Find third point to make isosceles triangle with a specific area

Question

Using points A(1,2) and B(-2,-2), find a third point, with a positive y-value, that makes ABC an isosceles triangle with area 10 units${^2}$.

I have found AB to be 5 and used this as $r^2$ below..

So using A as the centre of the circle, I have equation $(x-1)^2+(y-2)^2 = 25$

So I'd love to just pick a positive, arbitrary y-value and find x, but I need the triangle's area to be 10. But until I find the third piont, C(x,y), I'm not sure how to find base/height in order to fix the area.

Answer 1

My suggestion is to use the following: Your triangle is isosceles, so the hieght issued from the vertex C is the perpendicular bisector of the base AB. The midpoint of AB has the following coordinates I = ($-\frac{1}{2}$, 0). So the point C pass through the st line (D) passing through I and perpendicular to $(AB)$, and (D) has equation $y=\frac{-3}{4}x-\frac{3}{8}$. Note that $Ar=10= \frac{h\times b}{2}$. $b=base= AB= 5$ , so $h= 4$. So you have to find the point on (D) C=(x,y), such that distance form CI=4units. System Two equqtions two unkown simple to solve i.e. $$y=\frac{-3}{4}x-\frac{3}{8}$$ $$ (x+ \frac{1}{2})^2 + y^2=16 $$ May this help you .

Answer 2

Hint: Consider the line $\ell$ orthogonal to the segment $AB$ and passing through its midpoint. What can you tell about the triangle $\Delta ABC$ when $C$ is a point of $\ell$? Can you use this to find what you're looking for?

Answer 3

This is a tricky question because it did not specify which side should be considered as base.

Separating into 2 different cases, two sets of answers are therefore expected.

Case 1 (figure 1)

AB = 5, and M, the midpoint of AB is at $(\frac {-1}{2}, 0)$

$10 = \frac {5.CM}{2}$ yields $CM = 4$

$CM$ is the perpendicular bisector of $AB$ and its equation is $L: 4y = -3x – \frac {3}{2}$

C is at the intersection of L and J; where J is the circle whose equation is $(x + 0.5)^2 + y^2 = 4^2$

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Case 2 (figure 2)

Let $\angle ABC$ be $\theta$. Then, $10 = \frac {5^2 \sin \theta}{2}$ yields $\theta = \sin ^{-1} (\frac {4}{5}) = 53.13^0$

Also, C is on the circle whose equation is $(x + 2)^2 + (y + 2)^2 = 5^2$

The above two info together will determine the co-ordinate of C explicitly (although the answer is an ugly one).

The co-ordinates of C can also be found from the following facts:-

(1) $\angle ABK = … = \theta$, accidentally.

(2) $\sin \theta$, $\cos \theta$ and $\tan \theta$ are known simple fractions.

(3) KB, BH and HC are therefore can be found in terms of those simple fractions.

Answer 4

Using points $A(1,2)$ and $B(-2,-2)$, find a third point $C(C_x,C_y)$, with $C_y>0$, that makes ABC an isosceles triangle with area $s=10$ units$^2$.

A complete answer would be: there are five such points $(C_{00},C_{01},C_{02},C_{11},C^{\prime}_{02})$:

Let $AB=c,\ AC=b,\ BC=a$ and $h$ is the height from $C$ in the isosceles $\triangle ABC$.

Given the coordinates, we know that $c=5$ and from the formula $s=c h/2$, $h=4$.

Obviously, all third points $C_k$ of triangles with the base $AB$, the height $h$ and the area $s$ will be located on the two parallel lines below and above $AB$, $h$ units apart, we just need to sort out all the points that make any two sides of $\triangle ABC$ the same.

1) The first obvious choice is $a=b=\sqrt{89}/2\approx 4.717$.

Two other cases are symmetric:

2) $a=c=5$, $b$ is to find out,

3) $b=c=5$, $a$ is to find out.

To deal with cases 2) and 3), let's find possible length $a,b$ using a formula for the area in terms of the sides of triangle: \begin{align} s&=1/4\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}. \end{align}

For $a=c$, $b^4-4 c^2 b^2+16 s^2=0$ and we have two positive roots $b_0=2\sqrt{5}$, $b_1=4\sqrt{5}$.

This results in three different shapes of isosceles triangles and ten points:

$C_{00},C_{01},C_{02},C_{11},C_{12}$, $C^{\prime}_{00},C^{\prime}_{01},C^{\prime}_{02},C^{\prime}_{11},C^{\prime}_{12}$.

Five of them $(C_{00},C_{01},C_{02},C_{11},C^{\prime}_{02})$ have $y>0$.