+0  
 
+3
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avatar+322 

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

Best Answer 

 #2
avatar+7324 
+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
 #1
avatar+322 
+1

I don't know what it is!

Davis  Jun 3, 2017
 #2
avatar+7324 
+2
Best Answer

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
avatar+322 
+1

Thanks, this solved my question! :)

Davis  Jun 6, 2017

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