Question

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.

Answer 1

Answer:

x²+y²-6x-6y

Step-by-step explanation:

let the vertex c be (x, y)

the ends of hypotenuse of a right angle triangle

<apb=90

thn 72=x²+y²-6x-6y

Answer 2

Given triangle is right angled triangle.

Given ends of hypotenuse are: $(0,6), (6,0).$

length of the hypotenuse = $\sqrt{(0-6)^{2}+(6-0)^{2} } =\sqrt{36+36} =\sqrt{72}$

let the third vertex be: $(x,y)$

we know:

$hyp^{2} =opp^{2} +adj^{2}$

Therefore, we get:

$72=(x-0)^{2} +(y-6)^{2} +(x-6)^{2} +(y-0)^{2}$

$72=x^{2} +(y-6)^{2} +(x-6)^{2} +y^{2}$

$72=x^{2} +y^{2}+36-12y+x^{2} +36-12x+y^{2}$

$72=2x^{2} +2y^{2}+72 -12x-12y$

$2x^{2} +2y^{2}-12x-12y=0$

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

Therefore, The equation of the locus of its third vertex is: $x^{2} +y^{2} -6x-6y=0.$