# Find the Co-ordinates of the Point Equidistant from Three Given Points A(5,3), B(5, -5) and C(1,- 5). - Mathematics

Find the co-ordinates of the point equidistant from three given points A(5,3), B(5, -5) and C(1,- 5).

#### Solution

Let the required point be P (x, y). Then AP = BP = CP

That is, (AP)^2 = (BP)^2 = (cp)^2

This means(Ap)^2 = (BP)^2

⇒(x-5)^2 +(y-3)^2 = (x-5)^2 +(y+5)^2

⇒x^2-10x+25+y^2-6y +9 =x^2-10x +25+y^2 +10y+25

⇒x^2 -10x +y^2 -6y +34 =x^2 - 10x+y^2+10y+50

⇒x^2-10x +y^2-6y-x^2 +10x-y^2-10y = 50-34

⇒ -16y=16

⇒y=-16/16=-1

And (BP)^2 = (CP)^2

⇒(x-5)^2 +(y+5)^2 = (x-1)^2 +(y+5)^2

⇒ x^2 -10x +25 +y^2 +10y +25 = x^2 -2x +1 +y^2 +10y +25

⇒ x^2 -10x +y^2 +10y + 50= x^2 -2x +y^2 +10y +26

⇒x^2 -10x  +y^2 +10y -x^2 +2x - y^2 -10y = 26-50

⇒ -8x = -24

⇒ x = (-24)/(-8) = 3

Hence, the required point is (3, -1 ).

Concept: Coordinate Geometry
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#### APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 13