Question 181845
<font face="Times New Roman" size="+2">


Perpendicular lines have slopes that are negative reciprocals, that is:


*[tex \LARGE \text{          }\math L_1 \perp L_2 \ \ \Leftrightarrow\ \ m_1 = -\frac{1}{m_2}]


Since your given equation is already in slope-intercept form:


*[tex \LARGE \text{          }\math y = mx + b]


*[tex \LARGE \text{          }\math y = -\frac{2}{3}x + 4]


You can read the slope directly, i.e. *[tex \Large m_1 = -\frac{2}{3}]


Therefore any perpendicular line must have a slope:


*[tex \LARGE \text{          }\math m_2 = -\frac{1}{-\frac{2}{3}} = \frac{3}{2} ]


Now that you have the slope of the desired line along with the given point, *[tex \Large P_1(x_1,y_1) = (1, -4)], you can use the point-slope form of the equation of a line:


*[tex \LARGE \text{          }\math y - y_1 = m(x - x_1)]


where <i>m</i> is the slope and *[tex \Large (x_1,y_1)] is the given point to develop your desired equation:


*[tex \LARGE \text{          }\math y - (-4) = \frac{3}{2}(x - 1)]


Since your given equation was in slope-intercept form, you should rearrange your answer to that form as well.




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