Let S(0,p), T(6, -2) and O(0,0) are three vertices of a triangle STO.If ST = SO, Find the value of p please!

We can use the distance formula on this problem:

distance between  (x₁ , y₁)  and  (x₂ , y₂)   =   \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

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ST = SO

distance between point S and point T = distance between point S and point O

distance between  (0,p)  and  (6, -2)   =   distance between  (0,p)  and  (0,0)

\(\sqrt{(0-6)^2+(p--2)^2}=\sqrt{(0-0)^2+(p-0)^2}\)           Square both sides of this equation.

(0 - 6)² + (p - -2)²   =   (0-0)² + (p - 0)²

(-6)² + (p + 2)²   =   (0)² + (p)²

36 + p² + 4p + 4   =   p²

40 + 4p   =   0

4p   =   -40

p   = -10

I don't know what it is!

Thanks, this solved my question! :)