Question

The values of k for which the distance between the points A(k,2) and B(3, 4) is √8.
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Answer 1

$\bf \to \: \green{{ \underline{given \div }}}$

$\sf \to \: distance \: between \: the \: point \: a(k \: , \: 2) \: and \: b \: (3 \: , \: 4) \: is \: \sqrt{8}$

$\bf \to \: \orange{{ \underline{to \: find \: the \: value \: of \: k}}}$

$\bf \to \: \blue{{ \underline{step \: - \: by \: - \: step \: - \: explanation }}}$

$\sf \to \: points \: are \: \: a \: (k \: , \: 2) \: and \: b(3 \: , \: 4) \: is \: \sqrt{8} \\ \\ \sf \to \: by \: using \: the \: distance \: formula \\ \\ \sf \to \: d \: = \sqrt{( x_{1} \: - \: x_{2}) {}^{2} + \: ( y_{1} \: - \: y_{2}) {}^{2} } \\ \\ \sf \to \: \sqrt{8} = \sqrt{(k - 3) {}^{2} + (2 - 4) {}^{2} } \\ \\ \sf \to \: 8 = (k - 3) {}^{2} + ( - 2) {}^{2} \\ \\ \sf \to \: 8 = {k}^{2} + 9 - 6k \: + 4 \\ \\ \sf \to \: {k}^{2} - 6k + 13 - 8 = 0 \\ \\ \sf \to \: {k}^{2} - 6k \: + 5 = 0 \\ \\ \sf \to \: {k}^{2} - 5k - k \: + 5 = 0 \\ \\ \sf \to \: k \: (k - 5) - 1(k - 5) = 0 \\ \\ \sf \to \: (k - 1)(k - 5) = 0 \\ \\ \sf \to \: k \: = 1 \: and \: k \: = 5$

Answer 2

Given ,

The distance between the points A(k,2) and B(3, 4) is √8 units

We know that , the distance b/w Two points is given by

$\boxed{ \tt{Distance = \sqrt{ {( x_{2} - x_{1})}^{2} + {( y_{2} - y_{1})}^{2} } }}$

Thus ,

$\tt \mapsto \sqrt{8} = \sqrt{ {(3 - k)}^{2} + {(4 - 2)}^{2} }$

Squaring on both sides , we get

$\tt \mapsto 8 = {(3)}^{2} + {(k)}^{2} - 2(3)(k) + {(2)}^{2}$

$\tt \mapsto8 = 9 + {(k)}^{2} - 6k + 4$

$\tt \mapsto8 = {(k)}^{2} - 6k + 13$

$\tt \mapsto {(k)}^{2} - 6k + 5 = 0$

$\tt \mapsto {(k)}^{2} - 5k - k + 5 = 0$

$\tt \mapsto k(k - 5) - 1(k - 5) = 0$

$\tt \mapsto (k - 1)(k - 5) = 0$

$\tt \mapsto k = 1 \: \: or \: \: k = 5$

The value of k will be 1 or 5