A) 9:1
B) 1:3
C) 1:9
D) 3:1
Correct Answer: A
Solution :
Let (x, y) be foot of perpendicular drawn to the point \[({{x}_{1}},{{y}_{1}})\] on the line \[ax+by+c=0\] Relation :\[\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{-(a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}})}{{{a}^{2}}+{{b}^{2}}}\] Here \[({{x}_{1}},{{y}_{1}})=(0,0)\] given line is: \[3x+y-\lambda =0\]\[\frac{x-0}{3}=\frac{y-0}{1}=\frac{-((3\times 0)+(1\times 0)-\lambda )}{{{3}^{2}}+{{1}^{2}}}\] \[x=\frac{3\lambda }{10}and\,\,y=\frac{\lambda }{10}\] Hence foot of perpendicular\[P=\left( \frac{3\lambda }{10},\frac{\lambda }{10} \right)\] Line meets X-axis at\[A=\left( \frac{\lambda }{3},0 \right)\] and meets Y-axis at \[B=(0,\lambda )\] \[BP=\sqrt{{{\left( \frac{3\lambda }{10} \right)}^{2}}+{{\left( \frac{\lambda }{10}-\lambda \right)}^{2}}}\] \[\Rightarrow BP=\sqrt{\frac{9{{\lambda }^{2}}}{100}+\frac{81{{\lambda }^{2}}}{100}}\] \[\therefore BP=\sqrt{\frac{90{{\lambda }^{2}}}{100}}\] \[AP=\sqrt{{{\left( \frac{\lambda }{3}-\frac{3\lambda }{10} \right)}^{2}}+{{\left( 0-\frac{\lambda }{10} \right)}^{2}}}\] \[\Rightarrow AP=\sqrt{\frac{{{\lambda }^{2}}}{900}+\frac{{{\lambda }^{2}}}{100}}\] \[\therefore AP=\sqrt{\frac{{{\lambda }^{2}}}{900}+\frac{{{\lambda }^{2}}}{100}}\] \[\therefore AP=\sqrt{\frac{10{{\lambda }^{2}}}{900}}\] \[\therefore BP:AP=9:1\] Hence, correct option is \['A'\].
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