A) \[I\]
B) \[II\]
C) \[III\]
D) \[IV\]
Correct Answer: D
Solution :
[d] |
\[x=\frac{2\times 5+3\times 2}{2+3}\] \[y=\frac{2\times 2+3(-5)}{2+3}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,x=\frac{10+6}{5}=\frac{16}{5}=3.2\,\,\,\Rightarrow \,\,y=\frac{4-15}{5}=\frac{-11}{5}=-2.2\] Point \[P(3.2,\,-2.2)\] lies in IV quadrant. |
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