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October 18th, 2019, 11:21 PM  #1 
Newbie Joined: Aug 2018 From: US Posts: 6 Thanks: 0  Equation of a Line passes through.
Hi all, I’m taking a PreCal and I have totally forgotten this Algebra stuff. Have looked everywhere, but still my work is wrong. The question ask to look for the perpendicular line (in SlopeIntercept) that pass through (1,9) with the equation y=(4x)/3. The correct answer is y=3x12. Below is my work 
October 19th, 2019, 02:25 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 21,124 Thanks: 2332 
You are supposed to find the equation of a line that passes through the point (1, 9) and is perpendicular to the line with equation y = (4  x)/3. Perpendicular lines are parallel to the axes or have slopes whose product is 1. As the given line has slope 1/3, any line perpendicular to that line has slope 3. Hence the required line's equation, in pointslope form, is y = 9 + 3(x  1). Writing that in slopeintercept form gives y = 3x  12. 
October 21st, 2019, 01:23 AM  #3 
Newbie Joined: Aug 2018 From: US Posts: 6 Thanks: 0 
Why did you put “9” beside “+3(x1)” like that? I haven’t came across such equation y=9+3(x1) before. I tried intuitively instead with 9=3(x+1) but my answer was y=3x+21. Can you explain your method and why mine is wrong? Plz help, I got a test coming
Last edited by skipjack; October 21st, 2019 at 02:46 AM. Reason: to change x+1 to x1 in references to preceding post 
October 21st, 2019, 02:49 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 21,124 Thanks: 2332 
As you haven't shown every step (including any assumptions and reasoning) you used, I can't tell where exactly you went wrong. If a line of known slope $m$ passes through a point $(x_1,\, y_1)$, its slope at any other point $(x,\, y)$ on the line is also $m$, so $\displaystyle m = \frac{y  y_1}{x  x_1}$. That implies $y = y_1 + m(x  x_1)$, which also holds for $(x,\, y) = (x_1,\, y_1)$ and is called the pointslope form of the line's equation. For further information, see this article. 

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