Finding the equation of vertex of right angle triangle

[hb2325]

Points (6,0) and (O,8) are the endpoints of the hypotenuse of a
right-angled triangle, whose other vertex is at (x,y). What equation relates x
and y?Attempt at solution:

So What I am thinking is that the line will be perpendicular to the line with end points 6,0 and 0,8.

So The gradient of this line is 4/3, hence the gradient of the other line perpendicular to this one will be -3/4

So the equation relating X and y is y = -3/4 x + c ?

Now can I assume that given line is a perpendicular bisector to the hypotenuse and so interesects it at 3,4 (mid point) and hence substitue for c - which i get to 25/4

So y = -3/4 x + 25/4 , This kinda feels wrong and I would appreciate input.

Thanks.

 

[earboth]

Points A(6,0) and B(O,8) are the endpoints of the hypotenuse of a
right-angled triangle, whose other vertex is at (x,y). What equation relates x
and y?Attempt at solution:

So What I am thinking is that the line will be perpendicular to the line with end points 6,0 and 0,8.

So The gradient of this line is 4/3, hence the gradient of the other line perpendicular to this one will be -3/4

So the equation relating X and y is y = -3/4 x + c ?

Now can I assume that given line is a perpendicular bisector to the hypotenuse and so interesects it at 3,4 (mid point) and hence substitue for c - which i get to 25/4

So y = -3/4 x + 25/4 , This kinda feels wrong and I would appreciate input.

Thanks.

The locus of all vertices of the right angle is a circle whose diameter is the line segment AB.

1. Midpoint of $$\overline{AB}~\implies~M_{AB}(3,4)$$

2. Length of the radius. Use Pythagorean theorem to determine the diameter:

$$2r = \sqrt{8^2+6^2}=10~\implies~r = 5$$

3. Equation of the circle is:

$$(x-3)^2+(y-4)^2 = 25$$

EDIT: See following post, please!

 

[earboth]

Points A(6,0) and B(O,8) are the endpoints of the hypotenuse of a
right-angled triangle, whose other vertex is at (x,y). What equation relates x
and y?Attempt at solution:

So What I am thinking is that the line will be perpendicular to the line with end points 6,0 and 0,8.

So The gradient of this line is 4/3, hence the gradient of the other line perpendicular to this one will be -3/4

So the equation relating X and y is y = -3/4 x + c ?

Now can I assume that given line is a perpendicular bisector to the hypotenuse and so interesects it at 3,4 (mid point) and hence substitue for c - which i get to 25/4

So y = -3/4 x + 25/4 , This kinda feels wrong and I would appreciate input.

Thanks.

1. Draw a right triangle with AB as hypotenuse. The vertex of the right angle has the coordinates (x, y). $$|\overline{AB}| = 10$$

2. Use Pythagorean theorem to calculate the lengthes of the legs k and j of the right triangle:

$$k^2 = p^2+q^2 = x^2+(8-y)^2$$

$$j^2=s^2+t^2 = y^2+(x-6)^2$$

3. Since

$$k^2+j^2=100$$ ... you'll get

$$x^2+(8-y)^2 + y^2+(x-6)^2 = 100$$ ... expand the brackets:

$$x^2+64-16y+y^2 + y^2+x^2-12x+36 = 100$$

$$2x^2-12x+ 2y^2-16y = 0$$

$$x^2-6x+ y^2-8y = 0$$ ... complete the squares:

$$x^2-6x \color{red}{+ 9}+ y^2-8y\color{red}{+ 16} = 0\color{red}{+9+16}$$

$$(x-3)^2+(y-4)^2=5^2$$

This is the equation of a circle around (3,4) with radius r = 5