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

Distance

between two points $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and $\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated using the formula $\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } }$

Distance between $(3k,4)$ and $(2,k) = \sqrt { \left( 3k-2 \right) ^{ 2 }+\left( 4 - k \right) ^{ 2 } } = \sqrt { 9{k}^{2} + 4 - 12k + 16 + {k}^{2} - 8k } = \sqrt { 10{ k}^{ 2 } - 20k + 20 }$

Given, distance between the points $= 5\sqrt { 2 }$

$=> \sqrt { 10{ k }^{ 2 } - 20k + 20 } = 5\sqrt { 2 }$

$=> 10{ k }^{ 2 } - 20k + 20 = 25\times 2 = 50$

$=> { k }^{ 2 } - 2k + 2= 5$

$=> { k }^{ 2 } - 2k - 3 = 0$

$=> { k }^{ 2 } - 3k + k - 3 = 0$

$=> k(k - 3) + 1( k - 3) = 0$

$=> (k+ 1)(k - 3)= 0$

$=> k = -1,3$