Answer:
B) II and III only
Step-by-step explanation:
Coordinates of A = (-1,0)
Coordinates of P =(4,12)
Let (x,y) be the coordinates of B
Distance formula : =[tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
AP = BP
[tex]\sqrt{(4+1)^2+(12-0)^2}=\sqrt{(4-x)^2+(12-y)^2}[/tex]
[tex](4+1)^2+(12-0)^2=(4-x)^2+(12-y)^2[/tex]
[tex]169=16+x^2-8x+144+y^2-24y[/tex]
[tex]x^2-8x+y^2-24y-9=0[/tex]
I)(3/2,6)
[tex](\frac{3}{2})^2-8(\frac{3}{2})+6^2-24(6)-9=0\\-126.75 \neq 0[/tex]
II)(9,24)
[tex](9)^2-8(9)+(24)^2-24(24)-9=0[/tex]
0= 0
C)(-8,7)
[tex](-8)^2-8(-8)+(7)^2-24(7)-9=0[/tex]
0=0
So, II and III could be the coordinates of B
So Option B is true
II) II and III only
