Mixing
If $p,q,r$ be lengths of perpendiculars from vertices of triangle $ABC$ on any line, prove...
Let :
$$A:=(x_1,y_1),$$
$$B:=(x_2,y_2),$$
$$C:=(x_3,y_3)$$
be the vertices of the triangle $ABC$. Consider an arbitrary straight line in perpendicular form $xcos theta + ysin theta - t = 0$. Then the lengths of the perpendiculars from the vertices of the triangle are:
$$p=x_1cos theta + y_1sin theta - t,$$
$$q=x_2cos theta + y_2sin theta - t,$$
$$r=x_3cos theta + y_3sin theta - t.$$
Moreover, the lengths of the sides of the triangle are:
$$a^2=(x_3-x_2)^2+(y_3-y_2)^2,$$
$$b^2=(x_1-x_3)^2+(y_1-y_3)^2,$$
$$c^2=(x_2-x_1)^2+(y_2-y_1)^2.$$
Using the above values I tried to evaluate the LHS to prove the desired result, but that way is too tedious. Can anyone suggest me a better proof?
geometry euclidean-geometry analytic-geometry triangle area
asked Nov 21 '18 at 7:56
Awe Kumar Jha
38613
|
show 1 more comment
Let :
$$A:=(x_1,y_1),$$
$$B:=(x_2,y_2),$$
$$C:=(x_3,y_3)$$
be the vertices of the triangle $ABC$. Consider an arbitrary straight line in perpendicular form $xcos theta + ysin theta - t = 0$. Then the lengths of the perpendiculars from the vertices of the triangle are:
$$p=x_1cos theta + y_1sin theta - t,$$
$$q=x_2cos theta + y_2sin theta - t,$$
$$r=x_3cos theta + y_3sin theta - t.$$
Moreover, the lengths of the sides of the triangle are:
$$a^2=(x_3-x_2)^2+(y_3-y_2)^2,$$
$$b^2=(x_1-x_3)^2+(y_1-y_3)^2,$$
$$c^2=(x_2-x_1)^2+(y_2-y_1)^2.$$
Using the above values I tried to evaluate the LHS to prove the desired result, but that way is too tedious. Can anyone suggest me a better proof?
geometry euclidean-geometry analytic-geometry triangle area
asked Nov 21 '18 at 7:56
Awe Kumar Jha
38613
Here $∆$ is the area of the triangle.
– Awe Kumar Jha
Nov 21 '18 at 7:59
I think you are probably using "signed lengths," and if so, then this should be clearly stated in the problem statement. That is, you set one direction to be positive and the other negative. If the straight line cuts through the triangle and you are not using signed length, then I am afraid that the claim is false.
– Batominovski
Nov 21 '18 at 8:32
@Batominovski, the actual problem (as mentioned in the textbook) considers any straight line and tells nothing about whether the lengths are signed or not.
– Awe Kumar Jha
Nov 21 '18 at 8:57
Then, I suggest that you added "signed lengths" anyhow because the problem is false without using signed lengths.
– Batominovski
Nov 21 '18 at 8:59
2
@JeanMarie This is because the perpendiculars are not the altitudes of the triangle. Let me restate the problem. There is a triangle $ABC$ and there is a fixed straight line $l$. Project $A$, $B$, and $C$ orthogonally onto $l$ to get the points $X$, $Y$, and $Z$, respectively. Then, $p$, $q$, and $r$ are the signed lengths $AX$, $BY$, and $CZ$.
– Batominovski
Nov 21 '18 at 9:38
|
show 1 more comment
Let :
$$A:=(x_1,y_1),$$
$$B:=(x_2,y_2),$$
$$C:=(x_3,y_3)$$
be the vertices of the triangle $ABC$. Consider an arbitrary straight line in perpendicular form $xcos theta + ysin theta - t = 0$. Then the lengths of the perpendiculars from the vertices of the triangle are:
$$p=x_1cos theta + y_1sin theta - t,$$
$$q=x_2cos theta + y_2sin theta - t,$$
$$r=x_3cos theta + y_3sin theta - t.$$
Moreover, the lengths of the sides of the triangle are:
$$a^2=(x_3-x_2)^2+(y_3-y_2)^2,$$
$$b^2=(x_1-x_3)^2+(y_1-y_3)^2,$$
$$c^2=(x_2-x_1)^2+(y_2-y_1)^2.$$
Using the above values I tried to evaluate the LHS to prove the desired result, but that way is too tedious. Can anyone suggest me a better proof?
