Question

The equation to the line bisecting the join of $(3,-4)$ and $(5,2)$ and having its intercepts on the x-axis and the y-axis in the ratio $2:1$ is
A. $x+y-3=0$
B. $2x-y=0$
C. $x+2y=2$
D. $2x+y=7$

Answer 1

Hint: In this question, we are to find the equation of the line. Here it is given that the intercepts of the line are in the ratio of $2:1$. So, this condition helps us to evaluate the equation. The equation bisecting the line joining means the line passes through the mid-point of the given line joining.

Formula used: The midpoint of a line joining by $({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}})$ is
$=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$
The equation of a line (intercept form) is
$\dfrac{x}{a}+\dfrac{y}{b}=1$
An equation of a line passes through a point $({{x}_{1}},{{y}_{1}})$, then it is $a{{x}_{1}}+b{{y}_{1}}+c=0$

Complete step by step solution: Given that,
The ratio of the intercepts of the line is $2:1$
So,
The x-intercept is $2a$ and the y-intercepts is $a$
Then, the equation of the line is,
$\begin{align}
  & \dfrac{x}{2a}+\dfrac{y}{a}=1 \\
 & \Rightarrow x+2y=2a\text{ }...(1) \\
\end{align}$
The midpoint of the given line joining of $(3,-4)$ and $(5,2)$is
$\begin{align}
  & =\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right) \\
 & =\left( \dfrac{3+5}{2},\dfrac{-4+2}{2} \right) \\
 & =\left( \dfrac{8}{2},\dfrac{-2}{2} \right) \\
 & =(4,-1) \\
\end{align}$
Thus, the point $(4,-1)$lies on the line (1).
So, substituting it in the equation (1), we get
$\begin{align}
  & x+2y=2a \\
 & \Rightarrow 4+2(-1)=2a \\
 & \Rightarrow 2=2a \\
 & \therefore a=1 \\
\end{align}$
Thus, the x-intercept is $2a=2(1)=2$ and the y-intercept is $b=a=1$.
So, the equation of the line is
$\begin{align}
  & \dfrac{x}{2}+\dfrac{y}{1}=1 \\
 & \Rightarrow x+2y=2 \\
\end{align}$

Thus, Option (C) is correct.

Note: Here we need to remember that, the point that lies on line is to be substituted in the equation of the line to get the required value.