Question

The ends of hypotenuse of a right angled triangle are (0,6) (6,0) find the equation of the locus of its third vertex

Answer 1

The ends of hypotenuse of a right angled triangle are (0,6) (6,0)

The third vertex =(0,0)

The equation,
Y-0/x-0= 6-0/0-0
Y/x = 6
y = 6x

Answer 2

Answer:

Locus of vertex is a circle whose equation is, x2+y2−6x−6y=0.

Step-by-step explanation:

Given:  The ends of hypotenuse of a right angled triangle are (0,6) (6,0)

To Find: Equation of the locus of its third vertex

Solution:

Let assume, ΔABC will be right angled triangle.

whose hypotenuse is AC and angle B=90°.

co-ordinate A (6,0) , C (0,6).

Let co ordinate of B (x, y).

Slope of BA (m1)=(0-y)/(6-x)

Slope of BC (m2)=(6-y)/(0-x)

Angle between BA and BC is 90°

m1×m2= -1

(-y)/(6-x)×(6-y)/(-x)= -1

(6y-$y^{2}$)/(6x-$x^{2}$)=-1

-6x+$x^{2}$=6y-$y^{2}$

$x^{2} +y^{2}-6x-6y=0$,

Therefore locus of (x,y) is:-   $x^{2} + y^{2}$-6x-6y=0

Thus, the equation of the locus of its third vertex is x2 + y2 − 6x − 6y = 0.