equidistant points

**^_^Engineer_Adam^_^** · November 13th, 2006, 10:35 AM

Our topic is about distance formulas:

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

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

thank you very much

**earboth** · November 13th, 2006, 11:09 AM

Originally Posted by ^_^Engineer_Adam^_^

Our topic is about distance formulas:

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

Hello, Engineer Adam,

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:
$m=\frac{4-(-2)}{7-3}=\frac{3}{2}$

Therefore the perpendicular direction has the slope -2/3

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

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

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.

**Soroban** · November 13th, 2006, 12:35 PM

Hello, ^_^Engineer_Adam^_^!

Our topic is about distance formulas. . Really?

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

. . . . . $d \;=\;\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}$

1) If the point $P(x,3)$ is equidistant from $A(3,-2)$ and $B(7,4)$ find $x$

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

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

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

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

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

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

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

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

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

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

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