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.


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Answer 1

Answer:

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

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Answer 2

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Given triangle is right angled triangle.

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

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

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

we know:

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

Therefore, we get:

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

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

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

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

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

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

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