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October 18th, 2019, 11:21 PM   #1
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Equation of a Line passes through.

Hi all,
I’m taking a Pre-Cal 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 Slope-Intercept) that pass through (1,-9) with the equation y=(4-x)/3. The correct answer is y=3x-12. Below is my work
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October 19th, 2019, 02:25 AM   #2
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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 point-slope form, is y = -9 + 3(x - 1).

Writing that in slope-intercept form gives y = 3x - 12.
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October 21st, 2019, 01:23 AM   #3
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Why did you put “-9” beside “+3(x-1)” like that? I haven’t came across such equation y=-9+3(x-1) 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 x-1 in references to preceding post
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October 21st, 2019, 02:49 AM   #4
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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 point-slope form of the line's equation.

For further information, see this article.
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