Question 319171
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You weren't paying attention in class...


A line that passes through two points *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)], has a slope given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{y_1\ -\ y_2}{x_1\ -\ x_2} ]


Just plug in the numbers and do the arithmetic.


Parallel lines have identical slopes, that is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L_1\ \parallel\ L_2 \ \ \Leftrightarrow\ \ m_1\ =\ m_2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{0\ -\ 6}{-4\ -\ 0} \ =\ \frac{3}{2}]


So equation of *[tex \LARGE l_1]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 0\ =\ \frac{3}{2}(x\ -\ (-4))]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{3}{2}x\ +\ 6]


*[tex \LARGE l_2] has to go through the origin, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 0\ =\ \frac{3}{2}(x\ -\ 0)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{3}{2}x]


The red line is *[tex \LARGE l_1] and the green line is *[tex \LARGE l_2]


{{{drawing(
500, 500, -10, 10, -10, 10,
grid(1),
graph(
500, 500, -10, 10, -10, 10,
(3/2)x+6,(3/2)x))}}}



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