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Miscellaneous Examples · Example 16

Q.Show that the path of a moving point such that its distances from two lines 3x−2y=53x - 2y = 5 and 3x+2y=53x + 2y = 5 are equal is a straight line.

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Setting the distances from 3x−2y−5=03x-2y-5=0 and 3x+2y−5=03x+2y-5=0 equal and resolving the absolute value gives two straight lines, y=0y=0 and x=53x=\dfrac53 — proving the path is a straight line.

Step 1: Distance from a general point (x,y)(x,y) to each line

d1=∣3x−2y−5∣32+(−2)2=∣3x−2y−5∣13d_1=\frac{|3x-2y-5|}{\sqrt{3^2+(-2)^2}}=\frac{|3x-2y-5|}{\sqrt{13}}

d2=∣3x+2y−5∣32+22=∣3x+2y−5∣13d_2=\frac{|3x+2y-5|}{\sqrt{3^2+2^2}}=\frac{|3x+2y-5|}{\sqrt{13}}

Step 2: Apply the condition d1=d2d_1=d_2

Since the denominators are equal:

∣3x−2y−5∣=∣3x+2y−5∣|3x-2y-5|=|3x+2y-5|

Step 3: Resolve the modulus — two cases

Case (i):

3x−2y−5=3x+2y−5  ⟹  −4y=0  ⟹  y=03x-2y-5=3x+2y-5 \implies -4y=0 \implies y=0

This is a straight (horizontal) line.

Case (ii):

3x−2y−5=−(3x+2y−5)  ⟹  3x−2y−5=−3x−2y+5  ⟹  6x−10=0  ⟹  x=533x-2y-5=-(3x+2y-5) \implies 3x-2y-5=-3x-2y+5 \implies 6x-10=0 \implies x=\frac53

This is a straight (vertical) line. …

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