Question

A straight line moves so that the sum of the reciprocals of its intercepts on two perpendicular lines is constant, then the line passes through
A. A fixed point
B. A variable point
C. Origin
D. None of these

Answer 1

Hint: In this question, we are to find the point through which the given line passes. For this we use the equation of the line with the intercepts i.e., intercept form where the coordinate axes are the two perpendicular lines.

Formula Used: The equation of the line, that is passing through $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
Where $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ is said to be the slope of the line.
We can also write it in the point-slope form as
$y-{{y}_{1}}=m(x-{{x}_{1}})$
Consider the x-intercept as $a$ and the y-intercept as $b$. Then the equation of the line (intercept form) is $\dfrac{x}{a}+\dfrac{y}{b}=1$

Complete step by step solution: Consider the two perpendicular lines as the coordinate axes. So, the intercepts are
the x-intercept is $a$ and the y-intercept is $b$.
Then the equation of the line (intercept form) is
$\dfrac{x}{a}+\dfrac{y}{b}=1\text{ }...(1)$
As per the question, the sum of the reciprocals of the intercepts is a constant. Consider $k$ is a constant. So, the equation we can frame is
$\begin{align}
  & \dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{k} \\
 & \Rightarrow \dfrac{k}{a}\text{+}\dfrac{k}{b}\text{=1 }...(2) \\
\end{align}$
On comparing (1) and (2), we get
$(x,y)=(k,k)$
Therefore, the obtained result is a fixed point, we can say that the line passes through this fixed point.

Option ‘A’ is correct

Note: Here we may go wrong with the constant. We need to remember that, the sum of the reciprocals of intercepts is equal to the (reciprocal) constant. On simplifying and equating the obtained equations, we get a point which is a fixed point.