Question

The equation of a straight line passing through the point $(4,3)$ and making intercepts on the coordinate axes whose sum is $-1$ is
A. $\dfrac{x}{2}-\dfrac{y}{3}=1$ and $\dfrac{x}{-2}+\dfrac{y}{1}=1$
B. $\dfrac{x}{2}-\dfrac{y}{3}=-1$ and $\dfrac{x}{-2}+\dfrac{y}{1}=-1$
C. $\dfrac{x}{2}-\dfrac{y}{3}=1$ and $\dfrac{x}{2}+\dfrac{y}{1}=1$
D. $\dfrac{\pi }{3}$ and $\dfrac{x}{-2}+\dfrac{y}{1}=-1$

Answer 1

Hint: In this question, we are to find the equation of a straight line that is passing through the given point and making intercepts whose sum is $-1$. For this, we need to use the intercept form, and using the given condition i.e., the sum of the intercepts we can calculate the intercept values, and then we can frame the equation.

Formula used: The equation of a line with two intercepts on the $x$ and $y$ axes is
$\dfrac{x}{a}+\dfrac{y}{b}=1$
Where $a$ and $b$ are the intercepts on the $x$ and $y$ axes respectively.

Complete step by step solution: Given that the required equation makes intercepts.
So, consider the equation as
$\dfrac{x}{a}+\dfrac{y}{b}=1\text{ }...(1)$
It is given that; the sum of the intercepts equals $-1$.
i.e., $a+b=-1$
Then, we can rewrite
$\begin{align}
  & a+b=-1 \\
 & \Rightarrow b=-a-1 \\
\end{align}$
On substituting in (1),
\[\begin{align}
  & \dfrac{x}{a}+\dfrac{y}{b}=1 \\
 & \Rightarrow \dfrac{x}{a}+\dfrac{y}{-a-1}=1 \\
 & \Rightarrow \dfrac{x}{a}-\dfrac{y}{a+1}=1 \\
 & \Rightarrow x(a+1)-ay=a(a+1)\text{ }...(2) \\
\end{align}\]
Given that, equation (2) passes through the point $(4,3)$.
So, by substituting the point in (2), we get
$\begin{align}
  & x(a+1)-ay=a(a+1) \\
 & \Rightarrow 4(a+1)-a(3)={{a}^{2}}+a \\
 & \Rightarrow 4a+4-3a={{a}^{2}}+a \\
 & \Rightarrow {{a}^{2}}=4 \\
 & \therefore a=\pm 2 \\
\end{align}$
If $a=2$, we get the equation as
\[\begin{align}
  & \dfrac{x}{a}+\dfrac{y}{-a-1}=1 \\
 & \Rightarrow \dfrac{x}{2}+\dfrac{y}{-2-1}=1 \\
 & \Rightarrow \dfrac{x}{2}-\dfrac{y}{3}=1\text{ }...(3) \\
\end{align}\]
If $a=-2$, we get the equation as
\[\begin{align}
  & \dfrac{x}{a}+\dfrac{y}{-a-1}=1 \\
 & \Rightarrow \dfrac{x}{-2}+\dfrac{y}{-(-2)-1}=1 \\
 & \Rightarrow \dfrac{x}{-2}+\dfrac{y}{1}=1\text{ }...(4) \\
\end{align}\]
From (3) and (4), the required equations are $\dfrac{x}{2}-\dfrac{y}{3}=1$ and $\dfrac{x}{-2}+\dfrac{y}{1}=1$.

Thus, Option (A) is correct.

Note: Here we need to remember that, the sum of the intercepts is equal to . By this, we can substitute in the intercept form of the equation. Thus, we get the equations.