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If the Point P (2,2) is Equidistant from the Points a ( -2,K ) and B( -2k , -3) , Find K. Also, Find the Length of Ap. - Mathematics

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If the point P (2,2)  is equidistant from the points A ( -2,K ) and B( -2K , -3) , find k. Also, find the length of AP.

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Solution

As per the question, we have

AP = BP

`⇒sqrt((2+2)^2 +(2+k)^2) = sqrt(( 2+2k)^2 +(2+3)^2)`

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

⇒ 16+ 4 +k2 - 4k = 4+ 4k2 + 8k +25        (Squaring both sides) 

`⇒k^2 + 4k +3=0`

⇒ (k+1) (k+3) =0

⇒ k =-3, -1

Now for k = -1

`AP= sqrt ((2+2)^2 +(2-k)^2)`

`= sqrt((4)^2 +(2+1)^2)`

`= sqrt(16+9) = 5` units 

For k = -3 

`AP= sqrt(( 2+2)^2 +(2-k)^2)`

`= sqrt((4)^2+(2+3)^2)`

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

Hence, k= -1,-3; AP= 5 units for k=-1 and AP=`sqrt(41)` units for k=-3.

 

 

 

 

 

 

 

 

Concept: Coordinate Geometry
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APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 16

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