हिंदी

If the point C(–2, 3) is equidistant from the points A(3, –1) and B(x, 8), find the values of x. Also, find the distance between BC.

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प्रश्न

If the point C(–2, 3) is equidistant from the points A(3, –1) and B(x, 8), find the values of x. Also, find the distance between BC.

योग
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उत्तर

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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अध्याय 6: Coordinate Geometry - EXERCISE 6A [पृष्ठ ३१२]

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आर.एस. अग्रवाल Mathematics [English] Class 10
अध्याय 6 Coordinate Geometry
EXERCISE 6A | Q 15. | पृष्ठ ३१२
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