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Angelina_Jolie [31]
3 years ago
6

Find x so that the points (x,x+1), (x+2,x+3) and (x+3,2x+4) form a right-angled triangle.

Mathematics
1 answer:
azamat3 years ago
3 0

Let <em>a</em>, <em>b</em>, and <em>c</em> be vectors each starting at the origin and terminating at the points (<em>x</em>, <em>x</em> + 1), (<em>x</em> + 2, <em>x</em> + 3), and (<em>x</em> + 3, 2<em>x</em> + 4), respectively.

Then the vectors <em>a</em> - <em>b</em>, <em>a</em> - <em>c</em>, and <em>b</em> - <em>c</em> are vectors that point in directions parallel to each of the legs formed by the triangle with these points as its vertices.

If this triangle is to contain a right angle, then exactly one of these pairs of vectors must be orthogonal. In other words, one of the following must be true:

(<em>a</em> - <em>b</em>) • (<em>a</em> - <em>c</em>) = 0

<em>or</em>

(<em>a</em> - <em>b</em>) • (<em>b</em> - <em>c</em>) = 0

<em>or</em>

(<em>a</em> - <em>c</em>) • (<em>b</em> - <em>c</em>) = 0

We have

<em>a</em> - <em>b</em> = (<em>x</em>, <em>x</em> + 1) - (<em>x</em> + 2, <em>x</em> + 3) = (-2, -2)

<em>a</em> - <em>c</em> = (<em>x</em>, <em>x</em> + 1) - (<em>x</em> + 3, 2<em>x</em> + 4) = (-3, -<em>x</em> - 3)

<em>b</em> - <em>c</em> = (<em>x</em> + 2, <em>x</em> + 3) - (<em>x</em> + 3, 2<em>x</em> + 4) = (-1, -<em>x</em> - 1)

Case 1: If (<em>a</em> - <em>b</em>) • (<em>a</em> - <em>c</em>) = 0, then

(-2, -2) • (-3, -<em>x</em> - 3) = (-2)×(-3) + (-2)×(-<em>x</em> - 3) = 2<em>x</em> + 12 = 0   ==>   <em>x</em> = -6

which would make <em>a</em> - <em>c</em> = (-3, 3) and <em>b</em> - <em>c</em> = (-1, 5), and their dot product is not zero. Then the triangles vertices are at the points (-6, -5), (-4, -3), and (-3, -8).

Case 2: If (<em>a</em> - <em>b</em>) • (<em>b</em> - <em>c</em>) = 0, then

(-2, -2) • (-1, -<em>x</em> - 1) = (-2)×(-1) + (-2)×(-<em>x</em> - 1) = 2<em>x</em> + 4 = 0   ==>   <em>x</em> = -2

which would make <em>a</em> - <em>c</em> = (-3, -1) and <em>b</em> = (-1, 1), and their dot product is also not zero. The vertices are the points (-2, -1), (0, 1), and (1, 0).

Case 3: If (<em>a</em> - <em>c</em>) • (<em>b</em> - <em>c</em>) = 0, then

(-3, -<em>x</em> - 3) • (-1, -<em>x</em> - 1) = (-3)×(-1) + (-<em>x</em> - 3)×(-<em>x</em> - 1) = <em>x</em> ² + 4<em>x</em> + 6 = 0

but the solutions to <em>x</em> here are non-real, so we throw out this case.

So there are two possible values of <em>x</em> that make a right triangle, <em>x</em> = -6 and <em>x</em> = -2.

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