Question

The point of intersection of the lines $\dfrac{x}{a}+\dfrac{y}{b}=1$ and $\dfrac{x}{b}+\dfrac{y}{a}=1$ lies on the line
A. $x-y=0$
B. $(x+y)(a+b)=2ab$
C. $(lx+my)(a+b)=(l+m)ab$
D. All of these

Answer 1

Hint: In this question, we have to find the point of intersection of the given lines. For this, the direct formula we have for finding the point of intersection is applied. By substituting the obtained coordinates in the given equations, we get the equations of the lines that pass through this point.

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.
For calculating the point of intersection of the lines \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\] we use,
$\left( \dfrac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}-\dfrac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)$

Complete step by step solution: Given that,
The lines that intersect at a point are
$\dfrac{x}{a}+\dfrac{y}{b}=1\text{ }...(1)$
$\dfrac{x}{b}+\dfrac{y}{a}=1\text{ }...(2)$
We can write the above equations as
$\begin{align}
  & \Rightarrow bx+ay=ab \\
 & \Rightarrow bx+ay-ab=0\text{ }...(3) \\
\end{align}$
$\begin{align}
  & \Rightarrow ax+by=ab \\
 & \Rightarrow ax+by-ab=0\text{ }...(4) \\
\end{align}$
Then, their point of intersection is
$\begin{align}
  & =\left( \dfrac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}-\dfrac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right) \\
 & =\left( \dfrac{a(-ab)-b(-ab)}{b(b)-a(a)},\dfrac{(-ab)a-(-ab)b}{b(b)-a(a)} \right) \\
 & =\left( \dfrac{ab(b-a)}{(b-a)(b+a)},\dfrac{ab(b-a)}{(b-a)(b+a)} \right) \\
 & =\left( \dfrac{ab}{a+b},\dfrac{ab}{a+b} \right) \\
\end{align}$
To get the equation of the required line, the obtained point is substituted in the given equations. If any of them are satisfied with the point, then that equation is the required one.
So, the first equation, we have $x-y=0$
On substituting, we get
$\begin{align}
  & =x-y \\
 & =\dfrac{ab}{a+b}-\dfrac{ab}{a+b} \\
 & =0 \\
\end{align}$
Thus, the given lines lie on this line with the equation $x-y=0$.
The second equation we have $(x+y)(a+b)=2ab$
On substituting, we get
$\begin{align}
  & =(x+y)(a+b) \\
 & =\left( \dfrac{ab}{a+b}+\dfrac{ab}{a+b} \right)(a+b) \\
 & =\dfrac{2ab(a+b)}{(a+b)} \\
 & =2ab \\
\end{align}$
Thus, the given lines lie on this line with the equation $(x+y)(a+b)=2ab$.
The third equation we have $(lx+my)(a+b)=(l+m)ab$
On substituting, we get
$\begin{align}
  & =(lx+my)(a+b) \\
 & =\left( \dfrac{lab}{a+b}+\dfrac{mab}{a+b} \right)(a+b) \\
 & =\dfrac{ab(l+m)(a+b)}{(a+b)} \\
 & =(l+m)ab \\
\end{align}$
Thus, the given lines lie on this line with the equation $(lx+my)(a+b)=(l+m)ab$.
Therefore, all the given equations are true with the obtained point of intersection.

Option ‘D’ is correct

Note: Here we need to remember that the point of intersection of the given lines is also a point on the required line. So, if the equation is true for the point obtained then it will be the required equation.