The point P(2k, k) is equidistant from A(-2, 4) and B (7,-5).Find the value of k.​

Answer:k = 3Step-by-step explanation:Using the distance formulad = √ (x₂ - x₁ )² + (y₂ - y₁ )²with (x₁, y₁ ) = A (- 2, 4 ) and (x₂, y₂ ) = P (2k, k)AP = [tex]\sqrt{(2k+2)^2+(k-4)^2}[/tex]Repeatwith (x₁, y₁ ) = B (7, - 5) and P = (2k, k)BP = [tex]\sqrt{(2k-7)^2+(k+5)^2}[/tex]Given that AP = BP, then[tex]\sqrt{(2k+2)^2+(k-4)^2}[/tex] = [tex]\sqrt{(2k-7)^2+(k+5)^2}[/tex]Square both sides(2k + 2)² + (k - 4)² = (2k - 7)² + (k + 5)² ← expand factors on both sides4k² + 8k + 4 + k² - 8k + 16 = 4k² - 28k + 49 + k² + 10k + 25Simplify both sides by collecting like terms5k² + 20 = 5k² - 18k + 74 ( subtract 5k² from both sides )20 = - 18k + 74 ( subtract 74 from both sides )- 54 = - 18k ( divide both sides by - 18 )k = 3