Chapter 13, Problem 27RP

To determine

To find: The points and the quadrant in which the point that is on the line and is equidistant from the x axis and the y axis.

Answer to Problem 27RP

Therequired point is $\left(\frac{10}{7},-\frac{10}{7}\right)$ in the third quadrant and $\left(-\frac{10}{3},\frac{-10}{3}\right)$ in the fourth quadrant.

Explanation of Solution

Given:

The given equation is $2x-5y=10$ .

Calculation:

Consider the given equation is $2x-5y=10$ .

Since, the point is equidistant from both the axis then $y=x$ or $y=-x$ and the equation of the line is evaluated a,

  $\begin{array}{l}2x-5\left(x\right)=10\\ -3x=10\\ x=\frac{-10}{3}\end{array}$

Thus, the required point is $\left(-\frac{10}{3},\frac{-10}{3}\right)$ in the third quadrant.

The other solution is,

  $\begin{array}{l}2x-5\left(-x\right)=10\\ 7x=10\\ x=\frac{10}{7}\end{array}$

Thus, the required point is $\left(\frac{10}{7},-\frac{10}{7}\right)$ in the fourth quadrant.