The equation of the straight line joining the point $\left( a,b \right)$ to the point of
intersection of the lines $\dfrac{x}{a}+\dfrac{y}{b}=1$ and $\dfrac{x}{b}+\dfrac{y}{a}=1$ is
(a) ${{a}^{2}}y-{{b}^{2}}x=ab\left( a-b \right)$
(b) ${{a}^{2}}x+{{b}^{2}}y=ab\left( a+b \right)$
(c) ${{a}^{2}}y+{{b}^{2}}x=ab$
(d) ${{a}^{2}}x+{{b}^{2}}y=ab\left( a-b \right)$
Answer
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Hint: Solve the 2 line equations to find the point of intersection. Substitute these intersection points along with the given coordinate points back into the line equations.
The two equations given in the question are,
$\dfrac{x}{a}+\dfrac{y}{b}=1$ and $\dfrac{x}{b}+\dfrac{y}{a}=1$
The given equations can be rearranged as,
$\dfrac{x}{a}+\dfrac{y}{b}-1=0$ and $\dfrac{x}{b}+\dfrac{y}{a}-1=0$
The point of intersection of these two lines can be obtained by solving the equations and finding the values of \[x\] and \[y\]. Subtracting the equations,
\[\left( \dfrac{x}{a}+\dfrac{y}{b}-1 \right)-\left( \dfrac{x}{b}+\dfrac{y}{a}-1 \right)=0\]
Taking similar terms together,
\[\left( \dfrac{x}{a}-\dfrac{x}{b} \right)+\left( \dfrac{y}{b}-\dfrac{y}{a} \right)+(-1+1)=0\]
Taking out the common terms,
\[\begin{align}
& x\left( \dfrac{1}{a}-\dfrac{1}{b} \right)+y\left( \dfrac{1}{b}-\dfrac{1}{a} \right)=0 \\
& x\left( \dfrac{1}{a}-\dfrac{1}{b} \right)-y\left( \dfrac{1}{a}-\dfrac{1}{b} \right)=0 \\
\end{align}\]
Again, taking out the common term, we get,
\[\begin{align}
& \left( x-y \right)\left( \dfrac{1}{a}-\dfrac{1}{b} \right)=0 \\
& \left( x-y \right)=0 \\
& x=y \\
\end{align}\]
Now we can substitute \[x=y\] in one of the equations of the lines, say $\dfrac{x}{a}+\dfrac{y}{b}-1=0$,
$\begin{align}
& \dfrac{y}{a}+\dfrac{y}{b}-1=0 \\
& \Rightarrow y\left( \dfrac{1}{a}+\dfrac{1}{b} \right)-1=0 \\
& \Rightarrow y\left( \dfrac{1}{a}+\dfrac{1}{b} \right)=1 \\
\end{align}$
Taking the LCM,
$\begin{align}
& \Rightarrow y\left( \dfrac{b+a}{ab} \right)=1 \\
& \Rightarrow y=\dfrac{1}{\left( \dfrac{b+a}{ab} \right)} \\
& \Rightarrow y=\left( \dfrac{ab}{b+a} \right) \\
& \Rightarrow y=\left( \dfrac{ab}{a+b} \right) \\
\end{align}$
Since \[x=y\], we get the coordinates as$x=y=\left( \dfrac{ab}{a+b} \right)$. Therefore, we can write the coordinates of the point of intersection of the lines $\dfrac{x}{a}+\dfrac{y}{b}=1$ and $\dfrac{x}{b}+\dfrac{y}{a}=1$ as $\left[ \left( \dfrac{ab}{a+b} \right),\left( \dfrac{ab}{a+b} \right) \right]$.
