Mixing
Locus of vertice of equilateral triangle.
$begingroup$
Here is a question:
On two mutually perpendicular lines, two points $A$ and $B$ are taken such that $A$ is a variable point and $B$ is a fixed point. An equilateral triangle $ABC$ is completed. Find the locus of point $C$.
I'm getting a quadratic equation but the back answer is showing a linear equation. Can someone please help me with this question?
geometry vectors euclidean-geometry analytic-geometry geometric-transformation
edited Jan 26 at 12:28
Maria Mazur
47.9k1260120
asked Dec 1 '15 at 18:47
user294996user294996
111
$endgroup$
add a comment |
$begingroup$
Here is a question:
On two mutually perpendicular lines, two points $A$ and $B$ are taken such that $A$ is a variable point and $B$ is a fixed point. An equilateral triangle $ABC$ is completed. Find the locus of point $C$.
I'm getting a quadratic equation but the back answer is showing a linear equation. Can someone please help me with this question?
geometry vectors euclidean-geometry analytic-geometry geometric-transformation
edited Jan 26 at 12:28
Maria Mazur
47.9k1260120
asked Dec 1 '15 at 18:47
user294996user294996
111
$endgroup$
$begingroup$
What is the quadratic equation you get? Have you tried the simplest case, with $B=(1,0)$ and $A=(0,y)$?
$endgroup$
– Arthur
Dec 1 '15 at 18:58
add a comment |
$begingroup$
Here is a question:
On two mutually perpendicular lines, two points $A$ and $B$ are taken such that $A$ is a variable point and $B$ is a fixed point. An equilateral triangle $ABC$ is completed. Find the locus of point $C$.
I'm getting a quadratic equation but the back answer is showing a linear equation. Can someone please help me with this question?
geometry vectors euclidean-geometry analytic-geometry geometric-transformation
edited Jan 26 at 12:28
Maria Mazur
47.9k1260120
asked Dec 1 '15 at 18:47
user294996user294996
111
$endgroup$
Here is a question:
On two mutually perpendicular lines, two points $A$ and $B$ are taken such that $A$ is a variable point and $B$ is a fixed point. An equilateral triangle $ABC$ is completed. Find the locus of point $C$.
I'm getting a quadratic equation but the back answer is showing a linear equation. Can someone please help me with this question?
geometry vectors euclidean-geometry analytic-geometry geometric-transformation
geometry vectors euclidean-geometry analytic-geometry geometric-transformation
edited Jan 26 at 12:28
Maria Mazur
47.9k1260120
asked Dec 1 '15 at 18:47
user294996user294996
111
edited Jan 26 at 12:28
Maria Mazur
47.9k1260120
asked Dec 1 '15 at 18:47
user294996user294996
111
edited Jan 26 at 12:28
Maria Mazur
47.9k1260120
edited Jan 26 at 12:28
Maria Mazur
47.9k1260120
edited Jan 26 at 12:28
Maria Mazur
47.9k1260120
47.9k1260120
asked Dec 1 '15 at 18:47
user294996user294996
111
asked Dec 1 '15 at 18:47
user294996user294996
111
asked Dec 1 '15 at 18:47
user294996user294996
111
111
$begingroup$
What is the quadratic equation you get? Have you tried the simplest case, with $B=(1,0)$ and $A=(0,y)$?
$endgroup$
– Arthur
Dec 1 '15 at 18:58
add a comment |
$begingroup$
What is the quadratic equation you get? Have you tried the simplest case, with $B=(1,0)$ and $A=(0,y)$?
$endgroup$
– Arthur
Dec 1 '15 at 18:58
$begingroup$
What is the quadratic equation you get? Have you tried the simplest case, with $B=(1,0)$ and $A=(0,y)$?
$endgroup$
– Arthur
Dec 1 '15 at 18:58
$begingroup$
What is the quadratic equation you get? Have you tried the simplest case, with $B=(1,0)$ and $A=(0,y)$?
$endgroup$
– Arthur
Dec 1 '15 at 18:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let those lines be $p$ and $q$ (and suppose $B$ is on $q$).
We see that $C$ is rotation of $A$ for $60^{circ}$ around $B$.
