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Example 5 Find a point on the y−axis which is equidistant from the points A(6, 5) and B(– 4, 3). Given A(6, 5) & B(−4, 3) Since the required point is in y-axis, its x –coordinate will be zero Let Required point = C (0, a) As per question, point C is equidistant from A & B Hence, AC = BC Finding AC & BC separately Finding AC x1 = 6 , y1 = 5 x2 = 0 , y2 = a AC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 0 −6)2+(𝑎−5)2) = √((−6)2+(𝑎 −5)2) = √((6)2+(𝑎 −5)2) = √((6)2+ 𝑎2+52 −2(5)(𝑎) ) = √(36+ 𝑎2+25 −10𝑎) = √(𝒂𝟐 −𝟏𝟎𝒂+𝟔𝟏) Finding BC x1 = −4 , y1 = 3 x2 = 0 , y2 = a BC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 0 −(−4))2+(3 −𝑎)2) = √((0+4)2+(3 −𝑎)2) = √((4)2+(3 −𝑎)2) = √((4)2+ 𝑎2+32−2(3)(𝑎) ) = √(16+ 𝑎2+9−6𝑎) = √(𝒂𝟐−𝟔𝒂+𝟐𝟓) Now, AC = BC √(𝑎2 −10𝑎+61) = √(𝑎2−6𝑎+25) Squaring both sides (√(𝑎2 −10𝑎+61) " )2 = (" √(𝑎2−6𝑎+25))^2 a2 – 10a + 61 = a2 − 6a + 25 a2 – 10a − a2 + 6a = 25 – 61 −4a = −36 a = (−36)/(−4) a = 9 Hence the required point is C(0, a) = (0, 9)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.