Write an equation of the line that is perpendicular to the line 2x + 5y = -15 and contains the point (-8,3). Another one of these @Littlebird I'm going to try do it and see if I get it ;P
First I convert the standard form of 2x + 5x = -15 into y intercept form. I get: y = -2/5x - 3 Then I input the point (-8,3) 3 = -2/5(-8) + b Simplified 3 = 3 1/5 + b Subtracted 3 1/5 from both sides Got: -1/5 = b @Littlebird @DanJS Is this correct?
@wio @nincompoop @Jhannybean @Jesstho.-.
hi
You kept the slope the same
looks correct.
So you are using parallel line
OH YES, I knew I forgot something
Parallel slope to -2/5 is 2 1/5 right?
negative inverse
\[2x + 5y\] will be perpendicular to \[ -5x+2y \]
Oh, it asks for the perpendicular of the line, not the slope...
So I have to negative inverse everything o-o
Basically, when you have: \[ ax+by=c \]Then the perpendicular line will be a form like: \[ -bx+ay \]Does that make sense?
In our case, we have: \[ -5x+2y = \ldots \]To find out what \(\ldots\) is, we have to plug in the point they gave us.
(-8,3)
So we get: \[ -5(-8)+2(3)=\ldots \]
How did you get -5x + 2y from 2x + 5y = -15 Did you switch the variables around?
I thought you were supposed to simplify the line from 2x + 5y = -15 into slope intercept form
Well, the easiest way to explain it is with vectors, but you don't know vectors yet.
I mean, you could even use \(5x-2y\) and it would work as well
Since you're just multiplying by \(-1\).
Another way to explain it, other than vectors, is through rotations.
do you remember how to rotate a point?
no
\((x,y)\) rotated by \(90^\circ\) is going to be \((-y,x)\).
Basically \[ \color{red}{(x,y)}\to \color{blue}{(-y,x)} \]For our line: \[ 2\color{red}x + 5\color{red}y = -15 \to 2(\color{blue}{-y})+5(\color{blue}{x})=-15 \]This simplifies to:\[ 5x-2y=-15 \]However, this line doesn't necessarily go through our point, which is why I dropped the \(-15\).
|dw:1420611937471:dw|
Anyway, since: \[ -5(-8)+2(3)=-40+6=-34 \]Our line is given by: \[ -5x+2y=-34 \]Alternatively:\[ 5x-2y=34 \]Anyway, if you want, you can put it into slope intercept form.
@SarahLove<3 Does it make sense?
Yeah, the way it was taught for us was to use the point slope form, just remembered that.
Your way makes sense, perpendicular is basically a line that forms a right angle with the original line(90 degree angle) and that's what you showed in your picture with rotating a point.
It's a bit more complex though, from what I'm learning atm ;P haha
I got y = 5/2x + 23
I simplified yours and got y = -5/2x + 17
Point Slope Form Way: y - 3 = 5/2(x - -8) negative rules y - 3 = 5/2(x + 8) Property of Distribution y - 3 = 5/2x + 20 Added 3 to both sides y = 5/2x + 23
This was from me converting 2x + 5y = -15 into y = -2/5x - 3 Then converting the slope from -2/5x into 5/2x for the negative reciprocal. Then I did the Point Slope Form Way
@wio
I did the math wrong. \(-5(-8) +2(3) = 40+6=46\)
So I should have gotten:\[ -5x+2y=46\implies y=\frac 52x+23 \]
You were right, just checked my work
I missed something in my equation and I reworked it and I got the same thing as you
My way is a bit different, but it isn't complex.
Basically the trick is \(ax+by\to bx-ay\).
However, some people just like to use formulas and stuff.
\[ y=mx+b\implies y-mx=b\to x+my=b\implies y=-\frac 1m x+b \]
The whole "negative inverse" \(-1/m\), is actually a result of the other method I gave.
Wio so you mean I was right the whole time!?!?!? ;O
No, if you keep the slope the same it is a parallel line.
I got y = 5/2x + 23, WHEW
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