Answer
$k $ =$ \frac{1}{2}$ OR $0.5$
Work Step by Step
Distance 'd' between two points $(x_{1},y_{1}) $ and $(x_{2},y_{2})$ is given by-
d = $\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}} $
Now, Distance '$d_{1}$' between points $(k,k) $and $(-1,0)$ will be -
$d_{1}$ = $\sqrt{(k - (-1))^{2} + (k - 0)^{2}} $
i.e. $d_{1}$ = $\sqrt{(k + 1)^{2} + k^{2}} $
i.e. $d_{1}$ = $\sqrt{k^{2} +1 +2k + k^{2} } $
i.e. $d_{1}$ = $\sqrt{2k^{2} +2k +1 } $
Now, Distance '$d_{2}$' between points $(k,k) $and $(0,2)$ will be -
$d_{2}$ = $\sqrt{(k - 0)^{2} + (k - 2)^{2}} $
i.e. $d_{2}$ = $\sqrt{k^{2} + k^{2} - 4k +4 } $
i.e. $d_{2}$ = $\sqrt{2k^{2} - 4k +4 } $
Now According to problem-
$d_{1}$ = $d_{2}$
i.e. $\sqrt{2k^{2} +2k +1 } $ = $\sqrt{2k^{2} - 4k +4 } $
Squaring on both sides-
$2k^{2} +2k +1 $ = $2k^{2} - 4k +4 $
i.e. $2k +1 $ = $ - 4k +4 $ (Subtracting $2k^{2}$ from both sides)
i.e. $2k +4k $ = $ 4-1 $
i.e. $6k $ = $ 3$
i.e. $k $ = $ \frac{3}{6} = \frac{1}{2}$ OR $0.5$
i.e. $k $ =$ \frac{1}{2}$ OR $0.5$
