1Well that is the answer....! But why can't we use distance formula for that ???
If A=(3,4) and B is a variable point on the lines |x|=6 .If AB≤4 then the number of positions of B with integral co-ordinates is:
(a) 5 (b) 6
(c) 10 (d) 12
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10 Answers
62try doing this with graphs...
draw the graphs of |x|=6
and the equation of AB <=4 will be a circle...
Then u try.. if u dont get the solutino.. i will post it :)
1
33u can use..
(6-3)2+(y-4)2≤16
solve it to get -√7+4≤y≤√7+4
count no of integral y in this interval
so 5 points ..
x=-6 wont have any..
if u solve (-6-3)2+(y-4)2≤16 no soln..
62yup you can uma..
i will tell u the solution with that method too..
let the point on the line be (6, t)
the distance from 3,4 will be
root of
32+(t-4)2
so this will be ≤16
(t-4)2≤7
t-4≤√7 and t-4≥-√7
t=2,3,4,5,6
33B(6,y)
A(3,4)
apply distance formula
√(6-3)2+(y-4)2 ≤4
and solve it.. u will get a condition in y
-√7+4≤y≤√7+4
take approximate values..
1.__≤y≤6.__
so y=2,3,4,5,6
B: (6,2)
(6,3)
(6,4)
(6,5)
(6,6)