Point and a line

Suppose we have a line \(\displaystyle \ell:~ax+by+c=0\) and a point \(\displaystyle P:~(p,q)\)
The distance between \(\displaystyle P~\&~\ell\) is \(\displaystyle \dfrac{|ap+bq+c|}{\sqrt{a^2+b^2}}\)
 
You could use pka's formula but you can also do it from scratch.

OK, so you've got a line and a point. In case you didn't know, the distance between a point and a line is defined as the perpendicular distance between them. Draw a diagram.

That should help you think about what to do. Plan your attack on the problem. Can you suggest what to do first?
 
Harry_the_cat said:
You could use pka's formula but you can also do it from scratch.

OK, so you've got a line and a point. In case you didn't know, the distance between a point and a line is defined as the perpendicular distance between them. Draw a diagram.

That should help you think about what to do. Plan your attack on the problem. Can you suggest what to do first?
Click to expand...
How about using the distance formula?
 
Good. Ok, but to use the distance formula you need two points. How are you going to find the other point?
 
A line is a line. A point is a point. You can't convert a line to a point.

But you can find a point on the line. Not just any point but the point where you measure to (ie at right angles to the line).

Draw a line (L) and a point (P) NOT on the line. Now draw a line from the point to the line and perpendicular to it. Call that point on the line Q. PQ is the distance you want to find, right?

Now, what do you know about the line L and the line joining PQ?

(If you can respond quickly that would be great and I'll respond straight back.)
 
The shortest distance between a point and a line (which is what is meant by "the" distance between them) is along a line perpendicular to the line. The line is given by y= 6x+ 3 so has slope 6. A line perpendicular to that has slope -1/6. The line through (6, 2) with slope -1/6 is y= (-1/6)(x- 6)+ 2 so 6y= -x+ 6+ 12 or x+ 6y= 18. Where does that line intersect y= 6x+ 3? What is the distance from that point to (6, 2)?
 
HallsofIvy said:
The shortest distance between a point and a line (which is what is meant by "the" distance between them) is along a line perpendicular to the line. The line is given by y= 6x+ 3 so has slope 6. A line perpendicular to that has slope -1/6. The line through (6, 2) with slope -1/6 is y= (-1/6)(x- 6)+ 2 so 6y= -x+ 6+ 12 or x+ 6y= 18. Where does that line intersect y= 6x+ 3? What is the distance from that point to (6, 2)?
Click to expand...
d^2 = (0 - 6)^2 + (3-2)^2 = rt37
 
Yes, the foot of the perpendicular to the line Is (0, 3). Yes, the distance between (0, 3) and (6, 2), squared, is D^2= (0- 6)^2+ (3- 2)^2. But please do not write "= rt(37)"! D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37).
 
HallsofIvy said:
Yes, the foot of the perpendicular to the line Is (0, 3). Yes, the distance between (0, 3) and (6, 2), squared, is D^2= (0- 6)^2+ (3- 2)^2. But please do not write "= rt(37)"! D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37).
Click to expand...
Then how do I give explanations and justification for a problem like this?
 
rachelmaddie said:
Then how do I give explanations and justification for a problem like this?
Click to expand...
Well, just what I said in the post you quoted.
Yes, the foot of the perpendicular to the line Is (0, 3). Yes, the distance between (0, 3) and (6, 2), squared, is D^2= (0- 6)^2+ (3- 2)^2. But please do not write "= rt(37)"! D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37).
Click to expand...
That, you state that "the foot of the perpendicular to the line is (0, 3)" and that the distance "d^2 = (0 - 6)^2 + (3-2)^2 = rt37".

My only complaint was the last "=" is wrong. d^2 is NOT equal to "rt 37". d^2 is 37 and d itself is rt 37.
 
HallsofIvy said:
Well, just what I said in the post you quoted.

That, you state that "the foot of the perpendicular to the line is (0, 3)" and that the distance "d^2 = (0 - 6)^2 + (3-2)^2 = rt37".

My only complaint was the last "=" is wrong. d^2 is NOT equal to "rt 37". d^2 is 37 and d itself is rt 37.
Click to expand...

9)
The shortest distance between a point and a line (which is what is meant by "the" distance between them) is along a line perpendicular to the line. The line is given by y= 6x+ 3 so has slope 6. A line perpendicular to that has slope -1/6. The line through (6, 2) with slope -1/6 is y= (-1/6)(x- 6)+ 2 so 6y= -x+ 6+ 12 or x+ 6y= 18.
Use the distance formula to find the exact distance.
d^2 = (x2 - x1)^2 + (y2 - y1)^2
The foot of the perpendicular to the line Is (0, 3). So the distance between (0, 3) and (6, 2), squared, is D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37).

How is this?
 
rachelmaddie said:
Is there another method to finding the point?
Click to expand...
We are given the point. There is nothing to find.
The problem states: Given the point \(\displaystyle (6,2)\) and the line \(\displaystyle 6x-y=-3\) find the distance between them.
 
pka said:
We are given the point. There is nothing to find.
The problem states: Given the point \(\displaystyle (6,2)\) and the line \(\displaystyle 6x-y=-3\) find the distance between them.
Click to expand...
Is everything correct for my post #16?
 
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