Chapter 9.5, Problem 20PSB

To determine

To calculate: The Value of y.

Answer to Problem 20PSB

Value of y is $\frac{-1}{7}$ .

Explanation of Solution

Given information:

Here, we have given that $(5,y)$ is equidistant from $(1,4)$ and $(10,-3)$ .

Formula used: Distance formula: d = $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Calculation: As, $(5,y)$ is equidistant from $(1,4)$ and $(10,-3)$ ,

  $\Rightarrow$ Distance between $(5,y)$ and $(1,4)$ = Distance between $(5,y)$ and $(10,-3)$ .

  $\begin{array}{l}\sqrt{{(5-1)}^2+{(y-4)}^2}=\sqrt{{(5-10)}^2+{[y-(-3)]}^2}\\ \Rightarrow \sqrt{{(4)}^2+{(y-4)}^2}=\sqrt{{(-5)}^2+{(y-3)}^2}\\ \Rightarrow \sqrt{16+y^2-8y+16}=\sqrt{25+y^2+6y+9}\\ \Rightarrow \sqrt{y^2-8y+32}=\sqrt{y^2+6y+34}\end{array}$

Now, squaring both sides, we get,

  $\begin{array}{l}{y}^2-8y+32=y^2+6y+34\\ \Rightarrow -8y+32=6y+34\\ \Rightarrow -8y-6y=34-32\\ \Rightarrow -14y=2\\ \Rightarrow y=\frac{2}{-14}\\ \Rightarrow y=\frac{-1}{7}\end{array}$