Question

help!!! My exams are near​

Answer 1

$\huge\boxed{\ \ \ \ \ \ \ \ \ \ \ \huge\boxed{\ \ \ \ \ \ \ \ \ \ \ \bold{k= 3}\ \ \ \ \ \ \ \ \ \ \ }\ \ \ \ \ \ \ \ \ \ \ }$

$Distance between \ (5, k)\ and \ (2, 4) = Distance between \ (k,7)\ and \ (2,4) \\ \\ \sqrt{(5-2)^2+(k-4)^2}=\sqrt{(k-2)^2+(7-4)^2} \\ \\ \sqrt{3^2+k^2-8k+16}=\sqrt{k^2-4k+4+3^2} \\ \\ 9+k^2-8k+16=k^2-4k+4+9 \\ \\ 16-8k=4-4k \\ \\ 4(4-2k)=4(1-k) \\ \\ 4-2k=1-k \\ \\ 2k-k=4-1 \\ \\ k=\bold{3}$

$Hope this helps. Please mark it as the \ \bold{brainliest}. \\ \\ \\ \\ \\ Thank you. :-)$