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 [TEX]\overline{AB}~\implies~M_{AB}(3,4)[/TEX]

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

[TEX]2r = \sqrt{8^2+6^2}=10~\implies~r = 5[/TEX]

3. Equation of the circle is:

[TEX](x-3)^2+(y-4)^2 = 25[/TEX]

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). [TEX]|\overline{AB}| = 10[/TEX]

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

[TEX]k^2 = p^2+q^2 = x^2+(8-y)^2[/TEX]

[TEX]j^2=s^2+t^2 = y^2+(x-6)^2[/TEX]

3. Since

[TEX]k^2+j^2=100[/TEX] ... you'll get

[TEX]x^2+(8-y)^2 + y^2+(x-6)^2 = 100 [/TEX] ... expand the brackets:

[TEX]x^2+64-16y+y^2 + y^2+x^2-12x+36 = 100 [/TEX]

[TEX]2x^2-12x+ 2y^2-16y = 0 [/TEX]

[TEX]x^2-6x+ y^2-8y = 0 [/TEX] ... complete the squares:

[TEX]x^2-6x \color{red}{+ 9}+ y^2-8y\color{red}{+ 16} = 0\color{red}{+9+16} [/TEX]

[TEX](x-3)^2+(y-4)^2=5^2[/TEX]

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