Equation of a Line in Isosceles Triangle

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In summary, the problem is asking for the y-coordinate of point P in terms of x, given an isosceles right triangle with a hypotenuse of length 2 units and points C(1,0), A(1,0), and B(0,y). Using the Pythagorean theorem, the length of AB can be found to be \sqrt{2}, and the length of OB can be found to be \sqrt{2}-1. This information can be used to solve for the y-coordinate of B, and in turn, the y-coordinate of P.
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Jan Hill
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Homework Statement



Given an right angled isosceles triangle with hypoteneuse of 2 units. The hypoteneuse lies on the x-axis and is 2 units long. The points of the triangle are in terms of x and y as follows: C-1,0), A(1,0)and B(0,y). Inscribed in the triangle is a rectangle which has one point on the line AB of the triangle at point P(x, ?) I have to express the y co-ordinate of P in terms of x

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The Attempt at a Solution


I would like to find the equation of the line AB but I don't have the slope. I only have the one point A(1,0) but I think the isosceles nature of the triangle is the key to this. I don't know how that info though.
 
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  • #2
I didn't quite understand what you meant past the description of the triangle, but it looks like you just need to solve for the coordinate of the point B to help you figure it out.

First, let O be the origin (0, 0).

What is the distance of the line segment AB? (Hint, use the pythagorean theorem and the fact that the length of AB equal the length of BC).

Note that BOA defines another right triangle. Assuming you know the length of AB and the length of OA, you can use the pythagorean theorem again to find OB (and thus, the y-coordinate of B).

Hope that helps.
 
  • #3
The line from A to B and the line from C to B have the same length: x. Then, by the Pythagorean theorem, [itex]x^2+ x^2= 2x^2= 4[/itex] so that [itex]x= \sqrt{2}[/itex]. Now you know that the line from (0, 0) to (0, y) is a leg in a right triangle with the other leg of length 1 ((0,0) to A) and hypotenuse of length [itex]\sqrt{2}[/itex]. The Pythagorean theorem, again, \(\displaystyle y^2+ 1= 2\). That gives you point B.
 

FAQ: Equation of a Line in Isosceles Triangle

What is the equation of the line in an isosceles triangle?

The equation of a line in an isosceles triangle can be written in the form y = mx + b, where m is the slope of the line and b is the y-intercept. This equation describes the relationship between the x and y coordinates of points on the line.

How do you find the slope of a line in an isosceles triangle?

The slope of a line in an isosceles triangle can be found by using the formula m = (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are two points on the line. This formula calculates the change in y over the change in x, or rise over run.

What is the y-intercept in the equation of a line in an isosceles triangle?

The y-intercept in the equation of a line in an isosceles triangle is the point where the line crosses the y-axis. It is represented by the value b in the equation y = mx + b. The y-intercept is determined by the initial position of the line on the y-axis.

Can the equation of a line in an isosceles triangle be written in standard form?

Yes, the equation of a line in an isosceles triangle can be written in standard form, which is Ax + By = C. However, this form is not commonly used for lines in isosceles triangles because it does not explicitly show the slope and y-intercept.

How does the equation of a line in an isosceles triangle relate to the triangle's angles?

The slope of a line in an isosceles triangle is related to the triangle's angles through the tangent ratio, which is equal to the slope. The y-intercept is also related to the triangle's angles, as it represents the initial position of the line on the y-axis. The exact relationship depends on the specific values of the triangle's angles and the position of the line in relation to the triangle.

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