Point and a line

rachelmaddie said:
I’m not aloud to use that formula since it’s not part of my lesson.
Click to expand...

9.d^2=(x−6)^2+(y−2)^2, where y=6x+3. Or d^2=(x−6)2+(6x+1)^2. To minimize calculate derivative and set to 0. 2(x−6)+12(6x+1)=0 leading to x=0 and y=3 or d2=37.
 
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Rachelmaddie, There are several ways to do this problem.

1. The shortest way: Use pka's formula. But you said you can't because you haven't learnt that yet. That's ok. Personally, I don't like using formulae like that because it's just another thing to remember.

2. From scratch: That's what Halls and I were leading you to.
This involved
(i) finding the equation of the line perpendicular to the one you are given and which passes through the given point
(ii) finding the point of intersection of this line and the original line
(iii) then using the distance formula

3. Using calculus (as you have shown in your post #21). This involved finding the minimum distance between the general point (x, 6x+3) on the line and the point (6, 2). To complete your solution this way, you would need to justify it is indeed a minimum (eg first or second derivative test).

If you understand what you've written in post #21, then that's the way to go, simply because it is shorter and more efficient. (I was assuming you haven't studied calculus yet, so that's why I suggested the method I did.)

Whichever method you use you need to justify each step. Set it out fully and you should get full marks.
 
Harry_the_cat said:
Rachelmaddie, There are several ways to do this problem.

1. The shortest way: Use pka's formula. But you said you can't because you haven't learnt that yet. That's ok. Personally, I don't like using formulae like that because it's just another thing to remember.

2. From scratch: That's what Halls and I were leading you to.
This involved
(i) finding the equation of the line perpendicular to the one you are given and which passes through the given point
(ii) finding the point of intersection of this line and the original line
(iii) then using the distance formula

3. Using calculus (as you have shown in your post #21). This involved finding the minimum distance between the general point (x, 6x+3) on the line and the point (6, 2). To complete your solution this way, you would need to justify it is indeed a minimum (eg first or second derivative test).

If you understand what you've written in post #21, then that's the way to go, simply because it is shorter and more efficient. (I was assuming you haven't studied calculus yet, so that's why I suggested the method I did.)

Whichever method you use you need to justify each step. Set it out fully and you should get full marks.
Click to expand...
Can you please help me justify it?
 
Which method are you going to use?
Post your full solution and I will help you justify it. You need to justify each step you do.
 
Harry_the_cat said:
Which method are you going to use?
Post your full solution and I will help you justify it. You need to justify each step you do.
Click to expand...
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

d^2=(x−6)^2+(y−2)^2, where y=6x+3. Or d^2=(x−6)^2+(6x+1)^2. To minimize calculate derivative and set to 0. 2(x−6)+12(6x+1)=0 leading to x=0 and y=3 or d^2=37.


Then, 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, D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37)
 
[QUOTE="rachelmaddie, post: 477143, member: 74515"
Then, 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, D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37)[/QUOTE]
To: rachelmaddie, will you please please tell us i you are part of a real class in geometry or are you in one of those online courses?
I ask you this as someone who has taught courses in axiomatic geometry both undergraduate&graduate.
From the questions & replies that you have posted, I find to think you are in a real course.
I do not mean to imply that you are anyway at fault in this.
But if I were you and I paid for this course, I would demand a refund.
 
pka said:
[QUOTE="rachelmaddie, post: 477143, member: 74515"
Then, 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, D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37)
Click to expand...
To: rachelmaddie, will you please please tell us i you are part of a real class in geometry or are you in one of those online courses?
I ask you this as someone who has taught courses in axiomatic geometry both undergraduate&graduate.
From the questions & replies that you have posted, I find to think you are in a real course.
I do not mean to imply that you are anyway at fault in this.
But if I were you and I paid for this course, I would demand a refund.
[/QUOTE]
Yes, I’m in an online course. Can you please tell me if my work is correct for post #25?
 
rachelmaddie 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

d^2=(x−6)^2+(y−2)^2, where y=6x+3. Or d^2=(x−6)^2+(6x+1)^2. To minimize calculate derivative and set to 0. 2(x−6)+12(6x+1)=0 leading to x=0 and y=3 or d^2=37.


Then, 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, D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37)
Click to expand...
Please reread my post #22. Use either method 2 OR method 3 - you don't need to mix them as you have above.

Have you studied calculus? If not, ignore method 3 for the moment.

I have just copied and pasted your answer so it makes sense.
To justify your working you will need to address my comments in red.

