UPSC NDA Math Model Paper 6

Given: P(x,y) is equidistant from points A(3,4) and B(5,3)
As we know that, distance between two points A and B is given by: |AB|=√(x2−x1)2+(y2−y1)2
⇒|PA|=√(x−3)2+(y−4)2 .........(1)
⇒|PB|=√(x−5)2+(y−3)2 ........(2)
∵ |PA| = |PB|
⇒√(x−3)2+(y−4)2=√(x−5)2+(y−3)2
By squaring both the sides we get,
⇒(x−3)2+(y−4)2=(x−5)2+(y−3)2
⇒(x2+9−6x)+(y2+16−8y) =(x2+25−10x)+(y2+9−6y)
⇒4x−16=2y−7
⇒4x−2y=9
Hence, option C is the correct answer.