If two vertices of a triangle are (-2 , 3) and (5 , -1) , orthocentre lies at the origin and the centroid on the line X +Y = 7 , then find the third vertex of the triangle.

 

 

 

               

Let there be a triangle ABC where A(x,y),B(-2,3)and C(5,-1). O(0,0) be orthocentre.

Given is that centroid lies on the line X+Y=7.

Therefore, Coordinates of Centroid(X,Y) are:

       X=(x1+x2+x3)/3                Y=(y1+y2+y3)/3

         =(x-2+5)/3                        =(y+3-1)/3

         =(x+3)/3                           =(y+2)/3

Putting theses values in X+Y=7

(x+3)/3+(y+2)/3=7

or, x+3+y+2=21

or,x+y=16

Therefore, x=16-y.........eq-1

 

Let BE be line passing thru orthocentre and perpendicular to AC.

Slope of BE= -3-0/2-0 = -3/2

Slope of AC= -1-y/-5-x  =  -1-y/-5-16+y  ........[From eq 1]

                                  = -1-y/-21+y

Slope of BE*Slope of AC=-1

-3/2*(-1-y/-21+y)=-1

Get the value of y from here and substitute it in eq-1 to get value of x.

 

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