anonymous
  • anonymous
Write the equation of a line which passes through (3,5) and is perpendicular to y=(3/4)x-6 in slope-intercept form.
Mathematics
chestercat
  • chestercat
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e.mccormick
  • e.mccormick
First, do you know the method to find the slope of a perpendicular line?
anonymous
  • anonymous
y=mx+b?
anonymous
  • anonymous
|dw:1369021818640:dw|

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e.mccormick
  • e.mccormick
Not in this case. The perpendicular is the negative inverse of the slope of the given line. Do you know what that is?
anonymous
  • anonymous
|dw:1369021841094:dw|
anonymous
  • anonymous
|dw:1369021901093:dw|
anonymous
  • anonymous
Oh ok i always get those two confused.
shamim
  • shamim
if slope of ur given line is \[m _{1}\]
e.mccormick
  • e.mccormick
So, what would the slope of the perpendicular line be in this case?
shamim
  • shamim
then \[m _{1}=\frac{ 3 }{ 4 }\]
shamim
  • shamim
if the slope of a perpendicular line is \[m _{2}\]
anonymous
  • anonymous
Ok is the equation to find out what the perpendicular line y=mx+b or m=y^2 - y^1 / x^2-x^1?
shamim
  • shamim
then\[m _{1}m _{2}=-1\]
shamim
  • shamim
or\[\frac{ 3 }{ 4 }m _{2}=-1\]
e.mccormick
  • e.mccormick
We need the slope first. shamim has pointed out the slope of the original line. Do you know what is meant by the "negative inverse" of that?
anonymous
  • anonymous
The negative opposite?
shamim
  • shamim
\[m _{2}=\frac{ -4 }{ 3 }\]
e.mccormick
  • e.mccormick
Well, inverse is sort of opposite. If you mean inverse as in \(\frac{a}{b}\implies \frac{b}{a}\) is the inverse.
anonymous
  • anonymous
Alright, so in order to find the slope we use y=mx+b right?
e.mccormick
  • e.mccormick
To invert is to turn upside down or reverse position. That is where inverse comes from. Just some simple language, but they make it sound cryptic by tossing it into math. No, shamim went ahead and posted it. All you needed was the m part of the original line. Do you see the \(m_2\)?
anonymous
  • anonymous
Yes, i see. So next up would be to use the y^2-y^1 equation?
e.mccormick
  • e.mccormick
Yes.
anonymous
  • anonymous
So in this case (3,5) would be (x^1,y^1) while −4/3 would be (x^2,y^2)?
e.mccormick
  • e.mccormick
No, -4/3 is m. You just need that and one point.
e.mccormick
  • e.mccormick
\(y-y_1=m(x-x_1)\)
anonymous
  • anonymous
y-5 = -4/3(x-3) is that right so far?
e.mccormick
  • e.mccormick
Yes!
anonymous
  • anonymous
And that would be the final answer correct?
e.mccormick
  • e.mccormick
This became a bit of a posting mess and then the site went down, so here is a summary of what was done: You were given the line \(y=\frac{3}{4}x-6 \) and point \((3,5)\) with the instructions to find the slope-intercept equation of the perpendicular line through that point. First, you need the slope of the line being found. The perpendicular line has the negative inverse slope of the given line. To invert is to turn upside down or reverse position and negative is the mathematical changing of sign from \(+\) to \(-\) or \(-\) to \(+\). If you understand the concept, you flip it and make it the opposite sign. However, here is the mathematical way to find the slope of the perpendicular line: Start with the original line, \(y=\frac{3}{4}x-6 \). Form that you get \(m_1\). The negative inverse implies that if they were multiplied, the result would be -1. That is stated as: \(m_1m_2=-1\) and put what is known, \(m_1\) into that: \(\frac{3}{4}m_2=-1\) Then solve for \(m_2\) \(\frac{3}{4}m_2=-1\) \(\frac{4}{3}\cdot\frac{3}{4}m_2=\frac{4}{3}\cdot-1\) \(m_2=-\frac{4}{3}\) So \(m_2=-\frac{4}{3}\) becomes just m for the new formula, the point-slope version of a line. Point slope lets you use any point and the slope to find the formula of a line: \(y-y_1=m(x-x_1)\) \(y-y_1=-\frac{4}{3}(x-x_1)\) Then we toss in the point \((3,5)\): \(y-5=-\frac{4}{3}(x-3)\) That is a formula for the line, but they wanted a specific formula, the slope-intercept version. To get that, you need to solve for y. \(y-5=-\frac{4}{3}(x-3)\) \(y-5=-\frac{4}{3}x+4\) \(y-5+5=-\frac{4}{3}x+4+5\) \(y=-\frac{4}{3}x+9\) And there it is! The slope-intercept equation of the perpendicular line through the given point.

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