How to find a,b and c in terms of the semi perimeter (s) from this equation?
$begingroup$
I have an equation: $(s-a)/4=(s-b)/3=(s-c)/2$ , from this I'm supposed to get a,b and c (the sides of a triangle) in terms of s..
If I substitute $s=(a+b+c)/2$ in the given system of equations I get three other equations : $a+3b-2c=0$
$2a+b-c$ and $4a-3b+c=0$ . But still I don't see any quick way to find a,b,c in terms of s. Is the only way to get the them in terms of s, to actually solve for a,b and c and then get them in terms of s?
*by getting them in terms of s, I mean being able to write them as $a = (constant)*s$ etc..
linear-algebra systems-of-equations
edited Jan 2 at 12:55
Harry Peter
5,47911439
asked Dec 18 '18 at 8:02
karun mathewskarun mathews
245
$endgroup$
add a comment |
$begingroup$
I have an equation: $(s-a)/4=(s-b)/3=(s-c)/2$ , from this I'm supposed to get a,b and c (the sides of a triangle) in terms of s..
If I substitute $s=(a+b+c)/2$ in the given system of equations I get three other equations : $a+3b-2c=0$
$2a+b-c$ and $4a-3b+c=0$ . But still I don't see any quick way to find a,b,c in terms of s. Is the only way to get the them in terms of s, to actually solve for a,b and c and then get them in terms of s?
*by getting them in terms of s, I mean being able to write them as $a = (constant)*s$ etc..
linear-algebra systems-of-equations
edited Jan 2 at 12:55
Harry Peter
5,47911439
asked Dec 18 '18 at 8:02
karun mathewskarun mathews
245
$endgroup$
add a comment |
$begingroup$
I have an equation: $(s-a)/4=(s-b)/3=(s-c)/2$ , from this I'm supposed to get a,b and c (the sides of a triangle) in terms of s..
If I substitute $s=(a+b+c)/2$ in the given system of equations I get three other equations : $a+3b-2c=0$
$2a+b-c$ and $4a-3b+c=0$ . But still I don't see any quick way to find a,b,c in terms of s. Is the only way to get the them in terms of s, to actually solve for a,b and c and then get them in terms of s?
*by getting them in terms of s, I mean being able to write them as $a = (constant)*s$ etc..
linear-algebra systems-of-equations
edited Jan 2 at 12:55
Harry Peter
5,47911439
asked Dec 18 '18 at 8:02
karun mathewskarun mathews
245
$endgroup$
I have an equation: $(s-a)/4=(s-b)/3=(s-c)/2$ , from this I'm supposed to get a,b and c (the sides of a triangle) in terms of s..
If I substitute $s=(a+b+c)/2$ in the given system of equations I get three other equations : $a+3b-2c=0$
$2a+b-c$ and $4a-3b+c=0$ . But still I don't see any quick way to find a,b,c in terms of s. Is the only way to get the them in terms of s, to actually solve for a,b and c and then get them in terms of s?
*by getting them in terms of s, I mean being able to write them as $a = (constant)*s$ etc..
linear-algebra systems-of-equations
linear-algebra systems-of-equations
edited Jan 2 at 12:55
Harry Peter
5,47911439
asked Dec 18 '18 at 8:02
karun mathewskarun mathews
245
edited Jan 2 at 12:55
Harry Peter
5,47911439
asked Dec 18 '18 at 8:02
karun mathewskarun mathews
245
edited Jan 2 at 12:55
Harry Peter
5,47911439
edited Jan 2 at 12:55
Harry Peter
5,47911439
edited Jan 2 at 12:55
Harry Peter
5,47911439
5,47911439
asked Dec 18 '18 at 8:02
karun mathewskarun mathews
245
asked Dec 18 '18 at 8:02
karun mathewskarun mathews
245
asked Dec 18 '18 at 8:02
karun mathewskarun mathews
245
245
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If $frac a b=frac c d$ then $frac a b=frac c d=frac {a+c} {b+d}$ and similarly for equality for three fractions. Hence $frac {s-a} 4=frac {s-b} 3=frac {s-c} 2=frac {s-a+s-b+s-c} {4+3+2}=frac {3s-2s} 9$. Now you can get your expressions for $a,b,c$ easily.
answered Dec 18 '18 at 8:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
add a comment |
$begingroup$
Let
begin{align}
tfrac14(s-a)=tfrac13(s-b)=tfrac12(s-c)&=k
quadtext{for some }kinmathbb{R}
end{align}
Then we have a linear system of four equations in
four variables
begin{align}
left[
begin{matrix}
1&0&0&4 \
0&1&0&3\
0&0&1&2\
1&1&1&0
end{matrix}
right]
cdot
left[
begin{matrix}
a\b\c\k
end{matrix}
right]
&=
left[
begin{matrix}
s\s\s\2s
end{matrix}
right]
,
end{align}
which can be easily solved with any standard method
and has a solution in terms of $s$:
begin{align}
a &= tfrac59,s
,quad
b = tfrac23,s
,quad
c = tfrac79,s
,quad
k = tfrac19,s
.
