istance formula If A =(3,4) and B is a variable point on the line |x|=6.If AB is less than or equal

pasu shah

A =(3,4) and B is a variable point on the line |x|=6.If AB is less than or equal to 4 then the no. of positions of B with integral co-ordinates is:

a)5

b)10

c)12

d)6

Yagyadutt Mishra

Let us take the B point as (x,y) as of now......where |x| = 6 and no of solution of y will give the no of position of B.....

Now A is (3,4) B(x,y) and AB <=4

So root[ (x-3)^2 + (y-4)^2 ] <=4

squaring both side (because distance is always positive ...so we can square in inequality)

(x-3)^2 + (y-4)^2 <=16

so (y-4)^2 <= 16 -(x-3)^2

x = 6 then (y-4)^2 <= 16 - 9

so (y-4)^2 <= 7

or -root(7) <=y-4 <= root(7) => 4 -root(7)<= y <= 4 +root(7)

x=-6 then

(y-4)^2 < -ve term

Hence it is not possible...

So all the values of y in this range is the required value of no. of points of B...

Considering integral co-ordinate...

4 -root(7)<= y <= 4 +root(7) = > 1.3 <= y <= 6.7

Hence integral value of y is ..... 2,3 ,4 ,5,6

Hence finally 5 values of B is possible...

jay

This problem becomes very simple if u do it graphically. First u plot (3,4) n then x=6 , x=-6. Now clearly u will not get any solution for x=-6 because its shortest dist from (3,4) is 9. Now consider a point t(6,y) on x=6 on upward side and a point s(6,4) that is in horizontal level to given pt. So u will get a right angle triangle if u join all the three points. Now dis of given pt. to t is max.4 and dis betw given pt and s is 3 so by pythagores theorem u can get dis between s and t that is 2.6457 now we want integral value so there are 2 pt in upward direction to s such that its distance frm given pt is less than 4 similarly there will be 2 pts downward direction to s where distance is less than 4. So in total there are 2up pts+2down pts +1 points itself =5 pts.In coordinate geometry many problems could be solved by either ploating rough graph or by simple geomertr.

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