A) The same straight line u
B) Different straight line
C) It is not a straight line
D) None of these
Correct Answer: A
Solution :
\[u={{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0,v={{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\] and \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}=c\] (Let) Þ \[{{a}_{2}}=\frac{{{a}_{1}}}{c},{{b}_{2}}=\frac{{{b}_{1}}}{c},{{c}_{2}}=\frac{{{c}_{1}}}{c}\] Given that \[u+kv=0\] Þ \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}+k({{a}_{2}}x+{{b}_{2}}y+{{c}_{2}})=0\] Þ \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}+k\frac{{{a}_{1}}}{c}x+k\frac{{{b}_{1}}}{c}y+k\frac{{{c}_{1}}}{c}=0\] Þ \[{{a}_{1}}x\left( 1+\frac{k}{c} \right)+{{b}_{1}}y\left( 1+\frac{k}{c} \right)+{{c}_{1}}\left( 1+\frac{k}{c} \right)=0\] Þ \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0=u\].
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