So the locus of $C$ is two perpendicular lines (i.e. lines $p$ and $q$ rotated around $B$ for $60^{circ}$) and one of them (the map of $q$) goes through $B$.
answered Jan 25 at 20:55
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
add a comment |
$begingroup$
The midpoint between A and B is $$M=(frac{a}{2},frac{b}{2})$$ if $A=(a,0)$ and $B=(0,b)$
The direction normal to the line AB is $$vec{n} = (frac{b}{sqrt{a^2+b^2}},frac{a}{sqrt{a^2+b^2}})$$
The height of an equilateral triangle of side $ell = sqrt{a^2+b^2}$ is $$h = frac{sqrt{3}}{2} sqrt{a^2+b^2}$$
Point C is located a distance $h$ away from the midpoint M along the normal $vec{n}$
$$ C = M + vec{n} h = ( frac{a}{2}, frac{b}{2} ) + sqrt{3} ( frac{b}{2}, frac{a}{2} ) = ( frac{a+bsqrt{3}}{2}, frac{b+asqrt{3}}{2} ) $$
The locus is $C=(x,y)$ which from the above $a=2x - bsqrt{3}$ and
$$ y = frac{b+(2x - bsqrt{3})sqrt{3}}{2} $$
$$boxed{y = xsqrt{3} - b}$$
answered Dec 1 '15 at 19:19
ja72ja72
7,50712044
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let those lines be $p$ and $q$ (and suppose $B$ is on $q$).
We see that $C$ is rotation of $A$ for $60^{circ}$ around $B$.
So the locus of $C$ is two perpendicular lines (i.e. lines $p$ and $q$ rotated around $B$ for $60^{circ}$) and one of them (the map of $q$) goes through $B$.
answered Jan 25 at 20:55
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
add a comment |
$begingroup$
Let those lines be $p$ and $q$ (and suppose $B$ is on $q$).
We see that $C$ is rotation of $A$ for $60^{circ}$ around $B$.
So the locus of $C$ is two perpendicular lines (i.e. lines $p$ and $q$ rotated around $B$ for $60^{circ}$) and one of them (the map of $q$) goes through $B$.
answered Jan 25 at 20:55
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
add a comment |
$begingroup$
Let those lines be $p$ and $q$ (and suppose $B$ is on $q$).
We see that $C$ is rotation of $A$ for $60^{circ}$ around $B$.
So the locus of $C$ is two perpendicular lines (i.e. lines $p$ and $q$ rotated around $B$ for $60^{circ}$) and one of them (the map of $q$) goes through $B$.
answered Jan 25 at 20:55
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
Let those lines be $p$ and $q$ (and suppose $B$ is on $q$).
We see that $C$ is rotation of $A$ for $60^{circ}$ around $B$.
So the locus of $C$ is two perpendicular lines (i.e. lines $p$ and $q$ rotated around $B$ for $60^{circ}$) and one of them (the map of $q$) goes through $B$.
answered Jan 25 at 20:55
Maria MazurMaria Mazur
47.9k1260120
answered Jan 25 at 20:55
Maria MazurMaria Mazur
47.9k1260120
answered Jan 25 at 20:55
Maria MazurMaria Mazur
47.9k1260120
answered Jan 25 at 20:55
Maria MazurMaria Mazur
47.9k1260120
47.9k1260120
add a comment |
add a comment |
$begingroup$
The midpoint between A and B is $$M=(frac{a}{2},frac{b}{2})$$ if $A=(a,0)$ and $B=(0,b)$
The direction normal to the line AB is $$vec{n} = (frac{b}{sqrt{a^2+b^2}},frac{a}{sqrt{a^2+b^2}})$$
The height of an equilateral triangle of side $ell = sqrt{a^2+b^2}$ is $$h = frac{sqrt{3}}{2} sqrt{a^2+b^2}$$
