Question

Product of the intercepts of a straight line is 1 and the line passes through $\left( { - 12,1} \right)$ then its equation is
A) $2x + 25y = 1$
B) $x + 13y = 1$
C) $x + 16y = 4$
D) $16x + y = 4$

Answer 1

Hint:
Here, we will use the general equation of a line having $\left( {a,b} \right)$ as its intercepts. Substituting the given point (which satisfies the equation) in the general equation will help us to form a quadratic equation which when solved further, will give us the required intercepts. Then substituting them in the general equation will give us the required equation of the line.

Formula Used:
The general equation of a line is $\dfrac{x}{a} + \dfrac{y}{b} = 1$, where $a$ represents the intercept on the $x$ axis and $b$ represents the intercept on the $y$ axis.

Complete step by step solution:
Let the equation of the line be: $\dfrac{x}{a} + \dfrac{y}{b} = 1$
Now, according to the question,
Product of the intercepts of a straight line is 1.
$a \times b = 1$
Dividing both side by $b$, we get
$ \Rightarrow a = \dfrac{1}{b}$…………………..$\left( 1 \right)$
Now, it is also given that the line passes through $\left( { - 12,1} \right)$.
So, these points should satisfy the equation of the line as they are lying on it.
Therefore, substituting $x = - 12$ and $y = 1$ in the equation of the line, i.e. $\dfrac{x}{a} + \dfrac{y}{b} = 1$, we get,
$\dfrac{{ - 12}}{a} + \dfrac{1}{b} = 1$
Substituting the value of $a$from $\left( 1 \right)$, we get
$ \Rightarrow \dfrac{{ - 12}}{{\dfrac{1}{b}}} + \dfrac{1}{b} = 1$
$ \Rightarrow \dfrac{{ - 12b}}{1} + \dfrac{1}{b} = 1$
Taking LCM in the LHS, we get
$ \Rightarrow \dfrac{{ - 12{b^2} + 1}}{b} = 1$
Taking the denominator from the LHS to the RHS, we get,
$ \Rightarrow - 12{b^2} + 1 = b$
$ \Rightarrow 12{b^2} + b - 1 = 0$
Doing middle term split, we get
$ \Rightarrow 12{b^2} + 4b - 3b - 1 = 0$
$ \Rightarrow 4b\left( {3b + 1} \right) - 1\left( {3b + 1} \right) = 0$
Factoring out common terms, we get
$ \Rightarrow \left( {4b - 1} \right)\left( {3b + 1} \right) = 0$
Using the zero product property, we get
 $ \Rightarrow 4b - 1 = 0$
$ \Rightarrow b = \dfrac{1}{4}$
 Or
 $ \Rightarrow 3b + 1 = 0$
$ \Rightarrow b = \dfrac{{ - 1}}{3}$
Substituting these in equation $\left( 1 \right)$, we get
$a = 4$ when $b = \dfrac{1}{4}$
And $a = - 3$ when $b = \dfrac{{ - 1}}{3}$
Hence, substituting $\left( {4,\dfrac{1}{4}} \right)$ as the intercepts in the equation of the line, $\dfrac{x}{a} + \dfrac{y}{b} = 1$, we get
$\dfrac{x}{4} + \dfrac{{4y}}{1} = 1$
Taking LCM in the LHS, we get
$ \Rightarrow \dfrac{{x + 16y}}{4} = 1$
Multiplying both side by 4, we get
$ \Rightarrow x + 16y = 4$
Also, substituting $\left( { - 3,\dfrac{{ - 1}}{3}} \right)$ as the intercepts in the equation of the line, $\dfrac{x}{a} + \dfrac{y}{b} = 1$,
$ \Rightarrow \dfrac{x}{{ - 3}} + \dfrac{{3y}}{{ - 1}} = 1$
Taking LCM in the LHS, we get
$ \Rightarrow - \left( {\dfrac{{x + 9y}}{3}} \right) = 1$
Multiplying both side by $ - 3$, we get
$ \Rightarrow x + 9y = - 3$
Therefore, there are two lines whose product of the intercepts is 1 and pass through $\left( { - 12,1} \right)$.
The equations are: $x + 16y = 4$ and $x + 9y = - 3$
But, in this question, we are required to find only one such line.

Hence, option C is the correct answer.

Note:
In a straight line, the $x$ intercept is where a line crosses the $x$ axis and the $y$ intercept is where the line crosses the $y$ axis. The general equation of a straight line is $y = mx + c$, where, $m$ is the slope and $c$ is the constant respectively. But, if written in intercept form, then, the equation of the straight line is written as: $\dfrac{x}{a} + \dfrac{y}{b} = 1$.
This equation satisfies any point or coordinates lying on that particular straight line.