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Sequences And Series

Question
CBSEENMA11014980

Find the value of x, so that the point (6, 5, -3) is at a distance of 13 units from the point (x, -7, 0).

Solution

The given points are A (6, 5, -3) and B (x, -7, 0)
Given that                                   space space space open vertical bar AB close vertical bar space equals space 13
rightwards double arrow           square root of left parenthesis 6 minus straight x right parenthesis squared plus left parenthesis 5 plus 7 right parenthesis squared plus left parenthesis negative 3 minus 0 right parenthesis squared end root space equals space 13
Squaring both sides, we get
                 straight x squared plus 36 minus 12 straight x plus 144 plus 9 equals 169 space rightwards double arrow space straight x squared minus 12 straight x plus 20 equals 0
rightwards double arrow                       space space space left parenthesis straight x space minus 2 right parenthesis space left parenthesis straight x minus 10 right parenthesis space equals space 0    rightwards double arrow  x = 2, x = 10
Hence, the value of x is 2 or 10.