end{align}
answered Dec 18 '18 at 9:07
g.kovg.kov
6,1821818
$endgroup$
add a comment |
$begingroup$
A variation of Kavi Rama Murthy's answer. Denote the fractions by $k$:
$$(s-a)/4=(s-b)/3=(s-c)/2=k Rightarrow \
s-a=4k;\
s-b=3k;\
s-c=2k.$$
Add them to find $k$:
$$3s-2s=9k Rightarrow k=frac{s}{9}.$$
Hence:
$$s-a=frac{4s}{9} Rightarrow a=frac{5s}{9};\
s-b=frac{3s}{9} Rightarrow b=frac{2s}{3};\
s-c=frac{2s}{9} Rightarrow c=frac{7s}{9}.$$
answered Dec 18 '18 at 13:50
farruhotafarruhota
20.5k2739
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $frac a b=frac c d$ then $frac a b=frac c d=frac {a+c} {b+d}$ and similarly for equality for three fractions. Hence $frac {s-a} 4=frac {s-b} 3=frac {s-c} 2=frac {s-a+s-b+s-c} {4+3+2}=frac {3s-2s} 9$. Now you can get your expressions for $a,b,c$ easily.
answered Dec 18 '18 at 8:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
add a comment |
$begingroup$
If $frac a b=frac c d$ then $frac a b=frac c d=frac {a+c} {b+d}$ and similarly for equality for three fractions. Hence $frac {s-a} 4=frac {s-b} 3=frac {s-c} 2=frac {s-a+s-b+s-c} {4+3+2}=frac {3s-2s} 9$. Now you can get your expressions for $a,b,c$ easily.
answered Dec 18 '18 at 8:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
add a comment |
$begingroup$
If $frac a b=frac c d$ then $frac a b=frac c d=frac {a+c} {b+d}$ and similarly for equality for three fractions. Hence $frac {s-a} 4=frac {s-b} 3=frac {s-c} 2=frac {s-a+s-b+s-c} {4+3+2}=frac {3s-2s} 9$. Now you can get your expressions for $a,b,c$ easily.
answered Dec 18 '18 at 8:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
If $frac a b=frac c d$ then $frac a b=frac c d=frac {a+c} {b+d}$ and similarly for equality for three fractions. Hence $frac {s-a} 4=frac {s-b} 3=frac {s-c} 2=frac {s-a+s-b+s-c} {4+3+2}=frac {3s-2s} 9$. Now you can get your expressions for $a,b,c$ easily.
answered Dec 18 '18 at 8:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Dec 18 '18 at 8:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Dec 18 '18 at 8:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Dec 18 '18 at 8:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
62.9k42362
add a comment |
add a comment |
$begingroup$
Let
begin{align}
tfrac14(s-a)=tfrac13(s-b)=tfrac12(s-c)&=k
quadtext{for some }kinmathbb{R}
end{align}
Then we have a linear system of four equations in
four variables
begin{align}
left[
begin{matrix}
1&0&0&4 \
0&1&0&3\
0&0&1&2\
1&1&1&0
end{matrix}
right]
cdot
left[
begin{matrix}
a\b\c\k
end{matrix}
right]
&=
left[
begin{matrix}
s\s\s\2s
end{matrix}
right]
,
end{align}
which can be easily solved with any standard method
and has a solution in terms of $s$:
begin{align}
a &= tfrac59,s
,quad
b = tfrac23,s
,quad
c = tfrac79,s
,quad
k = tfrac19,s
.
end{align}
answered Dec 18 '18 at 9:07
g.kovg.kov
6,1821818
$endgroup$
add a comment |
$begingroup$
Let
begin{align}
tfrac14(s-a)=tfrac13(s-b)=tfrac12(s-c)&=k
quadtext{for some }kinmathbb{R}
end{align}
Then we have a linear system of four equations in
four variables
begin{align}
left[
begin{matrix}
1&0&0&4 \
0&1&0&3\
0&0&1&2\
1&1&1&0
end{matrix}
right]
cdot
left[
begin{matrix}
a\b\c\k
end{matrix}
right]
&=
left[
begin{matrix}
s\s\s\2s
end{matrix}
right]
,
end{align}
which can be easily solved with any standard method
and has a solution in terms of $s$:
begin{align}
a &= tfrac59,s
,quad
b = tfrac23,s
,quad
c = tfrac79,s
,quad
k = tfrac19,s
.
