# Misc 4 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 4 Find the coordinates of a point on y-axis which are at a distance of 52 from the point P (3, –2, 5). Since point is on y-axis its x & z coordinate is 0 Let the point on y-axis be A(0, a, 0) Given that Point A is at a distance of 52 from point P (3, – 2, 5) i.e. PA = 52 We know that Distance between two point (x1 y1 z1) & (x2 y2 z2) is D = x2−x12+y2−y12+z2 −z12 Calculating PA P (3, –2, 5) , A (0, a, 0) Here x1 = 3, y1 = – 2, z1 = 5 x2 = 0, y2 = a, z2 = 0 PA = 0−32+a−(−2)2+0 −52 52 = −32+a+22+0−52 52 = 9+𝑎+22+25 52 = 9+𝑎2+22+2𝑎2+25 52 = 9+𝑎2+4+4𝑎+25 52 = 𝑎2+4𝑎+9+4+25 52 = 𝑎2+4𝑎+38 (52)2 = 𝑎2+4𝑎+382 5 × 5 × 2 × 2 = 𝑎2+4𝑎+38 25 × 2 = 𝑎2+4𝑎+38 50 = 𝑎2+4𝑎+38 𝑎2+4𝑎+38 = 50 𝑎2+4𝑎+38 – 50 = 0 𝑎2+4𝑎− 12 = 0 𝑎2+6𝑎− 2a – 12 = 0 a(a + 6) – 2 (a + 6) = 0 (a – 2) (a + 6 ) = 0 So, a = 2 or a = – 6 Thus, the coordinates of the required point is (0, 2, 0) & (0, –6, 0)

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