Question

Given (6,1) and (x, -11), find all x such that the distance between these two points is 13.

Answer 1

Let:

`\begin{gathered} (x1,y1)=(6,1) \\ (x2,y2)=(x,-11) \end{gathered}`

Using the distance formula:

`\begin{gathered} d=√((x2-x1)^2+(y2-y1)^2) \\ \end{gathered}`

so:

`\begin{gathered} 13=√((x-6)^2+(-11-1)^2) \\ \end{gathered}`

Square both sides:

`169=(x-6)^2+144`

Expand (x - 6)²:

`\begin{gathered} 169=x^2-12x+36+144 \\ x^2-12x+180-169=0 \\ x^2-12x+11=0 \end{gathered}`

Factor:

The factor of 11 that sum to -12 are -11 and -1, so:

`\begin{gathered} (x-11)(x-1)=0 \\ so: \\ x=1 \\ or \\ x=11 \end{gathered}`

Answer:

x = 1 and x = 11