The equation of a line passing through two points \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\] is given by,
\[\begin{align}
& y-{{y}_{1}}=m(x-{{x}_{1}}) \\
& \Rightarrow y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\times (x-{{x}_{1}}) \\
\end{align}\]
So, the equation of the straight line joining the point $\left( a,b \right)$ and $\left[ \left( \dfrac{ab}{a+b} \right),\left( \dfrac{ab}{a+b} \right) \right]$ can be found as,
\[y-b=\left( \dfrac{\dfrac{ab}{a+b}-b}{\dfrac{ab}{a+b}-a} \right)\left( x-a \right)\]
Taking the LCM of the terms in the numerator and denominator,
\[y-b=\left( \dfrac{\dfrac{ab-b(a+b)}{a+b}}{\dfrac{ab-a(a+b)}{a+b}} \right)\left( x-a \right)\]
Opening the brackets and simplifying the terms,
\[\begin{align}
& y-b=\left( \dfrac{\dfrac{ab-ba-{{b}^{2}})}{a+b}}{\dfrac{ab-{{a}^{2}}-ab)}{a+b}} \right)\left( x-a \right) \\
& y-b=\left( \dfrac{\dfrac{-{{b}^{2}}}{a+b}}{\dfrac{-{{a}^{2}}}{a+b}} \right)\left( x-a \right) \\
& y-b=\left( \dfrac{{{b}^{2}}}{{{a}^{2}}} \right)\left( x-a \right) \\
& {{a}^{2}}(y-b)={{b}^{2}}(x-a) \\
& {{a}^{2}}y-{{a}^{2}}b={{b}^{2}}x-{{b}^{2}}a \\
& {{b}^{2}}x-{{a}^{2}}y={{b}^{2}}a-{{a}^{2}}b \\
& {{b}^{2}}x-{{a}^{2}}y=ab(b-a) \\
\end{align}\]
Looking at the given options, we can rewrite the obtained equation as
\[\begin{align}
& -{{b}^{2}}x+{{a}^{2}}y=-ab(b-a) \\
& {{a}^{2}}y-{{b}^{2}}x=ab(a-b) \\
\end{align}\]
Therefore, we get option (a) as the correct answer.
Note: The equation of a straight line passing through the point of intersection of two lines, say A and B can be written as \[A+\lambda B=0\]. Therefore, the equation of the straight line passing through the point of intersection of the given lines $\dfrac{x}{a}+\dfrac{y}{b}-1=0$ and $\dfrac{x}{b}+\dfrac{y}{a}-1=0$ can be written as, $\left( \dfrac{x}{a}+\dfrac{y}{b}-1 \right)+\lambda \left( \dfrac{x}{b}+\dfrac{y}{a}-1 \right)=0$. Since the required straight line passes through point $\left( a,b \right)$, we can substitute the coordinates in the above equation and find the value of $\lambda $. This method is lengthy as it has complex terms.
The two equations given in the question are,
$\dfrac{x}{a}+\dfrac{y}{b}=1$ and $\dfrac{x}{b}+\dfrac{y}{a}=1$
The given equations can be rearranged as,
$\dfrac{x}{a}+\dfrac{y}{b}-1=0$ and $\dfrac{x}{b}+\dfrac{y}{a}-1=0$
The point of intersection of these two lines can be obtained by solving the equations and finding the values of \[x\] and \[y\]. Subtracting the equations,
\[\left( \dfrac{x}{a}+\dfrac{y}{b}-1 \right)-\left( \dfrac{x}{b}+\dfrac{y}{a}-1 \right)=0\]
Taking similar terms together,
\[\left( \dfrac{x}{a}-\dfrac{x}{b} \right)+\left( \dfrac{y}{b}-\dfrac{y}{a} \right)+(-1+1)=0\]
Taking out the common terms,
\[\begin{align}
& x\left( \dfrac{1}{a}-\dfrac{1}{b} \right)+y\left( \dfrac{1}{b}-\dfrac{1}{a} \right)=0 \\
& x\left( \dfrac{1}{a}-\dfrac{1}{b} \right)-y\left( \dfrac{1}{a}-\dfrac{1}{b} \right)=0 \\
\end{align}\]
Again, taking out the common term, we get,
\[\begin{align}
& \left( x-y \right)\left( \dfrac{1}{a}-\dfrac{1}{b} \right)=0 \\
& \left( x-y \right)=0 \\
& x=y \\
\end{align}\]
Now we can substitute \[x=y\] in one of the equations of the lines, say $\dfrac{x}{a}+\dfrac{y}{b}-1=0$,
$\begin{align}
& \dfrac{y}{a}+\dfrac{y}{b}-1=0 \\
& \Rightarrow y\left( \dfrac{1}{a}+\dfrac{1}{b} \right)-1=0 \\
& \Rightarrow y\left( \dfrac{1}{a}+\dfrac{1}{b} \right)=1 \\
\end{align}$
Taking the LCM,
$\begin{align}
& \Rightarrow y\left( \dfrac{b+a}{ab} \right)=1 \\
& \Rightarrow y=\dfrac{1}{\left( \dfrac{b+a}{ab} \right)} \\
& \Rightarrow y=\left( \dfrac{ab}{b+a} \right) \\
& \Rightarrow y=\left( \dfrac{ab}{a+b} \right) \\
\end{align}$
Since \[x=y\], we get the coordinates as$x=y=\left( \dfrac{ab}{a+b} \right)$. Therefore, we can write the coordinates of the point of intersection of the lines $\dfrac{x}{a}+\dfrac{y}{b}=1$ and $\dfrac{x}{b}+\dfrac{y}{a}=1$ as $\left[ \left( \dfrac{ab}{a+b} \right),\left( \dfrac{ab}{a+b} \right) \right]$.
