1. ## equidistant points

Our topic is about distance formulas:

1.) If the point $\displaystyle (x,3)$ is equidistant from $\displaystyle (3,-2)$ and $\displaystyle (7,4)$ find $\displaystyle x$

2.) Find the point on the y - axis which is equidistant from $\displaystyle (-5,-2) and (3,2)$

thank you very much

2. ## nr. 1.) only

Our topic is about distance formulas:

1.) If the point $\displaystyle (x,3)$ is equidistant from $\displaystyle (3,-2)$ and $\displaystyle (7,4)$ find $\displaystyle x$...

the point you are looking for belongs to the perpendicular bisector of the two given points and the straight line parallel to the x-axis with y = 3.

1. Calculate the midpoint between the given points: M(5, 1)
2. Calculate the slope of the line connecting the given points:
$\displaystyle m=\frac{4-(-2)}{7-3}=\frac{3}{2}$

Therefore the perpendicular direction has the slope -2/3

3. Use point-slope-formula of a straight line:

$\displaystyle \frac{y-1}{x-5}=-\frac{2}{3} \ \text{and} \ y=3$

4. Plug in this value for y and solve for x. You should get x = 2

So the point you are looking for is P(2, 3)

EB

PS.: The 2nd problem is very similar to this one here. I leave for you.

Our topic is about distance formulas. . Really?

Here's an idea . . . How about using the Distance Formula?

. . . . . $\displaystyle d \;=\;\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}$
1) If the point $\displaystyle P(x,3)$ is equidistant from $\displaystyle A(3,-2)$ and $\displaystyle B(7,4)$ find $\displaystyle x$

Distance $\displaystyle \overline{PA} \:=\:\sqrt{(x-3)^2 + (3-[-2])^2} \:=\:\sqrt{x^2-6x + 34}$

Distance $\displaystyle \overline{PB}\:=\:\sqrt{(x-7)^2 + (3-4)^2}\:=\:\sqrt{x^2-14x + 50}$

Since $\displaystyle \overline{PA} = \overline{PB}$, we have: .$\displaystyle \sqrt{x^2-6x+34} \:=\:\sqrt{x^2-14x+50}$

Square both sides: .$\displaystyle x^2 - 6x + 34 \:=\:x^2-14x + 50$

And we have: .$\displaystyle 8x \:=\:16\quad\Rightarrow\quad \boxed{x\,=\,2}$

2) Find the point on the y - axis which is equidistant from $\displaystyle A(-5,-2)$ and $\displaystyle B(3,2)$

A point on the y-axis has the form: $\displaystyle P(0,\,y)$

Then: .$\displaystyle \begin{array}{cc}\overline{PA} \:= \:\sqrt{(0+5)^2+(y+2)^2} \:= \:\sqrt{y^2+4y+29} \\ \overline{PB} \:= \:\sqrt{(0-3)^2 + (y-2)^2} \:= \:\sqrt{y^2 - 4y + 13} \end{array}$

Hence: .$\displaystyle \sqrt{y^2 + 4y + 29} \:=\:\sqrt{y^2 - 4y + 13}\quad\Rightarrow\quad y^2 + 4y + 29 \:=\:y^2 - 4y + 13$

Therefore: .$\displaystyle 8y\:=\:-16\quad\Rightarrow\quad y \:=\:-2$ . . The point is $\displaystyle \boxed{(0,\,-2)}$