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

(A)(x, y) ย 

(B) (y, x)

(C) (x/2,y/2)ย 

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

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  1. Chapter 7 Class 10 Coordinate Geometry (Term 1)
  2. Serial order wise

Transcript

Question 18 The coordinates of the point which is equidistant from the three vertices of the โˆ† AOB as shown in the Fig. 7.1 is (x, y) (B) (y, x) (C) (๐‘ฅ/2,๐‘ฆ/2) (D) (๐‘ฅ/2,๐‘ฆ/2) Let Required point = P (p, q) Since point P is equidistant from A, B & C Hence, OP = AP = BP Now, OP =โˆš(( ๐‘ โˆ’0)^2+(๐‘žโˆ’0)^2 ) = โˆš(๐’‘^๐Ÿ+๐’’^๐Ÿ ) Finding AP AP = โˆš((๐‘ โˆ’0)^2+(๐‘žโˆ’2๐‘ฆ)^2 ) = โˆš(๐’‘^๐Ÿ+๐’’^๐Ÿ+๐Ÿ’๐’š^๐Ÿโˆ’๐Ÿ’๐’’๐’š) Finding BP BP = โˆš((๐‘ โˆ’2๐‘ฅ)^2+(๐‘žโˆ’0)^2 ) = โˆš(๐’‘^๐Ÿ+๐Ÿ’๐’™^๐Ÿโˆ’๐Ÿ’๐’‘๐’™+๐’’^๐Ÿ ) Now, OP = AP โˆš(๐‘^2+๐‘ž^2 ) = โˆš(๐‘^2+๐‘ž^2+4๐‘ฆ^2โˆ’4๐‘ž๐‘ฆ) Squaring both sides ๐‘^2+๐‘ž^2 = ๐‘^2+๐‘ž^2+4๐‘ฆ^2โˆ’4๐‘ž๐‘ฆ 0 = ๐Ÿ’๐’š^๐Ÿโˆ’๐Ÿ’๐’’๐’š 4๐‘ž๐‘ฆ=4๐‘ฆ^2 ๐‘ž=(4๐‘ฆ^2)/4๐‘ฆ ๐’’=๐’š Also, OP = BP โˆš(๐‘^2+๐‘ž^2 ) = โˆš(๐‘^2+4๐‘ฅ^2โˆ’4๐‘๐‘ฅ+๐‘ž^2 ) Squaring both sides ๐‘^2+๐‘ž^2 = ๐‘^2+4๐‘ฅ^2โˆ’4๐‘๐‘ฅ+๐‘ž^2 0 = ๐Ÿ’๐’™^๐Ÿโˆ’๐Ÿ’๐’‘๐’™ 4๐‘๐‘ฅ=4๐‘ฅ^2 ๐‘=(4๐‘ฅ^2)/4๐‘ฅ ๐’‘=๐’™ Thus, Coordinates of Point P = (p, q) = (x, y) So, the correct answer is (D)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.