Question 24 - CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

Last updated at Sept. 20, 2024 by Teachoo

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Question 24 Find the point(s) on the x-axis which is at a distance of √41 units from the point (8, -5).Let point A (8, −5) Since the required point is in x-axis, its y –coordinate will be zero Let Required point = B (x, 0) As per question, Distance AB = √41 √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2)=√41 √((𝒙−𝟖)𝟐+(𝟎 −(−𝟓) )𝟐)=√𝟒𝟏 √((𝑥−8)2+(0+5)2)=√41 √((𝑥−8)2+5^2 )=√41 √((𝒙−𝟖)𝟐+𝟐𝟓)=√𝟒𝟏 Squaring both sides (√((𝑥−8)2+25))^2=(√41)^2 (𝒙−𝟖)𝟐+𝟐𝟓=𝟒𝟏 (𝑥−8)2=41−25 (𝑥−8)2=16 (𝑥−8)2=4^2 Removing square root (𝒙−𝟖)=±𝟒 (𝒙−𝟖)=±𝟒 Solving So, required points are (12, 0) or (4, 0) (𝑥−8)=−4 𝑥=−4+8 𝒙=𝟒 (𝑥−8)=4 𝑥=4+8 𝒙=𝟏𝟐

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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