Question

the end of the hypotenuse of aright angled triangle are [0,6] and [6,0]. find the equation of the locus of its third vertex

Answer 1

Answer: Let ABC is a right angled whose hypotenuse

is AC , angle B =90° . A (0,6) , C (6,0) .

let coordinate of 3rd vertex B (h,k) .

slope of BA (m1) =(k-6)/(h-0).

slope of BC (m2) =(k-0)/(h-6).

BA and BC are perpendicular to each other.

m1×m2 = -1

[(k-6)/(h-0) ] ×[(k-0)/(h-6)] = - 1

k^2–6k = -h^2+6h

h^2+k^2–6(h+k) = 0 , therefore locus of (h,k)

x^2+y^2–6(x+y) = 0 . Answer.

(6,0)lie on x axis

(0,6)lie on y axis

∵it is a right angled triangle, its third vertex will be (0 , 0)

It is a isosceles triangle. If we draw a perpendicular from vertex (0 , 0)

to the line joining the vertices (6 , 0) and (0 , 6) , it will bisect the this line.

Hypotenuse is the diameter of the circumcircle. Hence (3, 3) is the

center of the circle.

Equation of the locus

(x−3)2+(y−3)2=32+32  

⟹x2−6x+32+y2−6y+32=32+32

⟹x2+y2−6x−6y=0