39first, let the lines represented by ax2+2hxy+by2 be
y=m1x+c and y=m2x+c.
And then try to work out.
its a fairly standard result.
show that orthocentre of triangle formed by lines ax2+2hxy+by2=0 and lx+my+n=1 is given by x/l =y/m =(a+b)/(am2-2lmh+bl2)
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3 Answers
39
62first, let the lines represented by ax2+2hxy+by2 be
y=m1x and y=m2x
Note: The constant term will be zero because both these have to pass thru the origin!!!
find the mid point of y=m1x with the line lx+my+n=1
And the line perpendicular to it...
There will be similar eqtn for y=m2x. (just by replacing m1 by m2)
Find the point of intersection fo these two lines! :)