An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from $(7 ,1 )$ to $(2 ,9 )$ and the triangle's area is $32$ , what are the possible coordinates of the triangle's third corner?

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Answer 1 · 2018-04-25T16:16:51

$(1825/178, 765/89) or( -223/178, 125/89)$

We relabel in standard notation: $b=c$ , $A(x,y)$ , $B(7,1),$ $C(2,9)$ . We have $text{area}=32$ .

The base of our isosceles triangle is $BC$ . We have

$a=|BC| =sqrt{5^2+8^2}=sqrt{89}$

The midpoint of $BC$ is $D=((7+2)/2, (1+9)/2) = (9/2, 5)$ . $BC$ 's perpendicular bisector goes through $D$ and vertex $A$ .

$h=AD$ is an altitude, which we get from the area:

$32 = \frac 1 2 a h = 1/2 \sqrt{89}\ h$

$h = 64/sqrt{89}$

The direction vector from $B$ to $C$ is

$C-B=(2-7,9-1)=(-5,8)$ .

The direction vector of its perpendiculars is $P=(8,5)$ , swapping the coordinates and negating one. Its magnitude must also be $|P|=sqrt{89}$ .

We need to go $h$ in either direction. The idea is:

$A = D \pm h P /|P|$

$A =(9/2,5) \pm (64/sqrt{89}) {(8,5)} / \sqrt{89}$

$A =(9/2,5) \pm 64/89 (8,5)$

$A = (9/2 +{ 8 (64)}/89, 5 + {5(64)}/89 ) or$ $A = (9/2 - { 8 (64)}/89 , 5 - {5(64)}/89)$

$A=(1825/178, 765/89) or A=( -223/178, 125/89)$

That's a bit messy. Is it right? Let's ask Alpha.

enter image source here

Great! Alpha verifies its isosceles and the area is $32.$ The other $A$ is right too.

An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (7 ,1 )(7 ,1 ) to (2 ,9 )(2 ,9 ) and the triangle's area is 32 32 , what are the possible coordinates of the triangle's third corner?