Find the coordinates of the point Q

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sigma

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Feb 19, 2006
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The question reads as:
Find the coordinates of the point Q on the line 2x + 3y = 7 that is closest to the point P = (4,4)

Here's the diagram.


I don't have any work yet. I'm really stuck with this question and don't really know where to begin.
 
The line 3x-2y=8 contains P and is perpendicular to the given line.
Where do the two lines intersect? That is Q.
The shortest distance from a point to a line is the ‘perpendicular distance’.
 
How did you get 3x-2y=8 to be the perpendicular line to the line that contains Q?
 
sigma said:
How did you get 3x-2y=8 to be the perpendicular line to the line that contains Q?
Click to expand...
You go to the textbook. Lookup what it means for two lines to be perpendicular.
You tell me what you find out!
You show me how to write the equation of a line through P perpendicular to the given line.
Only then will I give you more help.
 
Still can't figure out how you got that equation and trying to find help in my text book is like trying to find water on mars (textbooks can be so useless sometimes) but I think I figured something out.

So I labled the y intercept on the line 2x + 3y = 7 as A and the x intercept B. I established the coordinates of A and B using 2x+3y=7 and got (0, 7/3) for A and (7/2, 0) for B. Drew an arrow from A to B and solved that vector by doing B - A, so: AB=(7/2,0) - (0, 7/3) = (7/2, -7/3).

I then labled point Q (x,y) and because Q is on 2x+3y=7, I figured out y by solving for y, so y=7/3 -2x/3 and I subbed back into Q to get (x,y)=(x, 7/3 - 2x/3). I then drew an arrow PQ and solved that vector doing PQ=(x, 7/3 - 2x/3) - (4,4) = (x-4, 1-2x/3).

Now I know since Q is closest to P, I know that PQ is orthogonal to AB, so I computed the dot product of AB and PQ and set=0 to solve x. So:

(x-4, 1-2x/3) . (7/2, -7/3) = 0

7x/2 - 14 - 7/3 + 14x/9 = 0

7x/2 + 14x/9 = 49/3

91x/18 = 49/3

91x = 294

x = 42/13

Now, I plug that back into my corrdinates for Q (x, 7/3 - 2x/3) and solve to get:
(42/13, 7/39)

'Pant pant' So, how does that look? That equation you gave above didn't seem like it had to come into play.
 
This is very basic algebra and any algebra textbook should have it
Two lines are perpendicular if the product of their slopes is –1.
\(\displaystyle \L
\left\{ \begin{array}{l}
l_1 :\;2x + 3y = 7 \\
l_2 :\;3x - 2y = 8 \\
\end{array} \right.\quad ,\quad \left\{ \begin{array}{l}
m_1 = \frac{{ - 2}}{3} \\
m_2 = \frac{3}{2} \\
\end{array} \right.\quad \quad m_1 m_2 = - 1\).

These two lines are perpendicular and l<SUB>2</SUB> contains (4,4).
The two lines intersect at (2,1). That is the ‘closest point’, Q.
 
Thanks for the help, but I give up. I'm just getting confused and frustrated. I cannot figure out this question for the life of me and believe me I am trying my level best to do this work on my own without being spoon fed the answers but every time I try, I just get something that is far from correct, and I'm not learning off my mistakes, its just getting me more frustrated. I stare and stare and what help I am being given and I'm just not seeing how to do it. Believe me, I don't like it when I have to ask for help I like to figure things out on my own and usually I can and when I do, I learn it far better but for some reason, I'm being burnt bad here with this question. I spent hours looking in my textbooks (I have 4) without much success. I HATE GEOMETRY!!! I'm having a bad day. But still I really appreciate all the help. I don't want you to think you have wasted your time but I'm afraid to ask more questions (really, I'm not deliberately trying to poke and prod for answers and getting the work done for me, I'm that stumped on this question and I don't know how else to put it). I just don't know why this question is so hard for me.
 
You have 2x+3y=7. Solving for y we have, \(\displaystyle \frac{-2x}{3}+\frac{7}{3}\)

As pka said, perpendicular lines are negative reciprocals. The negative reciprocal of -2/3 is 3/2. See?.

You are given x=4 and y=4 as coordinates

\(\displaystyle y=mx+b\)

\(\displaystyle 4=\frac{3}{2}(4)+b\)

\(\displaystyle b=-2\)

So, the equation of your line is:

\(\displaystyle y=\frac{3}{2}x-2\)

Equate the two line equations and solve for x. y will follow.

\(\displaystyle \frac{-2x+7}{3}=\frac{3x-4}{2}\)

That's about all there is to it.

I hope you don't mind pka, but Sigma seemed desperate and on the verge of a breakdown...just kidding :wink:

What doesn't kill us makes us stronger.
 
Galactus, don’t mind at all
Sigma, let’s try a strictly vector approach.
Given any line px+qy+r=0 in R<SUP>2</SUP>. The direction vector of that line is <-q,p>.
You seem to understand that somehow orthogonality comes in to play.
So lets say Q:(a,b) is the closest point on 2x+3y=7 to the point P:(4,4).
Thus the vector <-3,2> is the direction vector for 2x+3y=7.
The vector <PQ>=<a-4,b-4>. How is that vector perpendicular to <-3,2>.
Well <a-4,b-4>·<-3,2>=0; or –3(a-4)+2(b-4)=0. (two vectors are perpendicular if their dot product is zero). That means that 3a-2b=4.
Because (a,b) is on 2x+3y=7, this is true 2a+3b=7.
Where have we seen this system, 3a-2b=4 & 2a+3b=7?
The solution is (2,1). That is the point Q.
 
Thanks galactus. And you are right, I am on the verge of a breakdown. How come my previous answer was so far off though? I thought if 2 vectors are orthogonal, you have to compute the dot product. That work I did is what a friend told me to do and its way off from this answer. I just thought a question like this would involve things like dot products and computing vectors. It pertains to a unit I'm covering right now and none of the concepts in that question are discussed in that unit (things like perpendicular lines having negative reciprocals of the slope, y=mx+b) I mean I have done all that stuff before and I know I should know it really well but I'm having a hard time remembering it and this question veered way off course of what unit its suppose to cover. The question just kinda kamikazed into materials that I learned a long time ago that they expect you to remember and not do any kind of refresher in any kind of examples in the unit. I'm just up against a faulty work book that gives questions to work on that they don't bother to teach. I feel like a basket case here.
 
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