A) 3 : 4 external
B) 3 : 4 internal
C) 1 : 4 internal
D) 4 : 3 internal
Correct Answer: B
Solution :
| [b] Let required ratio be k: 1. Then, coordinates of section point are \[\left( \frac{2k+1}{k+1},\frac{7k+3}{k+1} \right).\] But this point lies on \[3x+y-9=0\] \[\therefore \] \[3\,\left( \frac{2k+1}{k+1} \right)+\left( \frac{7k+3}{k+1} \right)-9=0\] \[\Rightarrow \] \[6k+3+7k+3-9k9=0\] \[\Rightarrow \] \[4k-3=0\] \[\Rightarrow \] \[k=\frac{3}{4}\] \[\therefore \]Required ratio\[=k:1=3/4:1=3:4\]internally |
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