geometry euclidean-geometry analytic-geometry triangle area
asked Nov 21 '18 at 7:56
Awe Kumar Jha
38613
Let :
$$A:=(x_1,y_1),$$
$$B:=(x_2,y_2),$$
$$C:=(x_3,y_3)$$
be the vertices of the triangle $ABC$. Consider an arbitrary straight line in perpendicular form $xcos theta + ysin theta - t = 0$. Then the lengths of the perpendiculars from the vertices of the triangle are:
$$p=x_1cos theta + y_1sin theta - t,$$
$$q=x_2cos theta + y_2sin theta - t,$$
$$r=x_3cos theta + y_3sin theta - t.$$
Moreover, the lengths of the sides of the triangle are:
$$a^2=(x_3-x_2)^2+(y_3-y_2)^2,$$
$$b^2=(x_1-x_3)^2+(y_1-y_3)^2,$$
$$c^2=(x_2-x_1)^2+(y_2-y_1)^2.$$
Using the above values I tried to evaluate the LHS to prove the desired result, but that way is too tedious. Can anyone suggest me a better proof?
geometry euclidean-geometry analytic-geometry triangle area
geometry euclidean-geometry analytic-geometry triangle area
asked Nov 21 '18 at 7:56
Awe Kumar Jha
38613
asked Nov 21 '18 at 7:56
Awe Kumar Jha
38613
edited Nov 21 '18 at 14:37
asked Nov 21 '18 at 7:56
Awe Kumar Jha
38613
asked Nov 21 '18 at 7:56
Awe Kumar Jha
38613
asked Nov 21 '18 at 7:56
Awe Kumar Jha
38613
38613
Here $∆$ is the area of the triangle.
– Awe Kumar Jha
Nov 21 '18 at 7:59
I think you are probably using "signed lengths," and if so, then this should be clearly stated in the problem statement. That is, you set one direction to be positive and the other negative. If the straight line cuts through the triangle and you are not using signed length, then I am afraid that the claim is false.
– Batominovski
Nov 21 '18 at 8:32
@Batominovski, the actual problem (as mentioned in the textbook) considers any straight line and tells nothing about whether the lengths are signed or not.
– Awe Kumar Jha
Nov 21 '18 at 8:57
Then, I suggest that you added "signed lengths" anyhow because the problem is false without using signed lengths.
– Batominovski
Nov 21 '18 at 8:59
2
@JeanMarie This is because the perpendiculars are not the altitudes of the triangle. Let me restate the problem. There is a triangle $ABC$ and there is a fixed straight line $l$. Project $A$, $B$, and $C$ orthogonally onto $l$ to get the points $X$, $Y$, and $Z$, respectively. Then, $p$, $q$, and $r$ are the signed lengths $AX$, $BY$, and $CZ$.
– Batominovski
Nov 21 '18 at 9:38
|
show 1 more comment
Here $∆$ is the area of the triangle.
– Awe Kumar Jha
Nov 21 '18 at 7:59
I think you are probably using "signed lengths," and if so, then this should be clearly stated in the problem statement. That is, you set one direction to be positive and the other negative. If the straight line cuts through the triangle and you are not using signed length, then I am afraid that the claim is false.
– Batominovski
Nov 21 '18 at 8:32
@Batominovski, the actual problem (as mentioned in the textbook) considers any straight line and tells nothing about whether the lengths are signed or not.
– Awe Kumar Jha
Nov 21 '18 at 8:57
Then, I suggest that you added "signed lengths" anyhow because the problem is false without using signed lengths.
– Batominovski
Nov 21 '18 at 8:59
2
@JeanMarie This is because the perpendiculars are not the altitudes of the triangle. Let me restate the problem. There is a triangle $ABC$ and there is a fixed straight line $l$. Project $A$, $B$, and $C$ orthogonally onto $l$ to get the points $X$, $Y$, and $Z$, respectively. Then, $p$, $q$, and $r$ are the signed lengths $AX$, $BY$, and $CZ$.
– Batominovski
Nov 21 '18 at 9:38
Here $∆$ is the area of the triangle.
– Awe Kumar Jha
Nov 21 '18 at 7:59
Here $∆$ is the area of the triangle.
– Awe Kumar Jha
Nov 21 '18 at 7:59
I think you are probably using "signed lengths," and if so, then this should be clearly stated in the problem statement. That is, you set one direction to be positive and the other negative. If the straight line cuts through the triangle and you are not using signed length, then I am afraid that the claim is false.
– Batominovski
Nov 21 '18 at 8:32
I think you are probably using "signed lengths," and if so, then this should be clearly stated in the problem statement. That is, you set one direction to be positive and the other negative. If the straight line cuts through the triangle and you are not using signed length, then I am afraid that the claim is false.
– Batominovski
Nov 21 '18 at 8:32
@Batominovski, the actual problem (as mentioned in the textbook) considers any straight line and tells nothing about whether the lengths are signed or not.
– Awe Kumar Jha
Nov 21 '18 at 8:57
@Batominovski, the actual problem (as mentioned in the textbook) considers any straight line and tells nothing about whether the lengths are signed or not.
– Awe Kumar Jha
Nov 21 '18 at 8:57
Then, I suggest that you added "signed lengths" anyhow because the problem is false without using signed lengths.
– Batominovski
Nov 21 '18 at 8:59
Then, I suggest that you added "signed lengths" anyhow because the problem is false without using signed lengths.
– Batominovski
Nov 21 '18 at 8:59
2
2
@JeanMarie This is because the perpendiculars are not the altitudes of the triangle. Let me restate the problem. There is a triangle $ABC$ and there is a fixed straight line $l$. Project $A$, $B$, and $C$ orthogonally onto $l$ to get the points $X$, $Y$, and $Z$, respectively. Then, $p$, $q$, and $r$ are the signed lengths $AX$, $BY$, and $CZ$.
– Batominovski
Nov 21 '18 at 9:38
@JeanMarie This is because the perpendiculars are not the altitudes of the triangle. Let me restate the problem. There is a triangle $ABC$ and there is a fixed straight line $l$. Project $A$, $B$, and $C$ orthogonally onto $l$ to get the points $X$, $Y$, and $Z$, respectively. Then, $p$, $q$, and $r$ are the signed lengths $AX$, $BY$, and $CZ$.
– Batominovski
Nov 21 '18 at 9:38
|
show 1 more comment
1 Answer
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Hint. Let $$f(u,v,w):=(v-w)^2(p-q)(p-r)+(w-u)^2(q-r)(q-p)+(u-v)^2(r-p)(r-q),.$$ Observe that
$$begin{align}f(u,v,w)&=big((v-w)pbig)^2+big((w-u)qbig)^2+big((u-v)rbig)^2\&phantom{aaaaa}+2,big((w-u)qbig),big((u-v)rbig)+2,big((u-v)rbig),big((v-w)pbig)+2,big((v-w)pbig),big((w-u)qbig),.end{align}$$
Therefore,
$$f(u,v,w)=Big(detbig(M(u,v,w)big)Big)^2,,$$ where $$M(u,v,w):=begin{bmatrix}1&p&u\1&q&v\1&r&wend{bmatrix},.$$
answered Nov 21 '18 at 9:03
Batominovski
33.9k33292
I think the parameters $u,v,w$ are the three vertices of the triangle.
– Awe Kumar Jha
Nov 21 '18 at 10:53
3
Not quite. Treat the straight line as the $x$-axis. Then the coordinates of the points $A$, $B$, and $C$ are $(u,p)$, $(v,q)$, and $(w,r)$, respectively. Then your expression equals $f(p,q,r)+f(u,v,w)$, but of course, $f(p,q,r)=0$.
– Batominovski
Nov 21 '18 at 11:19
I don't see how the parameters can be employed, for defining $v-w:=a$,$w-u:=b$,$u-v:=c$ gives an unexpected result , $a+b+c=0$. Moreover, we need to eliminate $p,q,r$ from the expression of $det A$.
– Awe Kumar Jha
Nov 21 '18 at 13:23
1
That is not what I did. You note that $a^2=(q-r)^2+(v-w)^2$, $b^2=(r-p)^2+(w-u)^2$, and $c^2=(p-q)^2+(u-v)^2$. And the absolute value of $detbig(M(u,v,w)big)$ is precisely twice the area of the triangle with vertices $A(u,p)$, $B(v,q)$, and $C(w,r)$. (I changed the name of the matrix to $M$ so that we don't confuse it with the point $A$.)
– Batominovski
Nov 21 '18 at 18:32
add a comment |
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Hint. Let $$f(u,v,w):=(v-w)^2(p-q)(p-r)+(w-u)^2(q-r)(q-p)+(u-v)^2(r-p)(r-q),.$$ Observe that
$$begin{align}f(u,v,w)&=big((v-w)pbig)^2+big((w-u)qbig)^2+big((u-v)rbig)^2\&phantom{aaaaa}+2,big((w-u)qbig),big((u-v)rbig)+2,big((u-v)rbig),big((v-w)pbig)+2,big((v-w)pbig),big((w-u)qbig),.end{align}$$
Therefore,
$$f(u,v,w)=Big(detbig(M(u,v,w)big)Big)^2,,$$ where $$M(u,v,w):=begin{bmatrix}1&p&u\1&q&v\1&r&wend{bmatrix},.$$
answered Nov 21 '18 at 9:03
Batominovski
33.9k33292
I think the parameters $u,v,w$ are the three vertices of the triangle.
– Awe Kumar Jha
Nov 21 '18 at 10:53
3
Not quite. Treat the straight line as the $x$-axis. Then the coordinates of the points $A$, $B$, and $C$ are $(u,p)$, $(v,q)$, and $(w,r)$, respectively. Then your expression equals $f(p,q,r)+f(u,v,w)$, but of course, $f(p,q,r)=0$.
– Batominovski
Nov 21 '18 at 11:19
I don't see how the parameters can be employed, for defining $v-w:=a$,$w-u:=b$,$u-v:=c$ gives an unexpected result , $a+b+c=0$. Moreover, we need to eliminate $p,q,r$ from the expression of $det A$.
– Awe Kumar Jha
Nov 21 '18 at 13:23
1
That is not what I did. You note that $a^2=(q-r)^2+(v-w)^2$, $b^2=(r-p)^2+(w-u)^2$, and $c^2=(p-q)^2+(u-v)^2$. And the absolute value of $detbig(M(u,v,w)big)$ is precisely twice the area of the triangle with vertices $A(u,p)$, $B(v,q)$, and $C(w,r)$. (I changed the name of the matrix to $M$ so that we don't confuse it with the point $A$.)
– Batominovski
Nov 21 '18 at 18:32
add a comment |
Hint. Let $$f(u,v,w):=(v-w)^2(p-q)(p-r)+(w-u)^2(q-r)(q-p)+(u-v)^2(r-p)(r-q),.$$ Observe that
$$begin{align}f(u,v,w)&=big((v-w)pbig)^2+big((w-u)qbig)^2+big((u-v)rbig)^2\&phantom{aaaaa}+2,big((w-u)qbig),big((u-v)rbig)+2,big((u-v)rbig),big((v-w)pbig)+2,big((v-w)pbig),big((w-u)qbig),.end{align}$$
Therefore,
$$f(u,v,w)=Big(detbig(M(u,v,w)big)Big)^2,,$$ where $$M(u,v,w):=begin{bmatrix}1&p&u\1&q&v\1&r&wend{bmatrix},.$$
answered Nov 21 '18 at 9:03
Batominovski
33.9k33292
I think the parameters $u,v,w$ are the three vertices of the triangle.
– Awe Kumar Jha
Nov 21 '18 at 10:53
3
Not quite. Treat the straight line as the $x$-axis. Then the coordinates of the points $A$, $B$, and $C$ are $(u,p)$, $(v,q)$, and $(w,r)$, respectively. Then your expression equals $f(p,q,r)+f(u,v,w)$, but of course, $f(p,q,r)=0$.
– Batominovski
Nov 21 '18 at 11:19
I don't see how the parameters can be employed, for defining $v-w:=a$,$w-u:=b$,$u-v:=c$ gives an unexpected result , $a+b+c=0$. Moreover, we need to eliminate $p,q,r$ from the expression of $det A$.
– Awe Kumar Jha
Nov 21 '18 at 13:23
1
That is not what I did. You note that $a^2=(q-r)^2+(v-w)^2$, $b^2=(r-p)^2+(w-u)^2$, and $c^2=(p-q)^2+(u-v)^2$. And the absolute value of $detbig(M(u,v,w)big)$ is precisely twice the area of the triangle with vertices $A(u,p)$, $B(v,q)$, and $C(w,r)$. (I changed the name of the matrix to $M$ so that we don't confuse it with the point $A$.)
