Find the value of k for which the distance between the points $A (3 k, 4)$ and $B (2, k)$ is $5 \sqrt{2}$ units.

k = - 1

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k = 3

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k = - 3

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k = 1

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Solution

The correct options are
A $k = - 1$
B $k = 3$
Distance
between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ can be calculated using the formula $\sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$
Distance between $(3 k, 4)$ and $(2, k) = \sqrt{{(3 k - 2)}^{2} + {(4 - k)}^{2}} = \sqrt{9 k^{2} + 4 - 12 k + 16 + k^{2} - 8 k} = \sqrt{10 k^{2} - 20 k + 20}$
Given, distance between the points $= 5 \sqrt{2}$
$=> \sqrt{10 k^{2} - 20 k + 20} = 5 \sqrt{2}$
$=> 10 k^{2} - 20 k + 20 = 25 \times 2 = 50$
$=> k^{2} - 2 k + 2 = 5$
$=> k^{2} - 2 k - 3 = 0$
$=> k^{2} - 3 k + k - 3 = 0$
$=> k (k - 3) + 1 (k - 3) = 0$
$=> (k + 1) (k - 3) = 0$
$=> k = - 1, 3$

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