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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
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)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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