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  • One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is A. (–1, –1) B. (2, 2) C. (–2, –2) D. (2, –2)

Q: One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is
A. (–1, –1)
B. (2, 2)
C. (–2, –2)
D. (2, –2)

Answers (1)

Let ABC be an equilateral triangle with vertex A (a,b) 

Let AD be perpendicular to BC and let (p,q) be the coordinates of D  

Given that the centroid P lies at the origin (0,0) 

We know that, the centroid of a triangle divides the median in the ratio 1:2  Now, using the section formula, we get 

\left ( x,y \right )=\left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\right )

\left ( 0,0 \right )=\left ( \frac{1*a+2*p}{1+2} , \frac{1*b+2*q}{1+2}\right )  

\left ( 0,0 \right )=\left ( \frac{a+2p}{3} , \frac{b+2q}{3}\right )

\frac{a+2p}{3}=0 and \frac{b+2q}{3}=0

 a+2p=0 and b+2q=0…A…                                        

 a+2p=b+2q  

  2p-2q=b-a …i

 It is given that BC=x+y-2=0 

 Since, the above equation passes through p,q 

  p+q-2=0  

Now we find the slope of line AP

   m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{b-0}{a-0}=\frac{b}{a}

Equation of line BC is x+y-2=0  

y= -x+2  

y=-(1)x+2  

Since the above equation is in y=mx+b form  

So, slope of line BC is mBC=-1   

Since both lines rae perpendicular \frac{b}{a}*\left ( -1 \right )=-1

 b=a 

Putting this value in equation (i )we get  2p-q=b-b=0 

 p=q   

 Now putting this value in equation (i) we get p+q-2=0  

 2p=2  

 p=1 

 q=1 

Putting the value of p and q in equation A , we get a+2*1=0   and b+2*1=0

a=-2 and b=-2 

So, the coordinates of vertex A (a,b) is (-2,-2)

Hence, the correct option is c

Posted by

infoexpert22

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