Question 198754
<font face="Garamond" size="+2">

Step 1:  Solve the given equation for *[tex \Large y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x - 3y = 1 \ \ \Rightarrow\ \ -3y = -2x + 1 \ \ \Rightarrow\ \ y = \frac{2}{3}x - \frac{1}{3}]


Now that the equation is in slope-intercept form you can see that the slope of the given line is *[tex \LARGE \frac{2}{3}].


Step 2: Use the following rule to determine the slope of the desired line:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L_1 \perp L_2 \ \ \Leftrightarrow\ \ m_1 = -\frac{1}{m_2} \text{ and } m_1, m_2 \neq 0]


In other words, find the negative reciprocal of the slope of the given line.


Step 3: Use the given point and the slope you determined in step 2 to substitute values into the point-slope form of the equation of a line:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y - y_1 = m(x - x_1) ]


Where *[tex \Large \left(x_1,y_1\right)] are the coordinates of the given point and *[tex \Large m] is the slope you derived in step 2.


Since the problem only asks for 'an equation', you really don't have to do anything else, although typically you should either present the equation in slope-intercept (*[tex \Large y = mx + b]) or standard (*[tex \Large ax + by = c]) form.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>