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Find the points on the y-axis, whose distances from the line  are 5 units.

Asked by Topperlearning User 20th October 2016, 6:26 AM
Answered by Expert
Answer:

The given line is

      8x + 6y = 48
Let any point on the y-axis be (0, y).
Perpendicular distance from (0, y) to the given line
open vertical bar fraction numerator 6 y minus 48 over denominator square root of 6 squared plus 8 squared end root end fraction close vertical bar equals 5
open vertical bar 6 y minus 48 close vertical bar equals 5 cross times 10 equals 50
T a k i n g space left parenthesis plus right parenthesis space s i g n comma space w e space g e t

6 y minus 48 equals 50

rightwards double arrow y equals 49 over 3
N o w space t a k i n g space left parenthesis minus right parenthesis space s i g n comma space w e space g e t

6 y minus 48 equals minus 50

rightwards double arrow y equals minus 1 third

T h e r e f o r e comma space t h e space p o i n t s space a r e space open parentheses 0 comma space 49 over 3 close parentheses space a n d space open parentheses 0 comma space minus 1 third close parentheses space.
Answered by Expert 20th October 2016, 8:26 AM
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