Question

If the point p(k-1 , 2) is equidistant from the points A(3 , k) and B( k , 5). Find the value of k .
( by using distance formula)

Answer 1

Answer:

Step-by-step explanation hope it helps you

Answer 2

Answer:

$Value\:of \:k=1 \:or \:k = 5$

Step-by-step explanation:

Given the point P(k-1,2) is equidistant from the points A(3,k) and B(k,5)

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The distance between two points

$A(x_{1},y_{1})\:and \:A(x_{1},y_{1})\\is \: \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

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$Here ,\: AP = BP$

$\implies AP^{2}=BP^{2}$

$\implies [(k-1)-3]^{2}+(2-k)^{2}=[(k-1)-k)]^{2}+(2-5)^{2}$

$\implies (k-4)^{2}+(2-k)^{2}=(-1)^{2}+(-3)^{2}$

$\implies k^{2}-8k+16+4-4k+k^{2}=1+9$

$\implies 2k^{2}-12k+20-10=0$

$\implies 2k^{2}-12k+10=0$

$\implies k^{2}-6k+5=0$

/* Splitting the middle term, we get

$\implies k^{2}-1k-5k+5=0$

$\implies k(k-1)-5(k-1)=0$

$\implies (k-1)(k-5)=0$

$\implies k-1=0\:or \:k-5=0$

$\implies k=1 \:or \:k = 5$

Therefore,

$Value\:of \:k=1 \:or \:k = 5$

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