Answer :
Two values of x such that are point (x, 5) is five units from the point (-1, 2) x=3 and x=-5.
using the distance formula we can find values of x.
[tex]d=\sqrt{(x_{2} -x_{1} )^2+(y_{2} -y_{1} )^2}[/tex]
here [tex]d=5 , x2=-1 , x1=x , y2=2 , y1=5[/tex]
putting all these values in the distance formula
[tex]5=\sqrt{(-1-x)^2+(2-5)^2}[/tex]
evaluating the given square and taking square on both sides we get
[tex]25=(x+1)^2+9\\\\16=(x+1)^2\\\\(x+1)=+4[/tex]and [tex](x+1)=-4[/tex]
hence we get two values of x using this distance formula and evaluating the quadratic equation in a single variable x.
[tex]x=4-1=3[/tex] and [tex]x=-4-1=-5[/tex]
x=3 and x=-5 are two values of x such that point (x, 5) is five units from the point (-1, 2).
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