end{align}
answered Dec 18 '18 at 9:07
g.kovg.kov
6,1821818
$endgroup$
add a comment |
$begingroup$
Let
begin{align}
tfrac14(s-a)=tfrac13(s-b)=tfrac12(s-c)&=k
quadtext{for some }kinmathbb{R}
end{align}
Then we have a linear system of four equations in
four variables
begin{align}
left[
begin{matrix}
1&0&0&4 \
0&1&0&3\
0&0&1&2\
1&1&1&0
end{matrix}
right]
cdot
left[
begin{matrix}
a\b\c\k
end{matrix}
right]
&=
left[
begin{matrix}
s\s\s\2s
end{matrix}
right]
,
end{align}
which can be easily solved with any standard method
and has a solution in terms of $s$:
begin{align}
a &= tfrac59,s
,quad
b = tfrac23,s
,quad
c = tfrac79,s
,quad
k = tfrac19,s
.
end{align}
answered Dec 18 '18 at 9:07
g.kovg.kov
6,1821818
$endgroup$
Let
begin{align}
tfrac14(s-a)=tfrac13(s-b)=tfrac12(s-c)&=k
quadtext{for some }kinmathbb{R}
end{align}
Then we have a linear system of four equations in
four variables
begin{align}
left[
begin{matrix}
1&0&0&4 \
0&1&0&3\
0&0&1&2\
1&1&1&0
end{matrix}
right]
cdot
left[
begin{matrix}
a\b\c\k
end{matrix}
right]
&=
left[
begin{matrix}
s\s\s\2s
end{matrix}
right]
,
end{align}
which can be easily solved with any standard method
and has a solution in terms of $s$:
begin{align}
a &= tfrac59,s
,quad
b = tfrac23,s
,quad
c = tfrac79,s
,quad
k = tfrac19,s
.
end{align}
answered Dec 18 '18 at 9:07
g.kovg.kov
6,1821818
answered Dec 18 '18 at 9:07
g.kovg.kov
6,1821818
answered Dec 18 '18 at 9:07
g.kovg.kov
6,1821818
answered Dec 18 '18 at 9:07
g.kovg.kov
6,1821818
6,1821818
add a comment |
add a comment |
$begingroup$
A variation of Kavi Rama Murthy's answer. Denote the fractions by $k$:
$$(s-a)/4=(s-b)/3=(s-c)/2=k Rightarrow \
s-a=4k;\
s-b=3k;\
s-c=2k.$$
Add them to find $k$:
$$3s-2s=9k Rightarrow k=frac{s}{9}.$$
Hence:
$$s-a=frac{4s}{9} Rightarrow a=frac{5s}{9};\
s-b=frac{3s}{9} Rightarrow b=frac{2s}{3};\
s-c=frac{2s}{9} Rightarrow c=frac{7s}{9}.$$
answered Dec 18 '18 at 13:50
farruhotafarruhota
20.5k2739
$endgroup$
add a comment |
$begingroup$
A variation of Kavi Rama Murthy's answer. Denote the fractions by $k$:
$$(s-a)/4=(s-b)/3=(s-c)/2=k Rightarrow \
s-a=4k;\
s-b=3k;\
s-c=2k.$$
Add them to find $k$:
$$3s-2s=9k Rightarrow k=frac{s}{9}.$$
Hence:
$$s-a=frac{4s}{9} Rightarrow a=frac{5s}{9};\
s-b=frac{3s}{9} Rightarrow b=frac{2s}{3};\
s-c=frac{2s}{9} Rightarrow c=frac{7s}{9}.$$
answered Dec 18 '18 at 13:50
farruhotafarruhota
20.5k2739
$endgroup$
add a comment |
$begingroup$
A variation of Kavi Rama Murthy's answer. Denote the fractions by $k$:
$$(s-a)/4=(s-b)/3=(s-c)/2=k Rightarrow \
s-a=4k;\
s-b=3k;\
s-c=2k.$$
Add them to find $k$:
$$3s-2s=9k Rightarrow k=frac{s}{9}.$$
Hence:
$$s-a=frac{4s}{9} Rightarrow a=frac{5s}{9};\
s-b=frac{3s}{9} Rightarrow b=frac{2s}{3};\
s-c=frac{2s}{9} Rightarrow c=frac{7s}{9}.$$
answered Dec 18 '18 at 13:50
farruhotafarruhota
20.5k2739
$endgroup$
A variation of Kavi Rama Murthy's answer. Denote the fractions by $k$:
$$(s-a)/4=(s-b)/3=(s-c)/2=k Rightarrow \
s-a=4k;\
s-b=3k;\
s-c=2k.$$
Add them to find $k$:
$$3s-2s=9k Rightarrow k=frac{s}{9}.$$
Hence:
$$s-a=frac{4s}{9} Rightarrow a=frac{5s}{9};\
s-b=frac{3s}{9} Rightarrow b=frac{2s}{3};\
s-c=frac{2s}{9} Rightarrow c=frac{7s}{9}.$$
answered Dec 18 '18 at 13:50
farruhotafarruhota
20.5k2739
answered Dec 18 '18 at 13:50
farruhotafarruhota
20.5k2739
answered Dec 18 '18 at 13:50
farruhotafarruhota
20.5k2739
answered Dec 18 '18 at 13:50
farruhotafarruhota
20.5k2739
20.5k2739
add a comment |
add a comment |
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