A(-1,4) and B(7,5) are two ends of the diameter of a circle . Find the equation of the locus of point C such that <ACB is a right angle .
Thanks so much !
hi
OR
Use the phythagoras theorem . Let c be (x,y)
$\displaystyle (x+1)^2+(4-y)^2+(7-x)^2+(5-y)^2=65$
then simplify it , you will find the locus of c to be a circle because the point C can be any point on the circumference since the angle in the semicircle is a right angle .