Finding Multiple Vertices of Right Triangles on a Given Line

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turophile

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Here's the statement of the problem:

Two of the vertices of a right triangle are (1, – 2) and (– 1, 0). Find x if the third vertex is (x, – 1).

I sketched the problem, and see that there are four right triangles that have a vertex on the line y = - 1. I found two of them so far:

Let l_1 be the line that includes (1, – 2) and (– 1, 0), and let m_1 be its slope. Then m_1 = (0 – (– 2))/(– 1 – 1) = 2/(– 2) = – 1, and the equation of l_1 is y = – x + x_0 ? – 2 = – 1 + x_0 ? x_0 = – 2 + 1 = – 1 ? y = – x – 1 ? y = – x – 1 ? x + y + 1 = 0.

Let l_2 be the line that includes (1, – 2) and (x, - 1) perpendicular to l_1, and let m_2 be its slope. Then m_2 = 1, and the equation of l_2 is y = x + x_0 ? – 2 = 1 + x_0 ? x_0 = – 2 – 1 = – 3 ? y = x – 3 ? - 1 = x – 3 ? x = 2.

Let l_3 be the line that includes (– 1, 0) and (x, - 1) perpendicular to l_1, and let m_3 be its slope. Then m_3 = 1, and the equation of l_3 is y = x + x_0 ? 0 = – 1 + x_0 ? x_0 = 1 ? y = x + 1 ? - 1 = x + 1 ? x = - 1 – 1 = - 2.

So two of the triangles have vertices at (- 2, - 1) and (2, - 1). I'm having more trouble finding with the other two vertices. I can see them on the graph, but I'm not sure how to set up the problem to solve for them. I am trying the Pythagorean Theorem but am getting stuck:

Let T = (1, – 2), U = (– 1, 0), and V = (x, – 1), and let TU be the hypotenuse of square triangle TUV. Then (d_TU)^2 = (d_TV)^2 + (d_UV)^2 ? sqrt((1 – (– 1))^2 + (– 2 – 0)^2) = sqrt((1 – x)^2 + (– 2 – (– 1))^2) + sqrt((– 1 – x)^2 + (0 – (– 1))^2) ? sqrt(2^2 + (– 2)^2) = sqrt(1 – 2x + x^2 + (– 1)^2) + sqrt(1 + 2x + x^2 + 1^2) ? sqrt(4 + 4) = sqrt(x^2 – 2x + 2) + sqrt(x^2 + 2x + 2) ? sqrt(8) = sqrt(x^2 – 2x + 2) + sqrt(x^2 + 2x + 2) ? ?
 
turophile said:
Here's the statement of the problem:

Two of the vertices of a right triangle are (1, – 2) and (– 1, 0). Find x if the third vertex is (x, – 1).

I sketched the problem, and see that there are four right triangles that have a vertex on the line y = - 1. I found two of them so far:

Let l_1 be the line that includes (1, – 2) and (– 1, 0), and let m_1 be its slope. Then m_1 = (0 – (– 2))/(– 1 – 1) = 2/(– 2) = – 1, and the equation of l_1 is y = – x + x_0 ? – 2 = – 1 + x_0 ? x_0 = – 2 + 1 = – 1 ? y = – x – 1 ? y = – x – 1 ? x + y + 1 = 0.

Let l_2 be the line that includes (1, – 2) and (x, - 1) perpendicular to l_1, and let m_2 be its slope. Then m_2 = 1, and the equation of l_2 is y = x + x_0 ? – 2 = 1 + x_0 ? x_0 = – 2 – 1 = – 3 ? y = x – 3 ? - 1 = x – 3 ? x = 2.

Let l_3 be the line that includes (– 1, 0) and (x, - 1) perpendicular to l_1, and let m_3 be its slope. Then m_3 = 1, and the equation of l_3 is y = x + x_0 ? 0 = – 1 + x_0 ? x_0 = 1 ? y = x + 1 ? - 1 = x + 1 ? x = - 1 – 1 = - 2.

So two of the triangles have vertices at (- 2, - 1) and (2, - 1). I'm having more trouble finding with the other two vertices. I can see them on the graph, but I'm not sure how to set up the problem to solve for them. I am trying the Pythagorean Theorem but am getting stuck:

Let T = (1, – 2), U = (– 1, 0), and V = (x, – 1), and let TU be the hypotenuse of square triangle TUV. Then (d_TU)^2 = (d_TV)^2 + (d_UV)^2 ? sqrt((1 – (– 1))^2 + (– 2 – 0)^2) = sqrt((1 – x)^2 + (– 2 – (– 1))^2) + sqrt((– 1 – x)^2 + (0 – (– 1))^2) ? sqrt(2^2 + (– 2)^2) = sqrt(1 – 2x + x^2 + (– 1)^2) + sqrt(1 + 2x + x^2 + 1^2) ? sqrt(4 + 4) = sqrt(x^2 – 2x + 2) + sqrt(x^2 + 2x + 2) ? sqrt(8) = sqrt(x^2 – 2x + 2) + sqrt(x^2 + 2x + 2) ? ?
Click to expand...

Assuming your work upto this point is correct:

\(\displaystyle \sqrt{(x+1)^2+1} \ \ = \ \ \sqrt{8} - \sqrt{(x-1)^2+1}\)

square both sides

\(\displaystyle (x+1)^2+1 \ \ = \ \ \ 8 + (x-1)^2+1 - 2\cdot \sqrt{8 \cdot (x-1)^2+8}\)

\(\displaystyle 8 - 4x \ \ = \ \ 4\cdot \sqrt{2 \cdot (x-1)^2+2}\)

Square both sides again - simplify and solve (and check for extraneous solution/s)
 
Thanks for the hint. I see my mistake. I think this is the solution for the two remaining vertices:

Let T = (1, – 2), U = (– 1, 0), and V = (x, – 1), and let TU be the hypotenuse of square triangle TUV. Then (d_TU)^2 = (d_TV)^2 + (d_UV)^2 ? (1 – (– 1))^2 + (– 2 – 0)^2 = (1 – x)^2 + (– 2 – (– 1))^2 + (– 1 – x)^2 + (0 – (– 1))^2 ? 4 + 4 = 1 – 2x + x^2 + (– 1)^2 + 1 + 2x + x^2 + 1^2 ? 8 = 1 – 2x + x^2 + 1 + 1 + 2x + x^2 + 1 ? 2x^2 + 4 – 8 = 0 ? 2x^2 = 4 ? x^2 = 2 ? x = ± sqrt(2).

So x = 2, - 2, sqrt(2), or - sqrt(2), which is what my sketch suggested.
 
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