Chapter 1.5, Problem 3CFU

To determine

To find: slope of line parallel and slope of line perpendicular to the given equation.

Answer to Problem 3CFU

Slope of line parallel $=-\frac{4}{3}$

Slope of perpendicular line $=\frac{3}{4}$

Explanation of Solution

Given information:

  $4x+3y+19=0$

Calculation

Equation of line of standard form is as follows:

  $Ax+By+C=0$

Equation of slope of parallel line is as follows:

  $m=-\frac{A}{B}$

On comparing the given equation with line of standard form, we get

  $\begin{array}{l}A=4\\ B=3\end{array}$

Substitute the values of A and B in the equation of slope of line, to get

  $m=-\frac{4}{3}$

Thus, slope of parallel lines is parallel to the given line also, i.e. $-\frac{4}{3}$ .

The slope of perpendicular line is opposite to the reciprocal of the original line, i.e.

  $\begin{array}{c}\perp m=-\frac{1}{m}\\ \perp m=-\left(\frac{1}{-\frac{4}{3}}\right)\left[\text{Substituting value of }m\right]\\ \perp m=-\left(-\frac{3}{4}\right)\\ \perp m=\frac{3}{4}\end{array}$

Therefore, the slope of perpendicular line is $\frac{3}{4}$ .