Method 2: (using analytical geometry)
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

d^2=(x−6)^2+(y−2)^2, where y=6x+3. Or d^2=(x−6)^2+(6x+1)^2. To minimize calculate derivative and set to 0. 2(x−6)+12(6x+1)=0 leading to x=0 and y=3 or d^2=37.

Then, 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). You will need to show how you got (0, 3).
So the distance between (0, 3) and (6, 2), squared, D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37)

Method 3: (using calculus)
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

d^2=(x−6)^2+(y−2)^2,
where y=6x+3.
Or d^2=(x−6)^2+(6x+1)^2.
To minimize calculate derivative and set to 0.
Derivative = 2(x−6)+12(6x+1)=0 leading to x=0 show how you got x=0
and y=3 show how you got y=3

You will also need to justify that this is a minimum distance.

or d^2=37. show how you got d^2 = 37

Then, 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, D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37)
 
Harry_the_cat said:
Please reread my post #22. Use either method 2 OR method 3 - you don't need to mix them as you have above.

Have you studied calculus? If not, ignore method 3 for the moment.

I have just copied and pasted your answer so it makes sense.
To justify your working you will need to address my comments in red.

Method 2: (using analytical geometry)
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

d^2=(x−6)^2+(y−2)^2, where y=6x+3. Or d^2=(x−6)^2+(6x+1)^2. To minimize calculate derivative and set to 0. 2(x−6)+12(6x+1)=0 leading to x=0 and y=3 or d^2=37.

Then, 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). You will need to show how you got (0, 3).
So the distance between (0, 3) and (6, 2), squared, D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37)

Method 3: (using calculus)
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

d^2=(x−6)^2+(y−2)^2,
where y=6x+3.
Or d^2=(x−6)^2+(6x+1)^2.
To minimize calculate derivative and set to 0.
Derivative = 2(x−6)+12(6x+1)=0 leading to x=0 show how you got x=0
and y=3 show how you got y=3

You will also need to justify that this is a minimum distance.

or d^2=37. show how you got d^2 = 37

Then, 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, D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37)
Click to expand...
All I know is that I’m suppose to use the distance formula so can you please help me with both methods?
 
Harry_the_cat said:
Please reread my post #22. Use either method 2 OR method 3 - you don't need to mix them as you have above.

Have you studied calculus? If not, ignore method 3 for the moment.

I have just copied and pasted your answer so it makes sense.
To justify your working you will need to address my comments in red.

Method 2: (using analytical geometry)
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

d^2=(x−6)^2+(y−2)^2, where y=6x+3. Or d^2=(x−6)^2+(6x+1)^2. To minimize calculate derivative and set to 0. 2(x−6)+12(6x+1)=0 leading to x=0 and y=3 or d^2=37.

Then, 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). You will need to show how you got (0, 3).
So the distance between (0, 3) and (6, 2), squared, D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37)

Method 3: (using calculus)
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

d^2=(x−6)^2+(y−2)^2,
where y=6x+3.
Or d^2=(x−6)^2+(6x+1)^2.
To minimize calculate derivative and set to 0.
Derivative = 2(x−6)+12(6x+1)=0 leading to x=0 show how you got x=0
and y=3 show how you got y=3

You will also need to justify that this is a minimum distance.

or d^2=37. show how you got d^2 = 37

Then, 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, D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37)
Click to expand...
rachelmaddie said:
All I know is that I’m suppose to use the distance formula so can you please help me with both methods?
Click to expand...
Disregard that. I’m not able to use calculus. Using analytical geometry how would I show how I got (0,3)?
 
You need to show how you find the point of intersection of the two lines you have. Hint: Simultaneous equations.
 
Ok. So, you have two lines which intersect. The original line y=6x+3 and the perpendicular line x+6y =18. Yes?
 
No, using what you have already done:

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.

Do you understand each step that you have written there?
 
Harry_the_cat said:
No, using what you have already done:

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.

Do you understand each step that you have written there?
Click to expand...
Yes I do
 
Ok so you have 2 lines which intersect. y=6x+3 and x+6y =18.

You need to solve them simultaneously.

Substitute y=6x+3 into x+6y=18 and you get:
x + 6(6x+3) = 18.

This is called "solving simultaneous equations by substitution". Have you learnt that?
 
Harry_the_cat said:
Ok so you have 2 lines which intersect. y=6x+3 and x+6y =18.

You need to solve them simultaneously.

Substitute y=6x+3 into x+6y=18 and you get:
x + 6(6x+3) = 18.

This is called "solving simultaneous equations by substitution". Have you learnt that?
Click to expand...
Is this part of the guide?
 
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