Chapter 16.1, Problem 11PS

To determine

To find: Theequations of the bisectors of the angle formed by the given equations.

Answer to Problem 11PS

The equations of the bisectors of the angle formed by the given equations are $x+y-8=0\text{or}3x-3y+2=0$ .

Explanation of Solution

Given information:

The given equationsare,

  $\begin{array}{l}x-2y+5=0\\ 2x-y-3=0\end{array}$

The distance from the point on the bisectorto the graph of the given equations are equal.

The equations of the bisectors of the angle formed by the given equations are calculated as,

  $\begin{array}{l}\frac{\left|a_{1}x+b_{1}y+c_1\right|}{\sqrt{a_1^2+b_1^2}}=\frac{\left|a_{2}x+b_{2}y+c_2\right|}{\sqrt{a_2^2+b_2^2}}\\ \frac{\left|1\left(x\right)+\left(-2\right)y+5\right|}{\sqrt{{\left(1\right)}^2+{\left(-2\right)}^2}}=\frac{\left|2\left(x\right)+\left(-1\right)y+\left(-3\right)\right|}{\sqrt{{\left(2\right)}^2+{\left(-1\right)}^2}}\\ \left|x-2y+5\right|=\left|2x-y-3\right|\\ \left(x-2y+5\right)=\left(2x-y-3\right)\text{or}\left(x-2y+5\right)=-\left(2x-y-3\right)\end{array}$

Simplify further,

  $x+y-8=0\text{or}3x-3y+2=0$

Thus, the equations of the bisectors of the angle formed by the given equations are $x+y-8=0\text{or}3x-3y+2=0$ .