Check if the Full-back J(5,-3) and centre-back I(-4,6) are equidistant from forward C(0,1) and if C is the mid-point of IJ.

Check if the Full-back J(5,-3) and centre-back I(-4,6) are equidistant - CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard

part 2 - Question 37 (ii) (Choice 2) - CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10
part 3 - Question 37 (ii) (Choice 2) - CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10

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We need to check if C is equidistant from J and I And C is mid-point of IJ Doing this one by one Check if C (0, 1) is equidistant from I (−4, 6) and J (5, −3) To prove if point C is equidistant from I & J We need to prove CI = CJ Since CI = CJ ∴ C is equidistant from I and J Finding CI CI = √(( −𝟒 −𝟎)𝟐+(𝟔−𝟏)𝟐) = √((−4)2+(5)2) = √(16+25 ) = √𝟒𝟏 Finding CJ CJ = √(( 𝟓 −𝟎)𝟐+(−𝟑−𝟏)𝟐) = √((5)2+(−4)2) = √(25+16 ) = √𝟒𝟏 Check if C (0, 1) is midpoint of I (−4, 6) and J (5, −3) If C is mid-point of IJ Coordinates of C = ((𝑥1 + 𝑥2)/2,(𝑦1 +𝑦2)/2) (0, 1) = ((−𝟒 + 𝟓)/𝟐,(𝟔 + (−𝟑))/𝟐) (0, 1) = (1/2,3/2) Since LHS ≠ RHS ∴ C is NOT the mid-point of IJ 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

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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