– Batominovski
Nov 21 '18 at 18:32
add a comment |
Hint. Let $$f(u,v,w):=(v-w)^2(p-q)(p-r)+(w-u)^2(q-r)(q-p)+(u-v)^2(r-p)(r-q),.$$ Observe that
$$begin{align}f(u,v,w)&=big((v-w)pbig)^2+big((w-u)qbig)^2+big((u-v)rbig)^2\&phantom{aaaaa}+2,big((w-u)qbig),big((u-v)rbig)+2,big((u-v)rbig),big((v-w)pbig)+2,big((v-w)pbig),big((w-u)qbig),.end{align}$$
Therefore,
$$f(u,v,w)=Big(detbig(M(u,v,w)big)Big)^2,,$$ where $$M(u,v,w):=begin{bmatrix}1&p&u\1&q&v\1&r&wend{bmatrix},.$$
answered Nov 21 '18 at 9:03
Batominovski
33.9k33292
Hint. Let $$f(u,v,w):=(v-w)^2(p-q)(p-r)+(w-u)^2(q-r)(q-p)+(u-v)^2(r-p)(r-q),.$$ Observe that
$$begin{align}f(u,v,w)&=big((v-w)pbig)^2+big((w-u)qbig)^2+big((u-v)rbig)^2\&phantom{aaaaa}+2,big((w-u)qbig),big((u-v)rbig)+2,big((u-v)rbig),big((v-w)pbig)+2,big((v-w)pbig),big((w-u)qbig),.end{align}$$
Therefore,
$$f(u,v,w)=Big(detbig(M(u,v,w)big)Big)^2,,$$ where $$M(u,v,w):=begin{bmatrix}1&p&u\1&q&v\1&r&wend{bmatrix},.$$
answered Nov 21 '18 at 9:03
Batominovski
33.9k33292
edited Nov 21 '18 at 18:34
answered Nov 21 '18 at 9:03
Batominovski
33.9k33292
answered Nov 21 '18 at 9:03
Batominovski
33.9k33292
answered Nov 21 '18 at 9:03
Batominovski
33.9k33292
33.9k33292
I think the parameters $u,v,w$ are the three vertices of the triangle.
– Awe Kumar Jha
Nov 21 '18 at 10:53
3
Not quite. Treat the straight line as the $x$-axis. Then the coordinates of the points $A$, $B$, and $C$ are $(u,p)$, $(v,q)$, and $(w,r)$, respectively. Then your expression equals $f(p,q,r)+f(u,v,w)$, but of course, $f(p,q,r)=0$.
– Batominovski
Nov 21 '18 at 11:19
I don't see how the parameters can be employed, for defining $v-w:=a$,$w-u:=b$,$u-v:=c$ gives an unexpected result , $a+b+c=0$. Moreover, we need to eliminate $p,q,r$ from the expression of $det A$.
– Awe Kumar Jha
Nov 21 '18 at 13:23
1
That is not what I did. You note that $a^2=(q-r)^2+(v-w)^2$, $b^2=(r-p)^2+(w-u)^2$, and $c^2=(p-q)^2+(u-v)^2$. And the absolute value of $detbig(M(u,v,w)big)$ is precisely twice the area of the triangle with vertices $A(u,p)$, $B(v,q)$, and $C(w,r)$. (I changed the name of the matrix to $M$ so that we don't confuse it with the point $A$.)
– Batominovski
Nov 21 '18 at 18:32
add a comment |
I think the parameters $u,v,w$ are the three vertices of the triangle.
– Awe Kumar Jha
Nov 21 '18 at 10:53
3
Not quite. Treat the straight line as the $x$-axis. Then the coordinates of the points $A$, $B$, and $C$ are $(u,p)$, $(v,q)$, and $(w,r)$, respectively. Then your expression equals $f(p,q,r)+f(u,v,w)$, but of course, $f(p,q,r)=0$.
– Batominovski
Nov 21 '18 at 11:19
I don't see how the parameters can be employed, for defining $v-w:=a$,$w-u:=b$,$u-v:=c$ gives an unexpected result , $a+b+c=0$. Moreover, we need to eliminate $p,q,r$ from the expression of $det A$.
