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.

 

[Mentallic]

The vertex definitely doesn't have to be perpendicular to the hypotenuse (and would this perpendicular line segment be going through the point (6,0) or (0,8)?).

You can have the two extra sides being parallel to the x and y-axis respectively, or you can flip the triangle around so the last point is on the other end of the hypotenuse.

So one of the criteria is that the two sides need to be perpendicular to each other (but not necessarily to the hypotenuse). So let's start with that. What equations can we form with the points (6,0), (0,8) and (x,y) given that the lines between (6,0), (x,y) and (0,8), (x,y) must be perpendicular to each other.

 

[hb2325]

Ok so gradient of the first line would be y/(x-6) and 2nd line would be y-8/x and their product should be -1? Or am I thinking about it the wrong way?

 

[DonAntonio]

Ok so gradient of the first line would be y/(x-6) and 2nd line would be y-8/x and their product should be -1? Or am I thinking about it the wrong way?

You're thinking it the correct way and you have the answer within reach.

DonAntonio

 

[hb2325]

Thanks So all I do is simplify the equation and get y^2 - 8y + x^2 - 6x = 0 and that's it?

 

[Mentallic]

Thanks So all I do is simplify the equation and get y^2 - 8y + x^2 - 6x = 0 and that's it?

Yes that's correct! But what does this equation represent? Maybe if you were to convert it into a more recognizable form...

Hint: [itex]x^2+ax=\left(x+\frac{a}{2}\right)^2-\frac{a^2}{4}[/itex]

 

[hb2325]

Circle?

 

[Mark44]

Yes, the equation you found is that of a circle whose center is at (3, 4) and whose radius is 5. The two possible points that could be the vertex of the right angle are on that circle.

Note that the circle equation relates the two vertex points, but other points on the circle define triangles that aren't right triangles.

 

[hb2325]

Thanks - so how I do I find the vertex of the right triangle? I'm kinda confused.

 

[Mark44]

Thanks - so how I do I find the vertex of the right triangle? I'm kinda confused.

There are two of them. If you have drawn a sketch of things, their locations are pretty obvious.

 

[hb2325]

Ah cool, I see it now, thanks a lot for your help!