# Example 5 - Chapter 7 Class 10 Coordinate Geometry (Term 1)

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

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) 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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Example 10

Example 11 Deleted for CBSE Board 2022 Exams

Example 12 Deleted for CBSE Board 2022 Exams

Example 13 Deleted for CBSE Board 2022 Exams

Example 14 Important Deleted for CBSE Board 2022 Exams

Example 15 Deleted for CBSE Board 2022 Exams

Chapter 7 Class 10 Coordinate Geometry (Term 1)

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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.