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)
