Question

a point P (2, - 1 ) is equidistant distance from the point A (k ,7) and the B (-3,k) find the value of K

Answer 1

PA =PB
SO

${(2 - k)}^{2} + {(7 + 1)}^{2} = \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {(2 + 3)}^{2} + {(k + 1)}^{2}$
4+k^2-4k+64=25+k^2+1+2k
=> 68-4k=26+2k
=> 68-26=6k
=> 42=6k
=> k=7

Answer 2

Here we goo ..

Since its equidistant it means it's the mid point...

So, by mid point theorem

For A

x=x1+x2/2

2=k+(-3)/2

2×2=k-3

4+3=k

k=7....answer

For B

y=y1+y2/2

-1=7+k/2

-1×2=7+k

-2=7+k

-2-7=k

-9=k....answer..


Hope it helps... (｡’▽’｡)♡