CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard
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CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard
Last updated at Dec. 13, 2024 by Teachoo
This question is similar to Chapter 7 Class 10 Coordinate Geometry - Examples
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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
