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The ends of the hypotenuse of a right angled triangle are (0,6) and (6,0). Find the equation of
the locus of its third vertex.
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Let ABC is a right angled triangle whose hypotenuse is AC and angle B=90°. co ordinate A (6,0) , C (0,6).
Let co ordinate of B (h,k).
Slope of BA (m1)=(0-k)/(6-h)
Slope of BC (m2)=(6-k)/(0-h)
Angle between BA and BC is 90°
m1×m2= -1
(-k)/(6-h)×(6-k)/(-h)= -1
(6k-k^2)/(6h-h^2)=-1
-6h+h^2=6k-k^2
h^2+k^2–6h-6k=0 , Therefore locus of (h,k) is:-
x^2+y^2–6x-6y=0 , answer
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