Chapter 1.3, Problem 113E

To determine

To find: relation between x and y such that $\left(x,y\right)$ is equidistant from points $\left(3,\frac{5}{2}\right)$ and $\left(-7,1\right)$ .

Answer to Problem 113E

  $80x+12y+139=0$

Explanation of Solution

Concept Used:

Distance Formula: Distance between the two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by:

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

Calculation:

It is given that the point $P\left(x,y\right)$ is equidistant from the points $A\left(3,\frac{5}{2}\right)$ and $B\left(-7,1\right)$ .

That is,

  $PA=PB$ ….. (1)

Now apply distance formula to find PA and PB

  $\begin{array}{c}PA=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(3-x\right)}^2+{\left(\frac{5}{2}-y\right)}^2}\end{array}$

  $\begin{array}{c}PB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-7-x\right)}^2+{\left(1-y\right)}^2}\end{array}$

Now substitute in equation (1),

  $\begin{array}{c}PA=PB\\ \sqrt{{\left(3-x\right)}^2+{\left(\frac{5}{2}-y\right)}^2}=\sqrt{{\left(-7-x\right)}^2+{\left(1-y\right)}^2}\\ {\left(3-x\right)}^2+{\left(\frac{5}{2}-y\right)}^2={\left(-7-x\right)}^2+{\left(1-y\right)}^2\end{array}$

  $\begin{array}{c}{x}^2-6x+9+y^2-5y+\frac{25}{4}=x^2+14x+49+y^2-2y+1\\ 20x+3y+\frac{139}{4}=0\\ 80x+12y+139=0\end{array}$

Conclusion:

The relationship x and y such that $\left(x,y\right)$ is equidistant from points $\left(3,\frac{5}{2}\right)$ and $\left(-7,1\right)$ is given by: $80x+12y+139=0$