if origin is the orthocentre ∆ABC where A(5,-1),B(-2,3) then the orthocentre of ∆OAC is

							
Orthocentre is the point of intersection of all altitudes of a triangle 

Let A=(x1,y1) B=(5,-1) and C=(-2,3) be the vertices of triangle 
Orthocentre, O=(0,0) 

Slope of BC = -4/7 
Altitude through A and perpendicular to BC is 
(y-y1) = 7/4(x-x1) 
--> 4y-4y1 = 7x-7x1 
this passes through orthocentre i.e., (0,0) 
therefore 4y1 = 7x1 ........................................... 


Slope of AC= (3-y1)/(-2-x1) 
Now, altitude through B and perpendicular to AC is 
y+1 = (x1+2)/(3-y1) [y-5] 
this passes through (0,0) 
therefore 5x1-y1+13=0 ....................................(2) 

On solving (1) and (2) 
we get (x1,y1) = (-4,-7) 

So A=(-4,-7) 

[OR] 

2)EASY PROCESS: 

Find orthocentre of given points(i.e., O,B,C) 
that is the orthocenter of triangle ABC 

Important Property: "If O is orthocentre of triangle ABC then the four points O,A,B,C are such that each point is orthocentre of triangle formed by other three points" 


Hint : If one of the vertices of a triangle is origin, with remaining vertices (x1,y1) and (x2,y2), 
then orthocentre is ( k{y2-y1} , k{x1-x2} ) 

where k=(x1x2 + y1y2) / (x1y2 - x2y1)
						

							
Dear student 
Ans foe the above question will be (-4,-7) , It is not mentioned in the above .
Good Luck Cheers