Question

If the point C ( - 2,3)  is equidistant form the points A (3, -1) and Bx (x ,8)  , find the value of x. Also, find the distance between BC

Answer 1

As per the question, we have

AC=BC

`⇒ sqrt((-2-3)^2 +(3+1)^2) = sqrt ((-2-x)^2 +(3-8)^2)`

`⇒sqrt((5)^2 +(4)^2 ) = sqrt(( x +2)^2 + (-5)^2)`

⇒ 25+16 = (x+2)² + 25        (Squaring both sides) 

⇒ 25+ 16 = (x +2)² +25 

⇒(x +2)² = 16 

`⇒ x +2 = +- 4`

` ⇒ x = -2 +-4=-2-4,-2 +4=-6,2`

Now,

`BC = sqrt((-2 - x)^2 +(3-8)^2`

`= sqrt((-2-2)^2 +(-5)`

`= sqrt((16+25)) = sqrt(41) `    units

Hence, x = 2 or - 6 and BC =` sqrt(41)` units .