Question

A point moves in such a way that it remains equidistant from each of the lines \(3x±2y= 5\). Then the path along which the point moves is?

• A. \(x=\dfrac{5}{3}\)
• B. \(y=\dfrac{5}{3}\)
• C. \(x=\dfrac{-5}{3}\)
• D. \(x=0\)
• E. \(y=\dfrac{-5}{3}\)

Answer 1

The Correct Option is A

Given that

The point is equidistant from the given two lines

i.e. ; 3x−2y−5=0 and 3x+2y−5=0

Now,

Let P(h, k) be a moving point such that it is equidistant from the lines 3x−2y−5=0 then

\(∣\dfrac{3h−2k−5}{(√9+4)}∣=∣\dfrac{3h+2k−5}{(√9+4)}∣\)

\(|3h−2k−5|=|3h+2k−5|\)

\(4k=0\) or \(6h−10=0\)

\(k=0\)  or \(3h=5\)

therefore, \(h=\dfrac{5}{3}\)

Hence, the path of the moving points are \(y=0\) or \(3x=5\) (\(x=\dfrac{5}{3}\)), which are straight lines    (Ans.)

Answer 2

To find the path of a point that remains equidistant from the lines \(3x + 2y = 5\) and \(3x - 2y = 5\), we equate the distances from the point \((x, y)\) to both lines.

The distance from \((x, y)\) to \(3x + 2y - 5 = 0\) is:

\[ \frac{|3x + 2y - 5|}{\sqrt{3^2 + 2^2}} \]

The distance to \(3x - 2y - 5 = 0\) is:

\[ \frac{|3x - 2y - 5|}{\sqrt{3^2 + (-2)^2}} \]

Setting them equal:

\[ |3x + 2y - 5| = |3x - 2y - 5| \]

This gives two cases:

1. \(3x + 2y - 5 = 3x - 2y - 5\) leads to \(y = 0\).
2. \(3x + 2y - 5 = - (3x - 2y - 5)\) simplifies to \(x = \frac{5}{3}\).

Only \(x = \frac{5}{3}\) appears in the options. Verifying:

For \(x = \frac{5}{3}\), the distances to both lines are:

\[ \frac{|2y|}{\sqrt{13}} \quad \text{and} \quad \frac{|2y|}{\sqrt{13}} \]

Thus, the correct path is \(x = \frac{5}{3}\), corresponding to option A.