Point C is located a distance $h$ away from the midpoint M along the normal $vec{n}$
$$ C = M + vec{n} h = ( frac{a}{2}, frac{b}{2} ) + sqrt{3} ( frac{b}{2}, frac{a}{2} ) = ( frac{a+bsqrt{3}}{2}, frac{b+asqrt{3}}{2} ) $$
The locus is $C=(x,y)$ which from the above $a=2x - bsqrt{3}$ and
$$ y = frac{b+(2x - bsqrt{3})sqrt{3}}{2} $$
$$boxed{y = xsqrt{3} - b}$$
answered Dec 1 '15 at 19:19
ja72ja72
7,50712044
$endgroup$
add a comment |
$begingroup$
The midpoint between A and B is $$M=(frac{a}{2},frac{b}{2})$$ if $A=(a,0)$ and $B=(0,b)$
The direction normal to the line AB is $$vec{n} = (frac{b}{sqrt{a^2+b^2}},frac{a}{sqrt{a^2+b^2}})$$
The height of an equilateral triangle of side $ell = sqrt{a^2+b^2}$ is $$h = frac{sqrt{3}}{2} sqrt{a^2+b^2}$$
Point C is located a distance $h$ away from the midpoint M along the normal $vec{n}$
$$ C = M + vec{n} h = ( frac{a}{2}, frac{b}{2} ) + sqrt{3} ( frac{b}{2}, frac{a}{2} ) = ( frac{a+bsqrt{3}}{2}, frac{b+asqrt{3}}{2} ) $$
The locus is $C=(x,y)$ which from the above $a=2x - bsqrt{3}$ and
$$ y = frac{b+(2x - bsqrt{3})sqrt{3}}{2} $$
$$boxed{y = xsqrt{3} - b}$$
answered Dec 1 '15 at 19:19
ja72ja72
7,50712044
$endgroup$
add a comment |
$begingroup$
The midpoint between A and B is $$M=(frac{a}{2},frac{b}{2})$$ if $A=(a,0)$ and $B=(0,b)$
The direction normal to the line AB is $$vec{n} = (frac{b}{sqrt{a^2+b^2}},frac{a}{sqrt{a^2+b^2}})$$
The height of an equilateral triangle of side $ell = sqrt{a^2+b^2}$ is $$h = frac{sqrt{3}}{2} sqrt{a^2+b^2}$$
Point C is located a distance $h$ away from the midpoint M along the normal $vec{n}$
$$ C = M + vec{n} h = ( frac{a}{2}, frac{b}{2} ) + sqrt{3} ( frac{b}{2}, frac{a}{2} ) = ( frac{a+bsqrt{3}}{2}, frac{b+asqrt{3}}{2} ) $$
The locus is $C=(x,y)$ which from the above $a=2x - bsqrt{3}$ and
$$ y = frac{b+(2x - bsqrt{3})sqrt{3}}{2} $$
$$boxed{y = xsqrt{3} - b}$$
answered Dec 1 '15 at 19:19
ja72ja72
7,50712044
$endgroup$
The midpoint between A and B is $$M=(frac{a}{2},frac{b}{2})$$ if $A=(a,0)$ and $B=(0,b)$
The direction normal to the line AB is $$vec{n} = (frac{b}{sqrt{a^2+b^2}},frac{a}{sqrt{a^2+b^2}})$$
The height of an equilateral triangle of side $ell = sqrt{a^2+b^2}$ is $$h = frac{sqrt{3}}{2} sqrt{a^2+b^2}$$
Point C is located a distance $h$ away from the midpoint M along the normal $vec{n}$
$$ C = M + vec{n} h = ( frac{a}{2}, frac{b}{2} ) + sqrt{3} ( frac{b}{2}, frac{a}{2} ) = ( frac{a+bsqrt{3}}{2}, frac{b+asqrt{3}}{2} ) $$
The locus is $C=(x,y)$ which from the above $a=2x - bsqrt{3}$ and
$$ y = frac{b+(2x - bsqrt{3})sqrt{3}}{2} $$
$$boxed{y = xsqrt{3} - b}$$
answered Dec 1 '15 at 19:19
ja72ja72
7,50712044
answered Dec 1 '15 at 19:19
ja72ja72
7,50712044
answered Dec 1 '15 at 19:19
ja72ja72
7,50712044
answered Dec 1 '15 at 19:19
ja72ja72
7,50712044
7,50712044
add a comment |
add a comment |
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$begingroup$
What is the quadratic equation you get? Have you tried the simplest case, with $B=(1,0)$ and $A=(0,y)$?
$endgroup$
– Arthur
Dec 1 '15 at 18:58