Chapter 1, Problem 51SGA

To determine

To find: the standard form of line equation.

Answer to Problem 51SGA

  $x+2y-9=0$

Explanation of Solution

Given information:

  $4x-2y+2=0$

Coordinates = $\left(1,4\right)$

Calculation:

Convert the equation in slope intercept form as shown below.

  $\begin{array}{l}2y=4x+2\\ y=\frac{4}{2}x+\frac{2}{2}\\ y=2x+1\end{array}$

A line perpendicular to line of equation, $4x-2y+2=0$ . So, the slope (m) is $2$ .

The slope of line A, perpendicular to $4x-2y+2=0$ is opposite reciprocal of the other line. So, the slope of line can be evaluated as follows:

  $\begin{array}{c}\perp m=-\frac{1}{m}\\ 2=-\frac{1}{m}\\ m=-\frac{1}{2}\end{array}$

Put the coordinates and slope value of line A, in the slope intercept equation and solve as follows:

  $\begin{array}{c}y=mx+b\\ 4=\left(-\frac{1}{2}\right)\left(1\right)+b\\ 4=\left(-\frac{1}{2}\right)+b\\ 4+\frac{1}{2}=b\\ b=\frac{8+1}{2}\\ b=\frac{9}{2}\end{array}$

Equations for converting the slope and y-intercept into standard form as follows:

  $\begin{array}{l}m=-\frac{A}{B}\text{ and }b=-\frac{C}{B}\\ -\frac{1}{2}=-\frac{A}{B}\text{ and }\frac{9}{2}=-\frac{C}{B}\left[\text{Put values of }m\text{ and }b\right]\\ \frac{1}{2}=\frac{A}{B}\text{ and }-\frac{9}{2}=\frac{C}{B}\left[\text{Simplify}\right]\end{array}$

The value of $A=1,B=2,\text{ and }C=-9$ .

Substitute the value of $A,B\text{ and }C$ in the standard form of line as shown below.

  $\begin{array}{l}Ax+By+C=0\\ 1x+2y-9=0\\ x+2y-9=0\end{array}$

Thus, the standard form of line is $x+2y-9=0$ .