Respuesta :

frika

Answer:

(14.1, 14.1)

Step-by-step explanation:

If the point (x,y) is on the line y=x, then it has coordinates (x,x).

Find the distances from this point to points (-6,2) and (10,-9):

  • distance from (x,x) to (-6,2): [tex]\sqrt{(x-(-6))^2+(x-2)^2}=\sqrt{(x+6)^2+(x-2)^2}[/tex]
  • distance from point (x,x) to (10,-9): [tex]\sqrt{(x-10)^2+(x-(-9))^2}=\sqrt{(x-10)^2+(x+9)^2}[/tex]

Equate these distances:

[tex]\sqrt{(x+6)^2+(x-2)^2}=\sqrt{(x-10)^2+(x+9)^2}[/tex]

Square this equation:

[tex](x+6)^2+(x-2)^2=(x-10)^2+(x+9)^2\\ \\x^2+12x+36+x^2-4x+4=x^2-20x+100+x^2+18x+81\\ \\8x+40=-2x+181\\ \\8x+2x=181-40\\ \\10x=141\\ \\x=14.1[/tex]

So, the coordinates are (14.1, 14.1)

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