Chapter 0, Problem 11RE

To determine

To find: The line is passing through the point $\left(-1,2\right)$ and perpendicular to $\frac{1}{2}x+\frac{1}{3}=1$ .

Answer to Problem 11RE

The obtained line is $3y-2x=8$ .

Explanation of Solution

Given Information:

The point is $\left(-1,2\right)$ and the line is $\frac{1}{2}x+\frac{1}{3}y=1$ .

Formula Used:

If the two line are perpendicular then their slope is defined as,

  $m_1{m}_2=-1$

The general form of the linear equations is $y-y_1=m\left(x-x_1\right)$ .

Where $m$ is representing slope.

Calculation:

Consider the given equations,

  $\frac{1}{2}x+\frac{1}{3}y=1$

Rewrite the given equation in general form.

  $\begin{array}{c}\frac{1}{2}x+\frac{1}{3}y=1\\ \frac{1}{3}y=-\frac{1}{2}x+1\\ y=-\frac{3}{2}x+3\end{array}$

Thus, the slope of the given equation is $-\frac{3}{2}$ .

Now, find the slope of that line which is perpendicular the given line.

  $\begin{array}{c}{m}_1{m}_2=-1\\ \left(-\frac{3}{2}\right)m_2=-1\\ m_2=\frac{2}{3}\end{array}$

Substitute the given point $\left(-1,2\right)$ and slope in the general equation to find the equation.

  $\begin{array}{c}y-2=\frac{2}{3}\left(x-\left(-1\right)\right)\\ 3y-6=2\left(x+1\right)\\ 3y=2x+2+6\\ 3y-2x=8\end{array}$

Therefore, the obtained equation is $3y-2x=8$ .