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

Answer: (A)

$x=\dfrac{5}{3}$

Answer 2

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.)