Question 1160870
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Since *[tex \Large l_1] "cuts the *[tex \Large y]-axis at 5", one of the points on the line is *[tex \Large (0,5)]. The other is given as *[tex \Large (-2,3)]. Knowing two points on a line allows you to compute the slope of the line:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_{l_1}\ =\ \frac{y_1\,-\,y_2}{x_1\,-\,x_2}\ =\ \frac{5\,-\,3}{0\,-\,(-2)}]


You can do your own arithmetic.


Given that *[tex \Large \overline{PS}] is parallel to *[tex \Large l_1], we know that the line containing *[tex \Large \overline{PS}] must have the same slope as *[tex \Large l_1].  And, since the line containing *[tex \Large \overline{PS}] must perforce pass through point *[tex \Large P], we can write the equation of the line containing *[tex \Large \overline{PS}] using the Point-Slope form of an equation of a line.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ m_{\overline{PS}}\(x\ -\ x_1\)\ =\ m_{l_1}\(x\ -\ x_1\)]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ m_{l_1}\(x\ -\ 1)\ + 2]

								

And you can simplify as needed.


But now you have a problem because the other three parts of your question cannot be answered specifically without additional information.  Point S could be anywhere on the line containing *[tex \Large \overline{PS}] with the exception of at point P itself.  You don't even specify if the vertices are labeled alphabetically clockwise or counterclockwise, so you can't tell whether *[tex \Large x_P\ >\ x_S] or *[tex \Large x_P\ <\ x_S].


See diagram:


*[illustration parallelogramPQRS.jpg]

								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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