}^{2}}+9y-3y-27=0\] \[\Rightarrow \] \[y\,\,(y+9)-3\,\,(y+9)=0\] \[\Rightarrow \] \[(y+9)(y-3)=0\] \[\Rightarrow \] \[y=-9\]or \[y=3\]\[\therefore \]\[y=-\,\,9\] or 3
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