Points of Triangle?

[dolina dahani]

Hi there, hope im making question in a right place. So the question is. Determine the point of isosceles triangle ABC if you know that its area P=8, the tip of triangle C(5,5) and its hight C1(1,1) [Its the point AB/2].
So what I have done is that I found length of AB=2*2^1/2 and I found length AC=34^1/2. So I tried to find the point A of AC1C where I tried to use pithegorean theorem so that the AC²=AC1²+C1C². So I replace C1C with its value 4*2^1/2, and then substitute AC with AC1+4*2^1/2 but i get some very weird solutions. The solutions from the book are A(2,0) B(0,2), . Sorry i cant provide you with the pic bcs my phone is so bad that the pics after uploading it here would be very bad. If there is maybe a different way of solving it Thank you alot

 

[Dr.Peterson]

The problem as you state it is incomplete. Apparently you are to find vertices A and B given the area P=8 (in English I'd expect P to be the perimeter, but I assume this is from another language!), the apex C(5,5), and the midpoint C₁ of the base AB (not the height, which is a distance). But it isn't at all clear exactly what you were given. (If it is exactly what I just said, then you couldn't find AB before finding C₁C.)

I can tell you that your work is at least consistent with the solution, and that the one specific mistake I can see is that AC is not equal to AC₁ + C₁C, as you seem to be saying, but rather, just as you had just said, by the Pythagorean theorem: AC² = (AC₁)² + (C₁C)².

So please state the problem exactly as given to you, translated completely.

 

[dolina dahani]

The problem as you state it is incomplete. Apparently you are to find vertices A and B given the area P=8 (in English I'd expect P to be the perimeter, but I assume this is from another language!), the apex C(5,5), and the midpoint C₁ of the base AB (not the height, which is a distance). But it isn't at all clear exactly what you were given. (If it is exactly what I just said, then you couldn't find AB before finding C₁C.)

I can tell you that your work is at least consistent with the solution, and that the one specific mistake I can see is that AC is not equal to AC₁ + C₁C, as you seem to be saying, but rather, just as you had just said, by the Pythagorean theorem: AC² = (AC₁)² + (C₁C)².

So please state the problem exactly as given to you, translated completely.

Ohh yeah sorry for the mistake I was looking at my work so the roots and squares cancelled out, sorry yes I meant that . And yeah those are the stuff given to me, Ill try to maybe translate the best that I can . And thanks for the asnwer
The Question:
Determine the foundations(points) of the base of an isosceles triangle ABC if its surface is known to be P = 8, Top C (5,5), and the angle bisector at the top intersects the base at point C1 (1,1)

Its obvious that the angle bisector is actually the hight since it is an isosceles triangle.

 

[Dr.Peterson]

The Question:
Determine the foundations(points) of the base of an isosceles triangle ABC if its surface is known to be P = 8, Top C (5,5), and the angle bisector at the top intersects the base at point C1 (1,1)

Much better!

As you say, we know that C₁C is not only the angle bisector, but also the altitude and the perpendicular bisector of AB. So AC₁C is a right triangle, one leg of which is C₁C, whose length (by the distance formula) is [MATH]4\sqrt{2}[/MATH]. What you haven't explicitly said, but may have done, is to find length AB by solving the area equation, [MATH]\frac{1}{2}(AB)(4\sqrt{2}) = 8[/MATH], so that [MATH]AB = 2\sqrt{2}[/MATH].

You also know that AB is perpendicular to C₁C, whose slope is 1, so the slope of AB is -1. Do you see that?

So we just have to find points A and B that are each [MATH]\sqrt{2}[/MATH] away from C₁ along the line with slope -1. I can think of several ways to do that.

I think this is a lot easier than what you seem to have been doing, using the distance from C rather than from C₁.

 

[dolina dahani]

Much better!

As you say, we know that C₁C is not only the angle bisector, but also the altitude and the perpendicular bisector of AB. So AC₁C is a right triangle, one leg of which is C₁C, whose length (by the distance formula) is [MATH]4\sqrt{2}[/MATH]. What you haven't explicitly said, but may have done, is to find length AB by solving the area equation, [MATH]\frac{1}{2}(AB)(4\sqrt{2}) = 8[/MATH], so that [MATH]AB = 2\sqrt{2}[/MATH].

You also know that AB is perpendicular to C₁C, whose slope is 1, so the slope of AB is -1. Do you see that?

So we just have to find points A and B that are each [MATH]\sqrt{2}[/MATH] away from C₁ along the line with slope -1. I can think of several ways to do that.

I think this is a lot easier than what you seem to have been doing, using the distance from C rather than from C₁.

Thanks, but I'm still unsure what to do. I found from slope of A to C1 that the x=2-y. I can't believe that I'm stuck here. Iv Done much complicated stuff but I can't solve this I'm literally going nuts right now...

 

[Dr.Peterson]

I just thought of it visually. How far is [MATH]\sqrt{2}[/MATH]? It's the diagonal of a unit square. So just follow the line one unit square in each direction.

To do it algebraically, you might intersect [MATH]y = 2-x[/MATH] with the circle [MATH](x-1)^2 + (y-1)^2 = 2[/MATH], which consists of all points [MATH]\sqrt{2}[/MATH] away from (1,1). Replace [MATH]y[/MATH] in the circle with [MATH]2-x[/MATH], and solve for [MATH]x[/MATH].