A) \[a=-1,\,b=3\]
B) \[a=-1,\,b=-3\]
C) \[a=0,\,b=0\]
D) \[a=1,\,b=-3\]
Correct Answer: D
Solution :
| Let \[P\left( 9a-2,-b \right)\]divides AS internally in the ratio 3 : 1. |
| By section formula, |
| \[9a-2=\frac{3\left( 8a \right)+1\left( 3a+1 \right)}{3+1}\] |
| [\[\because \]internal section formula, the coordinates of point P divides the line segment joining the point \[\left( {{x}_{1}},\,{{y}_{1}} \right)\]and \[\left( {{x}_{2}},\,{{y}_{2}} \right)\]in the ratio \[{{m}_{1}}:{{m}_{2}}\]internally is |
| \[\left. \left( \frac{{{m}_{2}}{{x}_{1}}+{{m}_{1}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}},\,\frac{{{m}_{2}}{{y}_{1}}+{{m}_{1}}{{y}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right) \right]\] |
| And \[-b=\frac{3\left( 5 \right)+1\left( -3 \right)}{3+1}\] |
| \[\Rightarrow \,\,\,9a-2=\frac{24a+3a+1}{4}\] |
| And \[-b=\frac{15-3}{4}\] |
| \[\Rightarrow \,\,\,9a-2=\frac{27a+1}{4}\] |
| And \[-b=\frac{12}{4}\] |
| \[\Rightarrow \,\,\,\,36a-8=27a+1\] and \[b=-3\] |
| \[\Rightarrow \,\,36a-27a-8-1=0\] |
| \[\Rightarrow \,\,\,9a-9=0\] |
| \[\therefore \,\,\,a=1\] |
| Hence, the required values of a and b are 1 and - 3. |
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