let A=(3,4) and B is a variable point on the line ∣x∣=6. if AB^(−) <4, then the number of position of B with integral coordinates is? please help!


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Question Number 98598 by MWSuSon last updated on 15/Jun/20

$l e t A = (3, 4) a n d B i s a v a r i a b l e p o i n t$ $o n t h e l i n e ∣ x ∣= 6 . i f \overset{―}{A B} < 4, t h e n t h e$ $n u m b e r o f p o s i t i o n o f B w i t h i n t e g r a l$ $c o o r d i n a t e s i s ?$ $p l e a s e h e l p!$
Commented bybobhans last updated on 15/Jun/20
$∣x∣ = 6 ⇒ { ((x = −6 or)),((x= 6)) :} case (1) let B(−6,y) then (√((y−4)^2 +81)) < 4 (y−4)^2 +81 < 16 ⇒y = ∅ case(2) let B(6,y) then (√((y−4)^2 +9)) < 4 (y−4)^2 −7 <0 ⇒(y−4−(√7))(y−4+(√7)) < 0 4 − (√7) < y < 4+(√7) 1.354 < y < 6.64 so position of B with integral coordinates where y ∈{2,3,4,5,6}$
$∣ x ∣ = 6 \Rightarrow {\begin{cases} x = - 6 o r \\ x = 6 \end{cases}$ $case (1) let B (- 6, y) then \sqrt{{(y - 4)}^{2} + 81} < 4$ ${(y - 4)}^{2} + 81 < 16 \Rightarrow y = \emptyset$ $case (2) let B (6, y) then \sqrt{{(y - 4)}^{2} + 9} < 4$ ${(y - 4)}^{2} - 7 < 0 \Rightarrow (y - 4 - \sqrt{7}) (y - 4 + \sqrt{7}) < 0$ $4 - \sqrt{7} < y < 4 + \sqrt{7}$ $1.354 < y < 6.64$ $so position of B with integral coordinates$ $where y \in {2, 3, 4, 5, 6}$
Commented byMWSuSon last updated on 15/Jun/20
Thanks a lot.
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