difficult coordinate geometry question

[ashish1354]

OA after some replies

[sanjana]

Since we know c(20,0) and A(-8,0) the distance between the 2 points is 20-(-8) = 28 (If the y coordinate is the same simply subtract the x coordinates to get the distance)

Therefore, AC=28
Given that : AC=BD,Hence BD=28.

Therefore we know that the y coordinate for B has to be 28,we can thus eliminate answers A,C,D

Now,the answer has to be (6,28) OR (14,28)

Now given that AB=DC
Therefore by distance formula,
(-8-x)^2+y^2=(x-20)^2
This reduces to 336-56x=y^2
when y=0,x=6
hence the point B(6,28) IMO.

[ashis979]

Started out the way you did, but then I got confused. Just to be clear, so you're saying, B(x,y) and D(x,0), correct?
y=28, that's pretty much a given since AC=BD, and like you said that distance is 28.
But CD is not equal to 20, now is it? Say the origin is O, therefore it is OC that is equal to 20. OA=8 and OC=20, that's the only way AC=28. There's that one piece OD that we need to take into account, yes? Length of OD is essentially x. So the equation should be:
(x+8)^2+28^2=(20-x)^2
...reduce to x= -8, but this can't happen since x(or the length of OD) can't be negative. Where am I wrong?

[nithi_mystics]

AC = 8 + 20 = 28
Since AC = BD, y coordinate of B is 28

We have only 2 options with y co-ordinate 28.

Now lets take CD = x. Therefore, OD = 20-x
O is the origin. And we need to find the x coordinate (thats 20-x )

In triangle ABD,

AB^2 = BD^2 + AD^2
CD^2 = 28^2 + (8+20-x)^2
x^2 = 28^2 + (28-x)^2

Solving this, we get x=14 ====> 20-x = 6

Hence (6,28) is the answer.