In a right triangle, the lengths of the segments connecting points of trisection of the hypotenuse to the vertex of the right triangle are 7 and 9. how long is the hypotenuse if the triangle?

  3√26

Explanation:

Suppose the right-angle vertex of the triangle is at coordinates (9, 0) and the end of the 9-unit segment at the hypotenuse is at coordinates (0, 0). Further suppose that the midpoint of the hypotenuse is at (x, y).

Then the distance from (x, y) to (9, 0) is 3 times the distance from (x, y) to (0, 0), and the point (2x, 2y) is on the circle of radius 7 centered at (9, 0).

These relationships give rise to two equations:

The first equation can be simplified to ...

  9x^2 +9y^2 = x^2 -18x +81 + y^2

  8x^2 +8y^2 = -18x +81 . . . . . . . . [eq A]

And the second equation expands to ...

  4x^2 -36x +81 + 4y^2 = 49

  4x^2 +4y^2 =36x -32 . . . . . . . . . [eq B]

Subtracting twice [eq B] from [eq A], we get ...

  (8x^2 +8y^2) -2(4x^2 +4y^2) = (-18x +81) -2(36x -32)

  0 = -90x +145 . . . .  simplify

  x = 29/18 . . . . . . . . .solve for x

Using [eq B], we can find the distance from the origin to (x, y) and the length of the hypotenuse, which is 6 times that.

  4(x^2 +y^2) = 36(29/18) -32) = 26 . . . . . substituting x into [eq B]

  2√(x^2 +y^2) = √26 . . . . . . . . . . . . . . . . square root

  6√(x^2 +y^2) = 3√26 . . . . . . . . . . . . . . . 6 times (0, 0) to (x, y)

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This solution relies on the fact that a right triangle can be inscribed in a semicircle, so the point where the 7- and 9-unit segments come together, the right angle, is on a circle whose radius is half the length of the hypotenuse. The point halfway between the other ends of those segments, the center of the hypotenuse, is the point we've written equations for.

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In the diagram, point D is (x, y), and point C is (2x, 2y) on the circle of radius 7 centered at A.

is only a line? i think so

2x^2+5x+17

your welcome. : )