+0  
 
+3
491
3
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!

 Jun 3, 2017

Best Answer 

 #2
avatar+7347 
+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

 Jun 4, 2017
 #1
avatar+322 
+1

I don't know what it is!

 Jun 3, 2017
 #2
avatar+7347 
+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! :)

 Jun 6, 2017

22 Online Users

avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.