(iii)What are the coordinates of the point on y axis equidistant from A and G?

 

[Sample Paper Class 10] What are the coordinates of the point on y - CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard

part 2 - Question 36 (iii) (Choice 1) - CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10
part 3 - Question 36 (iii) (Choice 1) - CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10
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Question 36 (iii) – Choice 1 What are the coordinates of the point on y axis equidistant from A and G? Here, A(-2,2) and G(-4,7) Since the required point is in y-axis, its x –coordinate will be zero Let Required point = Z (0, a) As per question, point Z is equidistant from A & B Hence, AZ = BZ Finding AC x1 = −2 , y1 = 2 x2 = 0 , y2 = a AC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 0 −(−2))2+(𝑎−2)2) = √((2)2+(𝑎−2)2) = √((2)2+ 𝑎2+22−2(2)(𝑎) ) = √(4+ 𝑎2+4−4𝑎) = √(𝒂𝟐−𝟒𝒂+𝟖) Finding BC x1 = −4 , y1 = 7 x2 = 0 , y2 = a BC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 0 −(−4))2+(7 −𝑎)2) = √((0+4)2+(7 −𝑎)2) = √((4)2+(7 −𝑎)2) = √((4)2+ 𝑎2+72−2(7)(𝑎) ) = √(16+ 𝑎2+49−14𝑎) = √(𝒂𝟐−𝟏𝟒𝒂+𝟔𝟓) Now, AC = BC √(𝑎2−4𝑎+8) = √(𝑎2−14𝑎+65) Squaring both sides (√(𝑎2−4𝑎+8) " )2 = (" √(𝑎2−14𝑎+65))^2 𝒂𝟐−𝟒𝒂+𝟖 = 𝒂𝟐−𝟏𝟒𝒂+𝟔𝟓 −4a + 8 = −14a + 65 −4a + 14a = 65 − 8 10a = 57 a = 57/10 a = 5.7 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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo