Mixing
Find the locus of the middle point of the intercept on the line y=x+c made by the lines 2x+3y=5 &...
$begingroup$
Here's my shot:
since the two lines are parallel, I figured that the middle point should be equidistant from the parallel lines,so using distance formula:-
$ frac{2x+3y-5}{sqrt{13}}=frac{2x+3y-8}{sqrt{13}}$
I get, $4x+6y-13=0$(considering one equation +ve and other -ve)
which seems to be the right answer,
but my question is, is there any other standard method to solve this question?
geometry analytic-geometry coordinate-systems
edited Jan 18 at 7:58
Sinπ
64511
asked Jan 18 at 7:17
Abdullah Abdullah
114
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add a comment |
$begingroup$
Here's my shot:
since the two lines are parallel, I figured that the middle point should be equidistant from the parallel lines,so using distance formula:-
$ frac{2x+3y-5}{sqrt{13}}=frac{2x+3y-8}{sqrt{13}}$
I get, $4x+6y-13=0$(considering one equation +ve and other -ve)
which seems to be the right answer,
but my question is, is there any other standard method to solve this question?
geometry analytic-geometry coordinate-systems
edited Jan 18 at 7:58
Sinπ
64511
asked Jan 18 at 7:17
Abdullah Abdullah
114
$endgroup$
add a comment |
$begingroup$
Here's my shot:
since the two lines are parallel, I figured that the middle point should be equidistant from the parallel lines,so using distance formula:-
$ frac{2x+3y-5}{sqrt{13}}=frac{2x+3y-8}{sqrt{13}}$
I get, $4x+6y-13=0$(considering one equation +ve and other -ve)
which seems to be the right answer,
but my question is, is there any other standard method to solve this question?
geometry analytic-geometry coordinate-systems
edited Jan 18 at 7:58
Sinπ
64511
asked Jan 18 at 7:17
Abdullah Abdullah
114
$endgroup$
Here's my shot:
since the two lines are parallel, I figured that the middle point should be equidistant from the parallel lines,so using distance formula:-
$ frac{2x+3y-5}{sqrt{13}}=frac{2x+3y-8}{sqrt{13}}$
I get, $4x+6y-13=0$(considering one equation +ve and other -ve)
which seems to be the right answer,
but my question is, is there any other standard method to solve this question?
geometry analytic-geometry coordinate-systems
geometry analytic-geometry coordinate-systems
edited Jan 18 at 7:58
Sinπ
64511
asked Jan 18 at 7:17
Abdullah Abdullah
114
edited Jan 18 at 7:58
Sinπ
64511
asked Jan 18 at 7:17
Abdullah Abdullah
114
edited Jan 18 at 7:58
Sinπ
64511
edited Jan 18 at 7:58
Sinπ
64511
edited Jan 18 at 7:58
Sinπ
64511
64511
asked Jan 18 at 7:17
Abdullah Abdullah
114
asked Jan 18 at 7:17
Abdullah Abdullah
114
asked Jan 18 at 7:17
Abdullah Abdullah
114
114
add a comment |
add a comment |
2 Answers
2
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votes
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The locus traced by all the midpoints is just the line in the middle: $2x+3y=frac{13}{2}$ or $4x+6y-13=0$.
answered Jan 18 at 7:33
Toby MakToby Mak
3,52311128
$endgroup$
add a comment |
$begingroup$
You need to solve the following system:
$$2x+3y=frac{5+8}{2}$$ and $$y=x+c.$$
answered Jan 18 at 7:33
Michael RozenbergMichael Rozenberg
105k1892198
$endgroup$
$begingroup$
Thanks, but I'm afraid it's wrong. The correct answer is only relevant to the first part.
$endgroup$
– Abdullah
Jan 22 at 7:33
$begingroup$
@Abdullah I don't agree with you. See please better what we need to find.
$endgroup$
– Michael Rozenberg
Jan 22 at 7:52
$begingroup$
to find the locus why would we need to solve the system of equations, that would give us the point right? Correct me if I'm wrong Michael, much appreciated.
$endgroup$
– Abdullah
Jan 22 at 9:29
$begingroup$
@Abdullah We need to find intersection points of the locus with the line $y=x+c$. See please better your problem.
$endgroup$
– Michael Rozenberg
Jan 22 at 10:57
$begingroup$
I see, but the answer doesn't matches it.
$endgroup$
– Abdullah
Jan 22 at 12:48
|
show 1 more comment
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The locus traced by all the midpoints is just the line in the middle: $2x+3y=frac{13}{2}$ or $4x+6y-13=0$.
answered Jan 18 at 7:33
Toby MakToby Mak
3,52311128
$endgroup$
add a comment |
$begingroup$
The locus traced by all the midpoints is just the line in the middle: $2x+3y=frac{13}{2}$ or $4x+6y-13=0$.
answered Jan 18 at 7:33
Toby MakToby Mak
3,52311128
$endgroup$
add a comment |
$begingroup$
The locus traced by all the midpoints is just the line in the middle: $2x+3y=frac{13}{2}$ or $4x+6y-13=0$.
answered Jan 18 at 7:33
Toby MakToby Mak
3,52311128
$endgroup$
The locus traced by all the midpoints is just the line in the middle: $2x+3y=frac{13}{2}$ or $4x+6y-13=0$.
answered Jan 18 at 7:33
Toby MakToby Mak
3,52311128
answered Jan 18 at 7:33
Toby MakToby Mak
3,52311128
answered Jan 18 at 7:33
Toby MakToby Mak
3,52311128
answered Jan 18 at 7:33
Toby MakToby Mak
3,52311128
3,52311128
add a comment |
add a comment |
$begingroup$
You need to solve the following system:
$$2x+3y=frac{5+8}{2}$$ and $$y=x+c.$$
answered Jan 18 at 7:33
Michael RozenbergMichael Rozenberg
105k1892198
$endgroup$
$begingroup$
Thanks, but I'm afraid it's wrong. The correct answer is only relevant to the first part.
$endgroup$
– Abdullah
Jan 22 at 7:33
$begingroup$
@Abdullah I don't agree with you. See please better what we need to find.
$endgroup$
– Michael Rozenberg
Jan 22 at 7:52
$begingroup$
to find the locus why would we need to solve the system of equations, that would give us the point right? Correct me if I'm wrong Michael, much appreciated.
$endgroup$
– Abdullah
Jan 22 at 9:29
$begingroup$
@Abdullah We need to find intersection points of the locus with the line $y=x+c$. See please better your problem.
$endgroup$
– Michael Rozenberg
Jan 22 at 10:57
$begingroup$
I see, but the answer doesn't matches it.
$endgroup$
– Abdullah
Jan 22 at 12:48
|
show 1 more comment
$begingroup$
You need to solve the following system:
$$2x+3y=frac{5+8}{2}$$ and $$y=x+c.$$
answered Jan 18 at 7:33
Michael RozenbergMichael Rozenberg
105k1892198
$endgroup$
$begingroup$
Thanks, but I'm afraid it's wrong. The correct answer is only relevant to the first part.
$endgroup$
– Abdullah
Jan 22 at 7:33
$begingroup$
@Abdullah I don't agree with you. See please better what we need to find.
$endgroup$
– Michael Rozenberg
Jan 22 at 7:52
$begingroup$
to find the locus why would we need to solve the system of equations, that would give us the point right? Correct me if I'm wrong Michael, much appreciated.
$endgroup$
– Abdullah
Jan 22 at 9:29
$begingroup$
@Abdullah We need to find intersection points of the locus with the line $y=x+c$. See please better your problem.
$endgroup$
– Michael Rozenberg
Jan 22 at 10:57
$begingroup$
I see, but the answer doesn't matches it.
