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If the Point P(K-1, 2) is Equidistant from the Points A(3,K) and B(K,5), Find the Value of K. - Mathematics

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Answer in Brief

If the point P(k-1, 2) is equidistant from the points A(3,k) and B(k,5), find the value of k.

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Solution 1

The given points are P(k-1, 2) , A(3,k) and B(k,5).

∵ AP =BP

∴ AP= BP2

`⇒ (k-1-3)^2 +(2-k)^2 = (k-1-k)^2 + (2-5)^2`

`⇒ (k-4)^2 +(2-k)^2 = (-1)^2 +(-3)^2`

`⇒ k^2 -8y +16+4+k^2 -4k =1+9`

`⇒ k^2 -6y +5 =0`

`⇒ (k-1)(k-5)=0`

`⇒  k=1 or k=5`

Hence , k=1 or k=5 

Solution 2

It is given that P(− 1, 2) is equidistant from the points A(3, k) and B(k, 5).
∴ AP = BP

\[\Rightarrow \sqrt{\left[ \left( k - 1 \right) - 3 \right]^2 + \left( 2 - k \right)^2} = \sqrt{\left[ \left( k - 1 \right) - k \right]^2 + \left( 2 - 5 \right)^2} \left( \text{ Distance formula }  \right)\]

\[ \Rightarrow \sqrt{\left( k - 4 \right)^2 + \left( 2 - k \right)^2} = \sqrt{\left( - 1 \right)^2 + \left( - 3 \right)^2}\]

Squaring on both sides, we get 

\[k^2 - 8k + 16 + 4 - 4k + k^2 = 10\]
\[ \Rightarrow 2 k^2 - 12k + 10 = 0\]
\[ \Rightarrow k^2 - 6k + 5 = 0\]
\[ \Rightarrow \left( k - 1 \right)\left( k - 5 \right) = 0\]

\[\Rightarrow k - 1 = 0 \text{ or k }  - 5 = 0\]

\[ \Rightarrow k = 1 or k = 5\]

Thus, the value of k is 1 or 5.

 
Concept: Coordinate Geometry
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APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.2 | Q 36 | Page 16
RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 4

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