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Coordinates of Right Triangles


Date: 06/25/97 at 15:27:15
From: Paige Wade
Subject: Right triangle

Find all possible values of k so that (-1,2), (-10,5), and (-4,k) are 
the vertices of a right triangle.

I don't even know where to start on this one.  I tried substituting 0 
for k into the distance formula, but I don't get a logical answer. Do 
I just leave k as k in the formula, or use another one?  I'm stumped.

Thanks,
Paige


Date: 06/28/97 at 15:22:40
From: Doctor Mark
Subject: Re: Right triangle

Paige, 

This is not an easy problem.

Here's one way to think about it. Let's call the points A (-1,2),   
B (-10,5), and C (-4, k). If these three points are to make a right 
triangle, then that triangle must have a hypotenuse. There are only 
three possibilities for the hypotenuse: the line AB, the line BC, or 
the line AC.

  If AB is the hypotenuse, then AC and CB are the legs.

  If BC is the hypotenuse, then AB and AC are the legs.

  If AC is the hypotenuse, then AB and BC are the legs.

In order for these sides to make up a right triangle, the sides of the 
triangle must satisfy the Pythagorean theorem. Since you know the 
coordinates of the points A, B, and C, you can use the distance 
formula (actually, the square of the distance, since that is what 
enters into the Pythagorean theorem) to write three sets of equations 
(one for each of the possibilities), then solve those equations for k.  
I'll do one of them and leave the two others to you:

Let AC be the hypotenuse.  Then: 

(AB)^2 is (-1-(-10))^2 +(2-5)^2 
(BC)^2 is (-10-(-4))^2 + (5-k)^2
(AC)^2 is (-1-(-4))^2 + ((2-k)^2

Use the Pythagorean formula for this triangle to find that k = 23.

In the second case, you again get a linear equation for k, and find 
that k = -7.

In the third case, you get a quadratic equation for k, and find that 
k = 8 or k = -1.

So the possible values of k are -1, -7, 8, or 23.  

Another approach would be to say that since the triangle is supposed 
to be a right triangle, the legs should be perpendicular. Since you 
can find the slopes of each of the lines which make up those legs, the 
demand that the lines be perpendicular (that the product of their 
slopes be -1) would again give you equations to solve for k.   

Altogether, an interesting problem.

-Doctor Mark,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Coordinate Plane Geometry
High School Geometry

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