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

The correct option is A

A fixed point

Explanation for the correct answer:

Step 1: Finding the intercepts

The equation of straight line in intercept form is

$\frac{x}{a}+\frac{y}{b}=1...\left(1\right)$

Here, $a$ is the $x$-intercept and $b$is the $y$-intercept.

The sum of the reciprocals of its intercept on two perpendicular lines is constant. So,

$$\begin{gathered} \Rightarrow \frac{1}{a}+\frac{1}{b}=\frac{1}{k} \\ \Rightarrow \frac{k}{a}+\frac{k}{b}=1...\left(2\right) \end{gathered}$$.

Step 2: Finding the point on the axes.

On comparing equation $1$ and $2$, we get

$\Rightarrow \frac{x}{a}+\frac{y}{b}=\frac{k}{a}+\frac{k}{b}$

Thus the equation $1$, passes through the point $(k,k)$

Therefore, the straight line passes through a fixed point.

Hence option (A), a fixed point is the correct answer.