– Awe Kumar Jha
Nov 21 '18 at 13:23
1
That is not what I did. You note that $a^2=(q-r)^2+(v-w)^2$, $b^2=(r-p)^2+(w-u)^2$, and $c^2=(p-q)^2+(u-v)^2$. And the absolute value of $detbig(M(u,v,w)big)$ is precisely twice the area of the triangle with vertices $A(u,p)$, $B(v,q)$, and $C(w,r)$. (I changed the name of the matrix to $M$ so that we don't confuse it with the point $A$.)
– Batominovski
Nov 21 '18 at 18:32
I think the parameters $u,v,w$ are the three vertices of the triangle.
– Awe Kumar Jha
Nov 21 '18 at 10:53
I think the parameters $u,v,w$ are the three vertices of the triangle.
– Awe Kumar Jha
Nov 21 '18 at 10:53
3
3
Not quite. Treat the straight line as the $x$-axis. Then the coordinates of the points $A$, $B$, and $C$ are $(u,p)$, $(v,q)$, and $(w,r)$, respectively. Then your expression equals $f(p,q,r)+f(u,v,w)$, but of course, $f(p,q,r)=0$.
– Batominovski
Nov 21 '18 at 11:19
Not quite. Treat the straight line as the $x$-axis. Then the coordinates of the points $A$, $B$, and $C$ are $(u,p)$, $(v,q)$, and $(w,r)$, respectively. Then your expression equals $f(p,q,r)+f(u,v,w)$, but of course, $f(p,q,r)=0$.
– Batominovski
Nov 21 '18 at 11:19
I don't see how the parameters can be employed, for defining $v-w:=a$,$w-u:=b$,$u-v:=c$ gives an unexpected result , $a+b+c=0$. Moreover, we need to eliminate $p,q,r$ from the expression of $det A$.
– Awe Kumar Jha
Nov 21 '18 at 13:23
I don't see how the parameters can be employed, for defining $v-w:=a$,$w-u:=b$,$u-v:=c$ gives an unexpected result , $a+b+c=0$. Moreover, we need to eliminate $p,q,r$ from the expression of $det A$.
– Awe Kumar Jha
Nov 21 '18 at 13:23
1
1
That is not what I did. You note that $a^2=(q-r)^2+(v-w)^2$, $b^2=(r-p)^2+(w-u)^2$, and $c^2=(p-q)^2+(u-v)^2$. And the absolute value of $detbig(M(u,v,w)big)$ is precisely twice the area of the triangle with vertices $A(u,p)$, $B(v,q)$, and $C(w,r)$. (I changed the name of the matrix to $M$ so that we don't confuse it with the point $A$.)
– Batominovski
Nov 21 '18 at 18:32
That is not what I did. You note that $a^2=(q-r)^2+(v-w)^2$, $b^2=(r-p)^2+(w-u)^2$, and $c^2=(p-q)^2+(u-v)^2$. And the absolute value of $detbig(M(u,v,w)big)$ is precisely twice the area of the triangle with vertices $A(u,p)$, $B(v,q)$, and $C(w,r)$. (I changed the name of the matrix to $M$ so that we don't confuse it with the point $A$.)
– Batominovski
Nov 21 '18 at 18:32
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Here $∆$ is the area of the triangle.
– Awe Kumar Jha
Nov 21 '18 at 7:59
I think you are probably using "signed lengths," and if so, then this should be clearly stated in the problem statement. That is, you set one direction to be positive and the other negative. If the straight line cuts through the triangle and you are not using signed length, then I am afraid that the claim is false.
– Batominovski
Nov 21 '18 at 8:32
@Batominovski, the actual problem (as mentioned in the textbook) considers any straight line and tells nothing about whether the lengths are signed or not.
– Awe Kumar Jha
Nov 21 '18 at 8:57
Then, I suggest that you added "signed lengths" anyhow because the problem is false without using signed lengths.
– Batominovski
Nov 21 '18 at 8:59
2
@JeanMarie This is because the perpendiculars are not the altitudes of the triangle. Let me restate the problem. There is a triangle $ABC$ and there is a fixed straight line $l$. Project $A$, $B$, and $C$ orthogonally onto $l$ to get the points $X$, $Y$, and $Z$, respectively. Then, $p$, $q$, and $r$ are the signed lengths $AX$, $BY$, and $CZ$.
– Batominovski
Nov 21 '18 at 9:38