# origin is the orthocentre of triangle ABC where A=(5,-1), B=(-2,3) , then the orthocentre of triangle OAC is (give two ways of solution 1- using circumcentre ,centroid and other centres.  2-using slope)

Arun
25750 Points
5 years ago
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)