Question

If $u={{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$, $v={{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ and $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$, then the curve $u+kv=0$ is
A. The same the straight line $u$
B. Different straight line
C. It is not a straight line
D. None of these

Answer 1

Hint: In this question, we are to find the resulting equation of the curve $u+kv=0$, on substituting the $u$ and $v$ values. Considering the given condition to a constant, we can simplify the given curve and extract the equation of a straight line from it.

Formula Used: The equation of the line, that is passing through $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
Where $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ is said to be the slope of the line.

Complete step by step solution: Given that,
A curve with the equation is $u+kv=0$
Where
$u={{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$
$v={{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$
It is also given that,
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$
Consider the above-given condition to a constant $c$. i.e.,
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}=c$
We can write
$\begin{align}
  & \dfrac{{{a}_{1}}}{{{a}_{2}}}=c\Rightarrow {{a}_{2}}=\dfrac{{{a}_{1}}}{c} \\
 & \dfrac{{{b}_{1}}}{{{b}_{2}}}=c\Rightarrow {{b}_{2}}=\dfrac{{{b}_{1}}}{c} \\
 & \dfrac{{{c}_{1}}}{{{c}_{2}}}=c\Rightarrow {{c}_{2}}=\dfrac{{{c}_{1}}}{c} \\
\end{align}$
Then, substituting all these in the given curve, we get
\[\begin{align}
  & u+kv=0 \\
 & ({{a}_{1}}x+{{b}_{1}}y+{{c}_{1}})+k({{a}_{2}}x+{{b}_{2}}y+{{c}_{2}})=0 \\
 & ({{a}_{1}}x+{{b}_{1}}y+{{c}_{1}})+k(\dfrac{{{a}_{1}}}{c}x+\dfrac{{{b}_{1}}}{c}y+\dfrac{{{c}_{1}}}{c})=0 \\
 & {{a}_{1}}x\left( 1+\dfrac{k}{c} \right)+{{b}_{1}}y\left( 1+\dfrac{k}{c} \right)+{{c}_{1}}\left( 1+\dfrac{k}{c} \right)=0 \\
\end{align}\]
\[\begin{align}
  & \Rightarrow \left( 1+\dfrac{k}{c} \right)\left( {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}} \right)=0 \\
 & \Rightarrow {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0=u \\
\end{align}\]
Therefore, the obtained straight line is the same as the given straight line $u$.

Option ‘A’ is correct

Note: Here we may go wrong while taking a constant for the given condition. By substituting the given values and the straight lines in the curve $u+kv=0$, we get the straight line the same as the given one.