Chapter 1.8, Problem 43E

To determine

To calculate: The point on y-axis that is equidistant from the points $\left(5,-5\right)$and $\left(1,1\right)$ .

Answer to Problem 43E

The point on y-axis that is equidistant from the points $\left(5,-5\right)$ and $\left(1,1\right)$ is $\left(0,-4\right)$ .

Explanation of Solution

Given information:

The two points $\left(5,-5\right)$ and $\left(1,1\right)$ from which the point on y-axis is equidistant.

Formula used:

Distance formula between two points $\text{X}\left(a_1,b_1\right)$ and $\text{Y}\left(a_2,b_2\right)$ is calculated as,

  $d\left(\text{XY}\right)=\sqrt{{\left(a_2-a_1\right)}^2+{\left(b_2-b_1\right)}^2}$

Calculation:

Consider the given two points $\left(5,-5\right)$ and $\left(1,1\right)$ .

Since, the point that is equidistant from the two points $\left(5,-5\right)$ and $\left(1,1\right)$ is on y-axis , so, its x-coordinate will be zero.

Let us assume the point on y-axis be $\text{P}\left(0,y\right)$ and name the given points as A and B respectively $\text{A}\left(5,-5\right)$ and $\text{B}\left(1,1\right)$ .

It is given that point P is equidistant from the points A and B, so, the distance between P and A is equal to the distance between P and B, which can be mathematically written as,

  $d\left(A,P\right)=d\left(B,P\right)$

Recall the distance formula between two points $\text{X}\left(a_1,b_1\right)$ and $\text{Y}\left(a_2,b_2\right)$ is calculated as,

  $d\left(\text{XY}\right)=\sqrt{{\left(a_2-a_1\right)}^2+{\left(b_2-b_1\right)}^2}$

Apply it,

  $\begin{array}{c}d\left(A,P\right)=d\left(B,P\right)\\ \sqrt{{\left(0-5\right)}^2+{\left(y+5\right)}^2}=\sqrt{{\left(0-1\right)}^2+{\left(y-1\right)}^2}\\ \sqrt{25+y^2+10y+25}=\sqrt{1+y^2-2y+1}\\ \sqrt{y^2+10y+50}=\sqrt{y^2-2y+2}\end{array}$

Square both sides of the equation as,

  $\begin{array}{c}{\left(\sqrt{y^2+10y+50}\right)}^2={\left(\sqrt{y^2-2y+2}\right)}^2\\ y^2+10y+50=y^2-2y+2\end{array}$

Cancel out the common terms and isolate $y$on left hand side and solve it further as,

  $\begin{array}{c}{y}^2+10y+50=y^2-2y+2\\ 10y+2y=2-50\\ 12y=-48\\ y=-4\end{array}$

Since, $y=-4$ , so, the point on y-axis is $\text{P}\left(0,-4\right)$ .

Thus, point on y-axis that is equidistant from the two points $\left(5,-5\right)$ and $\left(1,1\right)$ is $\left(0,-4\right)$ .