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.

  • $\begingroup$ By using circle with centre A and radius AB, I could find a point C on that circle such that AC = AB = radius. I was then hoping to find another equation to solve simultaneously with the circle equation to find C $\endgroup$ – Jack Apr 21 '15 at 13:56
  • $\begingroup$ @A.P.: Circles are the way to go if you want the length of AC to be equal to that of AB, or the length of BC to be equal to that of AB. You only get a straight line if you want the length of AC to be equal to that of BC. $\endgroup$ – Henry Apr 21 '15 at 14:05
  • $\begingroup$ @Henry Ah, I see what you mean. Still, it seems overly complicated to use quadratic equations when you could use linear equations... $\endgroup$ – A.P. Apr 21 '15 at 14:09

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 .

  • $\begingroup$ That's exactly it - thanks for the help $\endgroup$ – Jack Apr 21 '15 at 14:09
  • $\begingroup$ AB has a positive slope and D is perpendicular to it. Therefore, slope of D must be negative. $\endgroup$ – Mick Apr 22 '15 at 5:33
  • $\begingroup$ You are right @Mick Thank you , I will correct it . $\endgroup$ – Nizar Apr 22 '15 at 6:01

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?

  • $\begingroup$ You could tell that ABC is an isosceles triangle... So are you suggesting I find the mid-point of AB (call it M), make an equation for line L that is perpendicular to AB and passes through M, and use this to solve for C where MC is the height of the triangle and AB its base? $\endgroup$ – Jack Apr 21 '15 at 14:04
  • $\begingroup$ @Jack Yes, exactly! $\endgroup$ – A.P. Apr 21 '15 at 14:04
  • $\begingroup$ Excellent - thanks very much! :) $\endgroup$ – Jack Apr 21 '15 at 14:05

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.

enter image description here

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 $(x + 0.5)^2 + y^2 = 4^2$

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.

  • $\begingroup$ Nice answer! By the way, what did you use to make those diagrams? $\endgroup$ – A.P. Apr 22 '15 at 7:20
  • $\begingroup$ @A.P. I use Geogebra to make a rough sketch first. Copy it to the clipboard. Past the clipboard image to WORD. After re-sizing, copy the image to PC paintbrush to do the finishing touch. $\endgroup$ – Mick Apr 22 '15 at 8:47

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 $(C_{00},C_{01},C_{02},C_{11},C^{\prime}_{02})$:

enter image description here

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 $y>0$.


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