The equation of a line passing through two points \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\] is given by,
\[\begin{align}
& y-{{y}_{1}}=m(x-{{x}_{1}}) \\
& \Rightarrow y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\times (x-{{x}_{1}}) \\
\end{align}\]
So, the equation of the straight line joining the point $\left( a,b \right)$ and $\left[ \left( \dfrac{ab}{a+b} \right),\left( \dfrac{ab}{a+b} \right) \right]$ can be found as,
\[y-b=\left( \dfrac{\dfrac{ab}{a+b}-b}{\dfrac{ab}{a+b}-a} \right)\left( x-a \right)\]
Taking the LCM of the terms in the numerator and denominator,
\[y-b=\left( \dfrac{\dfrac{ab-b(a+b)}{a+b}}{\dfrac{ab-a(a+b)}{a+b}} \right)\left( x-a \right)\]
Opening the brackets and simplifying the terms,
\[\begin{align}
& y-b=\left( \dfrac{\dfrac{ab-ba-{{b}^{2}})}{a+b}}{\dfrac{ab-{{a}^{2}}-ab)}{a+b}} \right)\left( x-a \right) \\
& y-b=\left( \dfrac{\dfrac{-{{b}^{2}}}{a+b}}{\dfrac{-{{a}^{2}}}{a+b}} \right)\left( x-a \right) \\
& y-b=\left( \dfrac{{{b}^{2}}}{{{a}^{2}}} \right)\left( x-a \right) \\
& {{a}^{2}}(y-b)={{b}^{2}}(x-a) \\
& {{a}^{2}}y-{{a}^{2}}b={{b}^{2}}x-{{b}^{2}}a \\
& {{b}^{2}}x-{{a}^{2}}y={{b}^{2}}a-{{a}^{2}}b \\
& {{b}^{2}}x-{{a}^{2}}y=ab(b-a) \\
\end{align}\]
Looking at the given options, we can rewrite the obtained equation as
\[\begin{align}
& -{{b}^{2}}x+{{a}^{2}}y=-ab(b-a) \\
& {{a}^{2}}y-{{b}^{2}}x=ab(a-b) \\
\end{align}\]
Therefore, we get option (a) as the correct answer.
Note: The equation of a straight line passing through the point of intersection of two lines, say A and B can be written as \[A+\lambda B=0\]. Therefore, the equation of the straight line passing through the point of intersection of the given lines $\dfrac{x}{a}+\dfrac{y}{b}-1=0$ and $\dfrac{x}{b}+\dfrac{y}{a}-1=0$ can be written as, $\left( \dfrac{x}{a}+\dfrac{y}{b}-1 \right)+\lambda \left( \dfrac{x}{b}+\dfrac{y}{a}-1 \right)=0$. Since the required straight line passes through point $\left( a,b \right)$, we can substitute the coordinates in the above equation and find the value of $\lambda $. This method is lengthy as it has complex terms.
Last updated date: 22nd Sep 2023
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