Can you please explain what the “symmetric form” of a line signifies?
$begingroup$
This is the "symmetric form" of a line:
$$frac{x-x_1}{sintheta}=frac{y-y_1}{costheta}$$
where $theta$ is inclination of the line.
I know the equation, but it seems to me like the point-slope form of the line. Can you please tell me why it's called a symmetric form?
geometry
$endgroup$
add a comment |
$begingroup$
This is the "symmetric form" of a line:
$$frac{x-x_1}{sintheta}=frac{y-y_1}{costheta}$$
where $theta$ is inclination of the line.
I know the equation, but it seems to me like the point-slope form of the line. Can you please tell me why it's called a symmetric form?
geometry
$endgroup$
$begingroup$
Can you include the form in your question?
$endgroup$
– B.Swan
Nov 25 '18 at 7:04
$begingroup$
yep added the equation
$endgroup$
– Jyothi Krishna Gudi
Nov 25 '18 at 7:26
2
$begingroup$
Presumably it's called that because $(x, x_1)$ and $(y, y_1)$ play a nearly symmetric role. By contrast, for example, in slope-intercept form, i.e., $y = m x + b$, one variable is expressed in terms of the other.
$endgroup$
– Travis
Nov 25 '18 at 7:48
$begingroup$
It is called the "symmetric form" because the equation has has fractions on both side with corresponding values in the same places. The symmetry describes how the equation looks, not how the function behaves.
$endgroup$
– B.Swan
Nov 25 '18 at 7:52
1
$begingroup$
I would rather write it in the form $$(x-x_1)costheta= (y-y_1)sintheta.$$ That way I don't have to deal with attempts to divide by zero should the line be horizontal or vertical.
$endgroup$
– Jyrki Lahtonen
Nov 25 '18 at 8:55
add a comment |
$begingroup$
This is the "symmetric form" of a line:
$$frac{x-x_1}{sintheta}=frac{y-y_1}{costheta}$$
where $theta$ is inclination of the line.
I know the equation, but it seems to me like the point-slope form of the line. Can you please tell me why it's called a symmetric form?
geometry
$endgroup$
This is the "symmetric form" of a line:
$$frac{x-x_1}{sintheta}=frac{y-y_1}{costheta}$$
where $theta$ is inclination of the line.
I know the equation, but it seems to me like the point-slope form of the line. Can you please tell me why it's called a symmetric form?
geometry
geometry
edited Nov 25 '18 at 7:36
Blue
47.8k870152
47.8k870152
asked Nov 25 '18 at 6:52
Jyothi Krishna GudiJyothi Krishna Gudi
44
44
$begingroup$
Can you include the form in your question?
$endgroup$
– B.Swan
Nov 25 '18 at 7:04
$begingroup$
yep added the equation
$endgroup$
– Jyothi Krishna Gudi
Nov 25 '18 at 7:26
2
$begingroup$
Presumably it's called that because $(x, x_1)$ and $(y, y_1)$ play a nearly symmetric role. By contrast, for example, in slope-intercept form, i.e., $y = m x + b$, one variable is expressed in terms of the other.
$endgroup$
– Travis
Nov 25 '18 at 7:48
$begingroup$
It is called the "symmetric form" because the equation has has fractions on both side with corresponding values in the same places. The symmetry describes how the equation looks, not how the function behaves.
$endgroup$
– B.Swan
Nov 25 '18 at 7:52
1
$begingroup$
I would rather write it in the form $$(x-x_1)costheta= (y-y_1)sintheta.$$ That way I don't have to deal with attempts to divide by zero should the line be horizontal or vertical.
$endgroup$
– Jyrki Lahtonen
Nov 25 '18 at 8:55
add a comment |
$begingroup$
Can you include the form in your question?
$endgroup$
– B.Swan
Nov 25 '18 at 7:04
$begingroup$
yep added the equation
$endgroup$
– Jyothi Krishna Gudi
Nov 25 '18 at 7:26
2
$begingroup$
Presumably it's called that because $(x, x_1)$ and $(y, y_1)$ play a nearly symmetric role. By contrast, for example, in slope-intercept form, i.e., $y = m x + b$, one variable is expressed in terms of the other.
$endgroup$
– Travis
Nov 25 '18 at 7:48
$begingroup$
It is called the "symmetric form" because the equation has has fractions on both side with corresponding values in the same places. The symmetry describes how the equation looks, not how the function behaves.
$endgroup$
– B.Swan
Nov 25 '18 at 7:52
1
$begingroup$
I would rather write it in the form $$(x-x_1)costheta= (y-y_1)sintheta.$$ That way I don't have to deal with attempts to divide by zero should the line be horizontal or vertical.
$endgroup$
– Jyrki Lahtonen
Nov 25 '18 at 8:55
$begingroup$
Can you include the form in your question?
$endgroup$
– B.Swan
Nov 25 '18 at 7:04
$begingroup$
Can you include the form in your question?
$endgroup$
– B.Swan
Nov 25 '18 at 7:04
$begingroup$
yep added the equation
$endgroup$
– Jyothi Krishna Gudi
Nov 25 '18 at 7:26
$begingroup$
yep added the equation
$endgroup$
– Jyothi Krishna Gudi
Nov 25 '18 at 7:26
2
2
$begingroup$
Presumably it's called that because $(x, x_1)$ and $(y, y_1)$ play a nearly symmetric role. By contrast, for example, in slope-intercept form, i.e., $y = m x + b$, one variable is expressed in terms of the other.
$endgroup$
– Travis
Nov 25 '18 at 7:48
$begingroup$
Presumably it's called that because $(x, x_1)$ and $(y, y_1)$ play a nearly symmetric role. By contrast, for example, in slope-intercept form, i.e., $y = m x + b$, one variable is expressed in terms of the other.
$endgroup$
– Travis
Nov 25 '18 at 7:48
$begingroup$
It is called the "symmetric form" because the equation has has fractions on both side with corresponding values in the same places. The symmetry describes how the equation looks, not how the function behaves.
$endgroup$
– B.Swan
Nov 25 '18 at 7:52
$begingroup$
It is called the "symmetric form" because the equation has has fractions on both side with corresponding values in the same places. The symmetry describes how the equation looks, not how the function behaves.
$endgroup$
– B.Swan
Nov 25 '18 at 7:52
1
1
$begingroup$
I would rather write it in the form $$(x-x_1)costheta= (y-y_1)sintheta.$$ That way I don't have to deal with attempts to divide by zero should the line be horizontal or vertical.
$endgroup$
– Jyrki Lahtonen
Nov 25 '18 at 8:55
$begingroup$
I would rather write it in the form $$(x-x_1)costheta= (y-y_1)sintheta.$$ That way I don't have to deal with attempts to divide by zero should the line be horizontal or vertical.
$endgroup$
– Jyrki Lahtonen
Nov 25 '18 at 8:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is very much like the point-slope form of the line, which could be written
$$frac{y-y_1}{x-x_1}=m tag{1}$$
where $m$ is the line's slope. But, since $m = tantheta$, where $theta$ is the line's angle on inclination, and since $tantheta = sintheta/costheta$, the symmetric form
$$frac{x-x_1}{costheta} = frac{y-y_1}{sintheta} tag{2}$$
is (mostly) equivalent to point-slope form via cross-multiplication.
Separating the $x$ and $y$ elements "symmetrizes" the equation by isolating the line's horizontal "run" components ($x-x_1$ and $costheta$) from vertical "rise" components ($y-y_1$ and $sintheta$). The form is also symmetric in that now both horizontal ($theta = 0^circ$) as well as vertical ($theta=90^circ$) lines are problematic, since both directions lead to vanishing denominators. (That's why I used wrote "(mostly) equivalent" above: point-slope form is only problematic in one direction.)
It may seem a bit of form-over-substance here, but if we recognize $(costheta,sintheta)$ as the line's "direction vector", then this form generalizes nicely to higher dimensions. For instance, in $3$D, we have
$$frac{x-x_1}{a}=frac{y-y_1}{b}=frac{z-z_1}{c} tag{3}$$
where $(a,b,c)$ is a direction vector of the line (with "run", "rise", and ... um ... "rother" components).
By the way, since the fractions in $(2)$ (and $(3)$) are equal to each other, they're equal to a common ratio that we can call, say, $t$. That means we can re-express line equations by writing
$$x = x_1 + tcostheta qquad y = y_1 + tsintheta tag{4}$$
or, more compactly,
$$(x,y) = (x_1,y_1) + t,(costheta,sintheta) tag{5}$$
or, even-more-compactly,
$$p = p_1 + t,v tag{6}$$
This is the parametric form of the line ($t$ is the "parameter"). Here, too, there's conceptual separated, since the calculations are done component-wise, but we get to see those components in meaningful groups: variable point $p=(x,y)$, given point $p_1=(x_1,y_1)$, direction vector $v=(costheta,sintheta)$. This is effectively a "flattened", ratio-free version of the symmetric form.
It may be safe to say that one is far more likely to see $(5)$ or $(6)$ than $(2)$, since the non-ratio forms cleverly avoid the zero-denominator problem by not having denominators at all. (You might see $(2)$ or $(3)$ in a context where an author doesn't want to bother wasting a variable on an unused parameter.)
$endgroup$
add a comment |
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$begingroup$
This is very much like the point-slope form of the line, which could be written
$$frac{y-y_1}{x-x_1}=m tag{1}$$
where $m$ is the line's slope. But, since $m = tantheta$, where $theta$ is the line's angle on inclination, and since $tantheta = sintheta/costheta$, the symmetric form
$$frac{x-x_1}{costheta} = frac{y-y_1}{sintheta} tag{2}$$
is (mostly) equivalent to point-slope form via cross-multiplication.
Separating the $x$ and $y$ elements "symmetrizes" the equation by isolating the line's horizontal "run" components ($x-x_1$ and $costheta$) from vertical "rise" components ($y-y_1$ and $sintheta$). The form is also symmetric in that now both horizontal ($theta = 0^circ$) as well as vertical ($theta=90^circ$) lines are problematic, since both directions lead to vanishing denominators. (That's why I used wrote "(mostly) equivalent" above: point-slope form is only problematic in one direction.)
It may seem a bit of form-over-substance here, but if we recognize $(costheta,sintheta)$ as the line's "direction vector", then this form generalizes nicely to higher dimensions. For instance, in $3$D, we have
$$frac{x-x_1}{a}=frac{y-y_1}{b}=frac{z-z_1}{c} tag{3}$$
where $(a,b,c)$ is a direction vector of the line (with "run", "rise", and ... um ... "rother" components).
By the way, since the fractions in $(2)$ (and $(3)$) are equal to each other, they're equal to a common ratio that we can call, say, $t$. That means we can re-express line equations by writing
$$x = x_1 + tcostheta qquad y = y_1 + tsintheta tag{4}$$
or, more compactly,
$$(x,y) = (x_1,y_1) + t,(costheta,sintheta) tag{5}$$
or, even-more-compactly,
$$p = p_1 + t,v tag{6}$$
This is the parametric form of the line ($t$ is the "parameter"). Here, too, there's conceptual separated, since the calculations are done component-wise, but we get to see those components in meaningful groups: variable point $p=(x,y)$, given point $p_1=(x_1,y_1)$, direction vector $v=(costheta,sintheta)$. This is effectively a "flattened", ratio-free version of the symmetric form.
It may be safe to say that one is far more likely to see $(5)$ or $(6)$ than $(2)$, since the non-ratio forms cleverly avoid the zero-denominator problem by not having denominators at all. (You might see $(2)$ or $(3)$ in a context where an author doesn't want to bother wasting a variable on an unused parameter.)
$endgroup$
add a comment |
$begingroup$
This is very much like the point-slope form of the line, which could be written
$$frac{y-y_1}{x-x_1}=m tag{1}$$
where $m$ is the line's slope. But, since $m = tantheta$, where $theta$ is the line's angle on inclination, and since $tantheta = sintheta/costheta$, the symmetric form
$$frac{x-x_1}{costheta} = frac{y-y_1}{sintheta} tag{2}$$
is (mostly) equivalent to point-slope form via cross-multiplication.
Separating the $x$ and $y$ elements "symmetrizes" the equation by isolating the line's horizontal "run" components ($x-x_1$ and $costheta$) from vertical "rise" components ($y-y_1$ and $sintheta$). The form is also symmetric in that now both horizontal ($theta = 0^circ$) as well as vertical ($theta=90^circ$) lines are problematic, since both directions lead to vanishing denominators. (That's why I used wrote "(mostly) equivalent" above: point-slope form is only problematic in one direction.)
It may seem a bit of form-over-substance here, but if we recognize $(costheta,sintheta)$ as the line's "direction vector", then this form generalizes nicely to higher dimensions. For instance, in $3$D, we have
$$frac{x-x_1}{a}=frac{y-y_1}{b}=frac{z-z_1}{c} tag{3}$$
where $(a,b,c)$ is a direction vector of the line (with "run", "rise", and ... um ... "rother" components).
By the way, since the fractions in $(2)$ (and $(3)$) are equal to each other, they're equal to a common ratio that we can call, say, $t$. That means we can re-express line equations by writing
$$x = x_1 + tcostheta qquad y = y_1 + tsintheta tag{4}$$
or, more compactly,
$$(x,y) = (x_1,y_1) + t,(costheta,sintheta) tag{5}$$
or, even-more-compactly,
$$p = p_1 + t,v tag{6}$$
This is the parametric form of the line ($t$ is the "parameter"). Here, too, there's conceptual separated, since the calculations are done component-wise, but we get to see those components in meaningful groups: variable point $p=(x,y)$, given point $p_1=(x_1,y_1)$, direction vector $v=(costheta,sintheta)$. This is effectively a "flattened", ratio-free version of the symmetric form.
It may be safe to say that one is far more likely to see $(5)$ or $(6)$ than $(2)$, since the non-ratio forms cleverly avoid the zero-denominator problem by not having denominators at all. (You might see $(2)$ or $(3)$ in a context where an author doesn't want to bother wasting a variable on an unused parameter.)
$endgroup$
add a comment |
$begingroup$
This is very much like the point-slope form of the line, which could be written
$$frac{y-y_1}{x-x_1}=m tag{1}$$
where $m$ is the line's slope. But, since $m = tantheta$, where $theta$ is the line's angle on inclination, and since $tantheta = sintheta/costheta$, the symmetric form
$$frac{x-x_1}{costheta} = frac{y-y_1}{sintheta} tag{2}$$
is (mostly) equivalent to point-slope form via cross-multiplication.
Separating the $x$ and $y$ elements "symmetrizes" the equation by isolating the line's horizontal "run" components ($x-x_1$ and $costheta$) from vertical "rise" components ($y-y_1$ and $sintheta$). The form is also symmetric in that now both horizontal ($theta = 0^circ$) as well as vertical ($theta=90^circ$) lines are problematic, since both directions lead to vanishing denominators. (That's why I used wrote "(mostly) equivalent" above: point-slope form is only problematic in one direction.)
It may seem a bit of form-over-substance here, but if we recognize $(costheta,sintheta)$ as the line's "direction vector", then this form generalizes nicely to higher dimensions. For instance, in $3$D, we have
$$frac{x-x_1}{a}=frac{y-y_1}{b}=frac{z-z_1}{c} tag{3}$$
where $(a,b,c)$ is a direction vector of the line (with "run", "rise", and ... um ... "rother" components).
By the way, since the fractions in $(2)$ (and $(3)$) are equal to each other, they're equal to a common ratio that we can call, say, $t$. That means we can re-express line equations by writing
$$x = x_1 + tcostheta qquad y = y_1 + tsintheta tag{4}$$
or, more compactly,
$$(x,y) = (x_1,y_1) + t,(costheta,sintheta) tag{5}$$
or, even-more-compactly,
$$p = p_1 + t,v tag{6}$$
This is the parametric form of the line ($t$ is the "parameter"). Here, too, there's conceptual separated, since the calculations are done component-wise, but we get to see those components in meaningful groups: variable point $p=(x,y)$, given point $p_1=(x_1,y_1)$, direction vector $v=(costheta,sintheta)$. This is effectively a "flattened", ratio-free version of the symmetric form.
It may be safe to say that one is far more likely to see $(5)$ or $(6)$ than $(2)$, since the non-ratio forms cleverly avoid the zero-denominator problem by not having denominators at all. (You might see $(2)$ or $(3)$ in a context where an author doesn't want to bother wasting a variable on an unused parameter.)
$endgroup$
This is very much like the point-slope form of the line, which could be written
$$frac{y-y_1}{x-x_1}=m tag{1}$$
where $m$ is the line's slope. But, since $m = tantheta$, where $theta$ is the line's angle on inclination, and since $tantheta = sintheta/costheta$, the symmetric form
$$frac{x-x_1}{costheta} = frac{y-y_1}{sintheta} tag{2}$$
is (mostly) equivalent to point-slope form via cross-multiplication.
Separating the $x$ and $y$ elements "symmetrizes" the equation by isolating the line's horizontal "run" components ($x-x_1$ and $costheta$) from vertical "rise" components ($y-y_1$ and $sintheta$). The form is also symmetric in that now both horizontal ($theta = 0^circ$) as well as vertical ($theta=90^circ$) lines are problematic, since both directions lead to vanishing denominators. (That's why I used wrote "(mostly) equivalent" above: point-slope form is only problematic in one direction.)
It may seem a bit of form-over-substance here, but if we recognize $(costheta,sintheta)$ as the line's "direction vector", then this form generalizes nicely to higher dimensions. For instance, in $3$D, we have
$$frac{x-x_1}{a}=frac{y-y_1}{b}=frac{z-z_1}{c} tag{3}$$
where $(a,b,c)$ is a direction vector of the line (with "run", "rise", and ... um ... "rother" components).
By the way, since the fractions in $(2)$ (and $(3)$) are equal to each other, they're equal to a common ratio that we can call, say, $t$. That means we can re-express line equations by writing
$$x = x_1 + tcostheta qquad y = y_1 + tsintheta tag{4}$$
or, more compactly,
$$(x,y) = (x_1,y_1) + t,(costheta,sintheta) tag{5}$$
or, even-more-compactly,
$$p = p_1 + t,v tag{6}$$
This is the parametric form of the line ($t$ is the "parameter"). Here, too, there's conceptual separated, since the calculations are done component-wise, but we get to see those components in meaningful groups: variable point $p=(x,y)$, given point $p_1=(x_1,y_1)$, direction vector $v=(costheta,sintheta)$. This is effectively a "flattened", ratio-free version of the symmetric form.
It may be safe to say that one is far more likely to see $(5)$ or $(6)$ than $(2)$, since the non-ratio forms cleverly avoid the zero-denominator problem by not having denominators at all. (You might see $(2)$ or $(3)$ in a context where an author doesn't want to bother wasting a variable on an unused parameter.)
edited Nov 25 '18 at 15:02
answered Nov 25 '18 at 7:52
BlueBlue
47.8k870152
47.8k870152
add a comment |
add a comment |
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$begingroup$
Can you include the form in your question?
$endgroup$
– B.Swan
Nov 25 '18 at 7:04
$begingroup$
yep added the equation
$endgroup$
– Jyothi Krishna Gudi
Nov 25 '18 at 7:26
2
$begingroup$
Presumably it's called that because $(x, x_1)$ and $(y, y_1)$ play a nearly symmetric role. By contrast, for example, in slope-intercept form, i.e., $y = m x + b$, one variable is expressed in terms of the other.
$endgroup$
– Travis
Nov 25 '18 at 7:48
$begingroup$
It is called the "symmetric form" because the equation has has fractions on both side with corresponding values in the same places. The symmetry describes how the equation looks, not how the function behaves.
$endgroup$
– B.Swan
Nov 25 '18 at 7:52
1
$begingroup$
I would rather write it in the form $$(x-x_1)costheta= (y-y_1)sintheta.$$ That way I don't have to deal with attempts to divide by zero should the line be horizontal or vertical.
$endgroup$
– Jyrki Lahtonen
Nov 25 '18 at 8:55