Answer:
[tex](\frac{17}{18}, \frac{23}{6})[/tex]
Step-by-step explanation:
D1 = [tex]\sqrt{(x+3)^2 + (y-4)^2}[/tex]
D2 = [tex]\sqrt{x^2 + y^2}[/tex]
Then set D1 = D2, and substitute in y = 3x+1
[tex]\sqrt{(x+3)^2 + (3x-3)^2}[/tex] = [tex]\sqrt{x^2 + (3x+1)^2}[/tex]
[tex]x^2 +6x +9 +9x^2 - 18x + 9 = x^2 + 9x^2 + 6x + 1[/tex]
[tex]18x = 17[/tex]
x = 17/18
Then to find y, substitute back in x for y = 3x + 1
[tex]y = \frac{17}{18} * 3 + 1[/tex]
y = 23/6
