Verified answer

This is very simple.

As it is given that P(3,k) is equidistant from the two given pts. Apply distance formula

i.e. AP=BP

AP^2=BP^2

(5-3)^2+(-3-k)^2=(-3-3)^2+(1-k)^2

4+9+k^2+6k=36+k^2+1-2k

=>k=3

You need to solve

|| (3,k) - (5,-3) ||^2 = || (3,k) - (-3,1) ||^2

==>

2^2 + (k+3)^2 = 6^2 + (k-1)^2

Solve for k and you get the answer.

Note: there should be two k's, which means there are two solutions which give the same equidistance.

First discover f' of f(x). with any luck you comprehend a thank you to take the spinoff. this could be style of not elementary to describe here, yet you convey the flexibility of x down & multiply it by potential of your term, then subtract a million from the flexibility. 2x³. Your power is 3, so convey that down & multiply it by potential of the term. 3*2x³ = 6x³. Now subtract a million from the flexibility. 6x² Do an identical with -4x. the flexibility of x is a million, so once you multiply by potential of a million you nevertheless have -4x. Subract a million from the flexibility, you get -4x^0 = -4*a million = -4. So f'(x) = 6x² - 4. purely plug the three x's in. f'(-2) = 20 f'(0) = -4 f'(2) = 20