MHB Find Point in R2: $a(-5,2),b(12,9)$

[Amer]

Let $a(-5,2),b(12,9)$ be two points in $\mathbb{R^2}$ find a point $c\in \mathbb{R^2}$ such that

$ac = bc$ and
$ac \perp bc$

I have a long solution I made two equations ( I considered $c (x,y)$ )

$$ ac = bc $$
$$ \sqrt{ (x +5)^2 + (y-12)^2 } = \sqrt{(x -12)^2 + ( y - 9)^2 }$$

and

$$ \frac{y-9}{x-12} \cdot \frac{y-2}{x+5} = -1 $$

and solve them by substitution

is there any faster solution shorter ?
Thanks in advance.

 

[Opalg]

Let $a(-5,2),b(12,9)$ be two points in $\mathbb{R^2}$ find a point $c\in \mathbb{R^2}$ such that

$ac = bc$ and
$ac \perp bc$

I have a long solution I made two equations ( I considered $c (x,y)$ )

$$ ac = bc $$
$$ \sqrt{ (x +5)^2 + (y-12)^2 } = \sqrt{(x -12)^2 + ( y - 9)^2 }$$

and

$$ \frac{y-9}{x-12} \cdot \frac{y-2}{x+5} = -1 $$

and solve them by substitution

is there any faster solution shorter ?
Thanks in advance.

That solution looks fine (and not really too long). An alternative method would be to draw a diagram and do some geometry.

View attachment 4145​

The set of points equidistant from $a$ and $b$ is the perpendicular bisector of $ab$. So $c$ must lie on that line. In fact, there are two possible positions for $c$, I'll call them $c_1$ and $c_2$.

The other condition, $ac \perp bc$, implies that the points $a,c_1,b,c_2$ are the vertices of a square centred at the midpoint of $ab$, $(3.5,5.5)$. The vector from there to $b$ is $(12,9) - (3.5,5.5) = (8.5,3.5)$. A perpendicular vector of the same length is $\pm(-3.5,8.5)$. So the possible positions for $c$ are $(3.5,5.5) \pm (-3.5,8.5)$. That gives $c_1 = (0,14)$ and $c_2 = $(7,-3)$.

 

[Amer]

Great solution thanks :)