## The coordinates of the point which is equidistant from the three vertices of the ∆ AOB as shown in the Fig. 7.1 is

## (A)(x, y)

## (B) (y, x)

## (C) (x/2,y/2)

## (D) (x/2,y/2)

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Question 18 You are here

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Chapter 7 Class 10 Coordinate Geometry

Serial order wise

Last updated at April 16, 2024 by Teachoo

Question 18 The coordinates of the point which is equidistant from the three vertices of the ∆ AOB as shown in the Fig. 7.1 is (x, y) (B) (y, x) (C) (𝑥/2,𝑦/2) (D) (𝑥/2,𝑦/2) Let Required point = P (p, q) Since point P is equidistant from A, B & C Hence, OP = AP = BP Now, OP =√(( 𝑝 −0)^2+(𝑞−0)^2 ) = √(𝒑^𝟐+𝒒^𝟐 ) Finding AP AP = √((𝑝 −0)^2+(𝑞−2𝑦)^2 ) = √(𝒑^𝟐+𝒒^𝟐+𝟒𝒚^𝟐−𝟒𝒒𝒚) Finding BP BP = √((𝑝 −2𝑥)^2+(𝑞−0)^2 ) = √(𝒑^𝟐+𝟒𝒙^𝟐−𝟒𝒑𝒙+𝒒^𝟐 ) Now, OP = AP √(𝑝^2+𝑞^2 ) = √(𝑝^2+𝑞^2+4𝑦^2−4𝑞𝑦) Squaring both sides 𝑝^2+𝑞^2 = 𝑝^2+𝑞^2+4𝑦^2−4𝑞𝑦 0 = 𝟒𝒚^𝟐−𝟒𝒒𝒚 4𝑞𝑦=4𝑦^2 𝑞=(4𝑦^2)/4𝑦 𝒒=𝒚 Also, OP = BP √(𝑝^2+𝑞^2 ) = √(𝑝^2+4𝑥^2−4𝑝𝑥+𝑞^2 ) Squaring both sides 𝑝^2+𝑞^2 = 𝑝^2+4𝑥^2−4𝑝𝑥+𝑞^2 0 = 𝟒𝒙^𝟐−𝟒𝒑𝒙 4𝑝𝑥=4𝑥^2 𝑝=(4𝑥^2)/4𝑥 𝒑=𝒙 Thus, Coordinates of Point P = (p, q) = (x, y) So, the correct answer is (D)