Question

If the slope of line $\left( {\dfrac{x}{a} + \dfrac{y}{b} - 1} \right) + k\left( {\dfrac{x}{b} + \dfrac{y}{a} - 1} \right) = 0$ is -1, then the value of ‘$k$’ is
A) -1
B) 1
C) 0
D) 2

Answer 1

Hint:
In this question, we have the slope of the given line equation equals to -1 and we have to find the value of $k$. To find the value of $k$, firstly open the brackets of the given equation and then separate the terms consist of $x$, $y$ and constant and the equation will be in the form of $\alpha x + \beta y + c = 0$ and slope of line is given by $\dfrac{{ - \alpha }}{\beta }$. Now you can find the value of $k$ by taking a slope is equal to -1.

Complete step by step solution:
Let us see what is given to us? We are given with a line equation i.e.
$ \Rightarrow \left( {\dfrac{x}{a} + \dfrac{y}{b} - 1} \right) + k\left( {\dfrac{x}{b} + \dfrac{y}{a} - 1} \right) = 0$ …………(1)
And the slope of the line equation is -1 i.e.
$ \Rightarrow m = - 1$, where m is the slope of line. ………….(2)
We have to find the value of $k$.
To find the value of $k$, first of all open the brackets in (1) and we get,
\[
   \Rightarrow \dfrac{x}{a} + \dfrac{y}{b} - 1 + (k)\dfrac{x}{b} + (k)\dfrac{y}{a} - 1(k) = 0 \\
   \Rightarrow \dfrac{x}{a} + \dfrac{y}{b} - 1 + k\dfrac{x}{b} + k\dfrac{y}{a} - k = 0 \\
 \]
Now, separate the terms contains $x$, $y$ and constant as follows:
\[
   \Rightarrow \dfrac{x}{a} + k\dfrac{x}{b} + \dfrac{y}{b} + k\dfrac{y}{a} - 1 - k = 0 \\
   \Rightarrow x\left( {\dfrac{1}{a} + k\dfrac{1}{b}} \right) + y\left( {\dfrac{1}{b} + k\dfrac{1}{a}} \right) - \left( {1 + k} \right) = 0 \\
 \]
This equation comes in the form of $\alpha x + \beta y + c = 0$, then compare both the equations. We get,
$ \Rightarrow \alpha = \dfrac{1}{a} + k\dfrac{1}{b}$, ..…….(3)
$ \Rightarrow \beta = \dfrac{1}{b} + k\dfrac{1}{a}$ and ……….(4)
$ \Rightarrow c = - (k + 1)$
 The slope of the line is given by:
$ \Rightarrow m = - \dfrac{\alpha }{\beta }$
Compare it with equation (2), we get,
$ \Rightarrow - \dfrac{\alpha }{\beta } = - 1$
Cancel negative from both sides and we get,

$ \Rightarrow \dfrac{\alpha }{\beta } = 1$
Putting the values of (3) and (4) in the above equation, we get,
 $ \Rightarrow \dfrac{{\dfrac{1}{a} + k\dfrac{1}{b}}}{{\dfrac{1}{b} + k\dfrac{1}{a}}} = 1$
Now, taking the L.C.M of the terms in numerator and denominator we get,
 $ \Rightarrow \dfrac{1}{a} + k\dfrac{1}{b} = \dfrac{1}{b} + k\dfrac{1}{a}$
Taking terms contains $k$ on one side and constant on other side we get,
 \[ \Rightarrow \dfrac{1}{a} - \dfrac{1}{b} = k\dfrac{1}{a} - k\dfrac{1}{b}\]
Taking L.C.M of the terms we get,
 \[ \Rightarrow \dfrac{{b - a}}{{ab}} = k\left( {\dfrac{{b - a}}{{ab}}} \right)\]
By cancelling, \[\dfrac{{b - a}}{{ab}}\]from both sides, we get,
$ \Rightarrow k = 1$

Hence, option ${2}^{\text{nd}}$is the correct answer.

Note:
You can also solve this question by putting values from options one by one and check whether slope is -1 or not.
By taking option 1, the equation becomes
$ \Rightarrow \left( {\dfrac{x}{a} + \dfrac{y}{b} - 1} \right) - \left( {\dfrac{x}{b} + \dfrac{y}{a} - 1} \right) = 0$
$
   \Rightarrow \dfrac{x}{a} - \dfrac{x}{b} + \dfrac{y}{b} - \dfrac{y}{a} = 0 \\
   \Rightarrow \left( {\dfrac{1}{a} - \dfrac{1}{b}} \right)x + \left( {\dfrac{1}{b} - \dfrac{1}{a}} \right)y = 0 \\
 $
Here you can see that $x$ and $y$, they are opposite of each other, hence the slope is 1. Which is not correct.
Similarly, we will check with option ${2}^{\text{nd}}$and the slope will come out to be -1, which is true. Hence option ${2}^{\text{nd}}$is correct.