Question

write the coordinates of point P on x axis which is equidistant from the point A(-2,0) B(6,0)​

Answer 1

i also want to know, it was in yesterday's exam...  : ) : )

Answer 2

Solution:

Given points are  A9(-2,0) and B(6,0)

∵ on X axis the point P will be (t,0)

we have to find t =?

According to the question these points are equidistant

therefore AP = BP now using Distance formula we get

AP = √ (x2-x1) +(y2-y1)

here A(-2,0) and P(t.0)

x1 = -2 , x2 = t and y1 = 0 and y2 = 0

therefore AP = √t²+4+4t +0

similarly

For BP , here x1 = 6 x2 = t and y1 =0 y2 = 0

therefore BP = √t²+36-12t

now BP = AP  , Squaring both side we get

t²+4+4t = t²+36+12t

⇒ -36+4 = 12t- 4t = 8t

⇒8t = -32

∴ t = -4

hence the co-ordinate of P will be (-4,0)