$endgroup$
– Abdullah
Jan 22 at 12:48
|
show 1 more comment
$begingroup$
You need to solve the following system:
$$2x+3y=frac{5+8}{2}$$ and $$y=x+c.$$
answered Jan 18 at 7:33
Michael RozenbergMichael Rozenberg
105k1892198
$endgroup$
You need to solve the following system:
$$2x+3y=frac{5+8}{2}$$ and $$y=x+c.$$
answered Jan 18 at 7:33
Michael RozenbergMichael Rozenberg
105k1892198
answered Jan 18 at 7:33
Michael RozenbergMichael Rozenberg
105k1892198
answered Jan 18 at 7:33
Michael RozenbergMichael Rozenberg
105k1892198
answered Jan 18 at 7:33
Michael RozenbergMichael Rozenberg
105k1892198
105k1892198
$begingroup$
Thanks, but I'm afraid it's wrong. The correct answer is only relevant to the first part.
$endgroup$
– Abdullah
Jan 22 at 7:33
$begingroup$
@Abdullah I don't agree with you. See please better what we need to find.
$endgroup$
– Michael Rozenberg
Jan 22 at 7:52
$begingroup$
to find the locus why would we need to solve the system of equations, that would give us the point right? Correct me if I'm wrong Michael, much appreciated.
$endgroup$
– Abdullah
Jan 22 at 9:29
$begingroup$
@Abdullah We need to find intersection points of the locus with the line $y=x+c$. See please better your problem.
$endgroup$
– Michael Rozenberg
Jan 22 at 10:57
$begingroup$
I see, but the answer doesn't matches it.
$endgroup$
– Abdullah
Jan 22 at 12:48
|
show 1 more comment
$begingroup$
Thanks, but I'm afraid it's wrong. The correct answer is only relevant to the first part.
$endgroup$
– Abdullah
Jan 22 at 7:33
$begingroup$
@Abdullah I don't agree with you. See please better what we need to find.
$endgroup$
– Michael Rozenberg
Jan 22 at 7:52
$begingroup$
to find the locus why would we need to solve the system of equations, that would give us the point right? Correct me if I'm wrong Michael, much appreciated.
$endgroup$
– Abdullah
Jan 22 at 9:29
$begingroup$
@Abdullah We need to find intersection points of the locus with the line $y=x+c$. See please better your problem.
$endgroup$
– Michael Rozenberg
Jan 22 at 10:57
$begingroup$
I see, but the answer doesn't matches it.
$endgroup$
– Abdullah
Jan 22 at 12:48
$begingroup$
Thanks, but I'm afraid it's wrong. The correct answer is only relevant to the first part.
$endgroup$
– Abdullah
Jan 22 at 7:33
$begingroup$
Thanks, but I'm afraid it's wrong. The correct answer is only relevant to the first part.
$endgroup$
– Abdullah
Jan 22 at 7:33
$begingroup$
@Abdullah I don't agree with you. See please better what we need to find.
$endgroup$
– Michael Rozenberg
Jan 22 at 7:52
$begingroup$
@Abdullah I don't agree with you. See please better what we need to find.
$endgroup$
– Michael Rozenberg
Jan 22 at 7:52
$begingroup$
to find the locus why would we need to solve the system of equations, that would give us the point right? Correct me if I'm wrong Michael, much appreciated.
$endgroup$
– Abdullah
Jan 22 at 9:29
$begingroup$
to find the locus why would we need to solve the system of equations, that would give us the point right? Correct me if I'm wrong Michael, much appreciated.
$endgroup$
– Abdullah
Jan 22 at 9:29
$begingroup$
@Abdullah We need to find intersection points of the locus with the line $y=x+c$. See please better your problem.
$endgroup$
– Michael Rozenberg
Jan 22 at 10:57
$begingroup$
@Abdullah We need to find intersection points of the locus with the line $y=x+c$. See please better your problem.
$endgroup$
– Michael Rozenberg
Jan 22 at 10:57
$begingroup$
I see, but the answer doesn't matches it.
$endgroup$
– Abdullah
Jan 22 at 12:48
$begingroup$
I see, but the answer doesn't matches it.
$endgroup$
– Abdullah
Jan 22 at 12:48
|
show 1 more comment
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