Finding Coordinates for a Triangle

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To find the coordinates of the third vertex of an isosceles triangle when two corners and the height are known, it's essential to clarify whether the known length refers to a side or the base. If the base is provided, two solutions exist for the third vertex, located symmetrically above and below the base. The midpoint of the base can be calculated, and the slope of the altitude is the negative reciprocal of the base's slope. By setting the distance from the midpoint to the third vertex equal to the height, a quadratic equation can be formed to find the coordinates. This method allows for solving the problem effectively using the relationships between the coordinates and the triangle's properties.
powp
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Hello All,

How do you find one set of corrdinates of a trangle when you know two corners and the lengths of the sides?

Is it possible?? It seems that there would be two solutions.

Thanks

P
 
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do you mean "side" or "sides"? since it makes a big difference here.

if you mean "side", and if you mean some characteristic triangle (isoscoles, right, etc.), then yes there will be 2 solutions.

if you mean "sides" then there will clearly be one unique solution of finding the third point.

please clarify you question so that we can help you where you are having a hard time.
 
I know the length between the two point and I know the height of the triange and since it is a isoscoles trianges I can find the other sides.
 
powp said:
I know the length between the two point and I know the height of the triange and since it is a isoscoles trianges I can find the other sides.

In that case, the problem depends strongly upon whether the side you are given is the base.

If you are given the coordinates of the two end points of the base, then, of course, you could calculate the length but you don't really need that. I am going to assume that you are given the coordinates of the two points of the base,(x0,y0) and (x1,y1), and the height of an isosceles triangle.

The altitude of an isosceles triangle passes through the center of the base and is perpendicular to it. Knowing (x0,y0) and (x1,y1), you can find the slope of the line through those two points: \frac{y_1-y_0}{x_1-x_0). The slope of the line on which the altitude lies is negative reciprocal of that: \frac{x_0-x_1}{y_1-y_0}. Of course, the midpoint of the base is \(\frac{x_0+x_1}{2},\frac{y_0+y_1}{2}\). The equation of the line through that midpoint having that slope gives you an equation connecting x and y for any point on that line. Use it to make the formula for distance from the midpoint equal to the given height a single quadratic equation for x (or y) and solve.

Yes, there will be two solutions on opposite sides of the base.
 
I am not getting this. Can you please help a bit more. I have two unknown and one equation who do I solve?
 
You don't have two unknowns. The distance from (x,y) to the midpoint equals the given height- that's you one equation. However, you also have the equation of the line (x,y) lies on. That is y= mx+ b for m and b you can calculate. Replace y by mx+b and solve for x. After you find x, you can calculate y= mx+b.
 
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