This question is similar to Chapter 7 Class 10 Coordinate Geometry - Examples

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https://www.teachoo.com/1748/527/Example-5---Find-a-point-on-y-axis-which-is-equidistant/category/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

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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