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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
Solution
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 .
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