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)
Made by
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.
Hi, it looks like you're using AdBlock :(
Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription.