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# 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!

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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!

Davis  Jun 3, 2017

#2
+5931
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We can use the distance formula on this problem:

distance between  (x, y1)  and  (x, 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

hectictar  Jun 4, 2017
Sort:

#1
+321
+1

I don't know what it is!

Davis  Jun 3, 2017
#2
+5931
+2

We can use the distance formula on this problem:

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

____________________________________________________________

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

hectictar  Jun 4, 2017
#3
+321
+1

Thanks, this solved my question! :)

Davis  Jun 6, 2017

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