Question

The equation of the line passing through the point of intersection of the lines $\frac{x}{a}+\frac{y}{b}=1$and $\frac{x}{b}+\frac{y}{a}=1$ and having slope $0$ is

• A. $x=a$
• B. $y=b$
• C. $y=\frac{ab}{(a+b)}$
• D. None of these

Answer 1

The correct option is C

$y=\frac{ab}{(a+b)}$

Explanation of the correct answer.

Point of intersection of line:

$\frac{x}{a}+\frac{y}{b}=1$

$\Rightarrow$$bx+ay=ab$……..$\left(1\right)$

$\frac{x}{b}+\frac{y}{a}=1$

$\Rightarrow$$ax+by=ab$……..$\left(2\right)$

Now, $\left[b\times \left(1\right)\right]-\left[a\times \left(2\right)\right]$

$b^{2}x+aby-a^{2}x-aby=a{b}^2-a^{2}b$

$\Rightarrow$ $x(b^2-a^2)=a{b}^2-a^{2}b$

$\Rightarrow$ $x(b-a)(b+a)=ab(b-a)$

$\Rightarrow$ $x=\frac{ab}{(a+b)}$

Substitute the value of $x$ in $\left(2\right)$.

$a\left(\frac{ab}{a+b}\right)+by=ab$

$\Rightarrow$ $by=ab-\frac{a^{2}b}{a+b}$

$\Rightarrow$ $y=a-\frac{a^2}{(a+b)}$

$\Rightarrow$ $y=\frac{a^2+ab-a^2}{(a+b)}$

$\Rightarrow$ $y=\frac{ab}{(a+b)}$

Hence intersection point of the given equations is $\left(\frac{ab}{a+b},\frac{ab}{a+b}\right)$.

Equation of the line is given as,

$\left(y-y_1\right)=m\left(x-x_1\right)$ $\left[Where{x}_1=\frac{ab}{a+b},y_1=\frac{ab}{a+b},m=0\right]$

$\Rightarrow$$y-\frac{ab}{(a+b)}=0$

$\Rightarrow$ $y=\frac{ab}{(a+b)}$

Hence, Option $\left(\mathrm{C}\right)$ is the correct answer.