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

Last updated at Aug. 16, 2021 by Teachoo

Examples

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Example 5 Important You are here

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Example 7 Important

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Example 9 Important

Example 10 Important

Example 11 Deleted for CBSE Board 2023 Exams

Example 12 Deleted for CBSE Board 2023 Exams

Example 13 Important Deleted for CBSE Board 2023 Exams

Example 14 Important Deleted for CBSE Board 2023 Exams

Example 15 Deleted for CBSE Board 2023 Exams

Chapter 7 Class 10 Coordinate Geometry

Serial order wise

Last updated at Aug. 16, 2021 by Teachoo

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)