Respuesta :
Let:
[tex]\begin{gathered} (x1,y1)=(6,1) \\ (x2,y2)=(x,-11) \end{gathered}[/tex]Using the distance formula:
[tex]\begin{gathered} d=\sqrt{(x2-x1)^2+(y2-y1)^2} \\ \end{gathered}[/tex]so:
[tex]\begin{gathered} 13=\sqrt{(x-6)^2+(-11-1)^2} \\ \end{gathered}[/tex]Square both sides:
[tex]169=(x-6)^2+144[/tex]Expand (x - 6)²:
[tex]\begin{gathered} 169=x^2-12x+36+144 \\ x^2-12x+180-169=0 \\ x^2-12x+11=0 \end{gathered}[/tex]Factor:
The factor of 11 that sum to -12 are -11 and -1, so:
[tex]\begin{gathered} (x-11)(x-1)=0 \\ so: \\ x=1 \\ or \\ x=11 \end{gathered}[/tex]Answer:
x = 1 and x = 11
