Corresponding Point for a Glide Reflection
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I was wondering if there was an efficient method that could solve these types of problems.
Here is the problem:
Plot the points K = (0,0), L = (7,-1), M = (9,3), P = (6,7), Q = (10,5), and R = (1,2). You will see that the triangles KLM and RPQ are congruent. Find coordinates for the point in triangle KLM that corresponds to (3,4) in triangle RPQ.
analytic-geometry transformation
asked Jan 18 at 2:56
rainrain
105
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add a comment |
$begingroup$
I was wondering if there was an efficient method that could solve these types of problems.
Here is the problem:
Plot the points K = (0,0), L = (7,-1), M = (9,3), P = (6,7), Q = (10,5), and R = (1,2). You will see that the triangles KLM and RPQ are congruent. Find coordinates for the point in triangle KLM that corresponds to (3,4) in triangle RPQ.
analytic-geometry transformation
asked Jan 18 at 2:56
rainrain
105
$endgroup$
add a comment |
$begingroup$
I was wondering if there was an efficient method that could solve these types of problems.
Here is the problem:
Plot the points K = (0,0), L = (7,-1), M = (9,3), P = (6,7), Q = (10,5), and R = (1,2). You will see that the triangles KLM and RPQ are congruent. Find coordinates for the point in triangle KLM that corresponds to (3,4) in triangle RPQ.
analytic-geometry transformation
asked Jan 18 at 2:56
rainrain
105
$endgroup$
I was wondering if there was an efficient method that could solve these types of problems.
Here is the problem:
Plot the points K = (0,0), L = (7,-1), M = (9,3), P = (6,7), Q = (10,5), and R = (1,2). You will see that the triangles KLM and RPQ are congruent. Find coordinates for the point in triangle KLM that corresponds to (3,4) in triangle RPQ.
analytic-geometry transformation
analytic-geometry transformation
asked Jan 18 at 2:56
rainrain
105
asked Jan 18 at 2:56
rainrain
105
asked Jan 18 at 2:56
rainrain
105
asked Jan 18 at 2:56
rainrain
105
asked Jan 18 at 2:56
rainrain
105
105
add a comment |
add a comment |
2 Answers
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What helps is to know the general form for an isometry of the Euclidean plane:
begin{align*}f(x,y) &= (x,y) begin{pmatrix} a & b \ c & d end{pmatrix} + (e,f) \
&= (ax+cy+e,bx+dy+f)
end{align*}
such that the matrix $begin{pmatrix} a & b \ c & d end{pmatrix}$ is is orthonormal, meaning that $a^2+b^2=1$, $c^2+d^2=1$ and $ac+bd=0$. In addition to these three equations, you get three more equations $f(K)=R$, $f(L)=P$, $f(M)=Q$ by plugging in the coordinates of $K,L,M,R,P,Q$. You can then solve for the six unknowns $a,b,c,d,e,f$. Substituting those back into the formula for $f(x,y)$, you can then compute $f(3,4)$.
answered Jan 18 at 4:04
Lee MosherLee Mosher
49.3k33685
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Working in homogeneous coordinates, the image of $(3,4)$ under the affine map that sends $triangle{RPQ}$ to $triangle{LMK}$ is (writing all of the coordinates as column vectors) $$begin{bmatrix}K&L&M \ 1&1&1 end{bmatrix} begin{bmatrix}R&P&Q\1&1&1end{bmatrix}^{-1}begin{bmatrix}3\4\1end{bmatrix}.$$ However, there’s a much simpler solution to this particular problem.
Once you’ve plotted these points as instructed, observe that $(3,4)$ lies on the segment $overline{PR}$ (you can easily verify this algebraically). Since $overline{PR}congoverline{LK}$, you just have to find the point on the latter segment that’s the same distance from $L$ as $(3,4)$ is from $P$. Equivalently, solve for $t$ in $(1-t)P+tR=(3,4)$ and plug that value into $(1-t)L+tK$.
answered Jan 19 at 2:04
amdamd
30k21050
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
What helps is to know the general form for an isometry of the Euclidean plane:
begin{align*}f(x,y) &= (x,y) begin{pmatrix} a & b \ c & d end{pmatrix} + (e,f) \
&= (ax+cy+e,bx+dy+f)
end{align*}
such that the matrix $begin{pmatrix} a & b \ c & d end{pmatrix}$ is is orthonormal, meaning that $a^2+b^2=1$, $c^2+d^2=1$ and $ac+bd=0$. In addition to these three equations, you get three more equations $f(K)=R$, $f(L)=P$, $f(M)=Q$ by plugging in the coordinates of $K,L,M,R,P,Q$. You can then solve for the six unknowns $a,b,c,d,e,f$. Substituting those back into the formula for $f(x,y)$, you can then compute $f(3,4)$.
answered Jan 18 at 4:04
Lee MosherLee Mosher
49.3k33685
$endgroup$
add a comment |
$begingroup$
What helps is to know the general form for an isometry of the Euclidean plane:
begin{align*}f(x,y) &= (x,y) begin{pmatrix} a & b \ c & d end{pmatrix} + (e,f) \
&= (ax+cy+e,bx+dy+f)
end{align*}
such that the matrix $begin{pmatrix} a & b \ c & d end{pmatrix}$ is is orthonormal, meaning that $a^2+b^2=1$, $c^2+d^2=1$ and $ac+bd=0$. In addition to these three equations, you get three more equations $f(K)=R$, $f(L)=P$, $f(M)=Q$ by plugging in the coordinates of $K,L,M,R,P,Q$. You can then solve for the six unknowns $a,b,c,d,e,f$. Substituting those back into the formula for $f(x,y)$, you can then compute $f(3,4)$.
answered Jan 18 at 4:04
Lee MosherLee Mosher
49.3k33685
$endgroup$
add a comment |
$begingroup$
What helps is to know the general form for an isometry of the Euclidean plane:
begin{align*}f(x,y) &= (x,y) begin{pmatrix} a & b \ c & d end{pmatrix} + (e,f) \
&= (ax+cy+e,bx+dy+f)
end{align*}
such that the matrix $begin{pmatrix} a & b \ c & d end{pmatrix}$ is is orthonormal, meaning that $a^2+b^2=1$, $c^2+d^2=1$ and $ac+bd=0$. In addition to these three equations, you get three more equations $f(K)=R$, $f(L)=P$, $f(M)=Q$ by plugging in the coordinates of $K,L,M,R,P,Q$. You can then solve for the six unknowns $a,b,c,d,e,f$. Substituting those back into the formula for $f(x,y)$, you can then compute $f(3,4)$.
answered Jan 18 at 4:04
Lee MosherLee Mosher
49.3k33685
$endgroup$
What helps is to know the general form for an isometry of the Euclidean plane:
begin{align*}f(x,y) &= (x,y) begin{pmatrix} a & b \ c & d end{pmatrix} + (e,f) \
&= (ax+cy+e,bx+dy+f)
end{align*}
such that the matrix $begin{pmatrix} a & b \ c & d end{pmatrix}$ is is orthonormal, meaning that $a^2+b^2=1$, $c^2+d^2=1$ and $ac+bd=0$. In addition to these three equations, you get three more equations $f(K)=R$, $f(L)=P$, $f(M)=Q$ by plugging in the coordinates of $K,L,M,R,P,Q$. You can then solve for the six unknowns $a,b,c,d,e,f$. Substituting those back into the formula for $f(x,y)$, you can then compute $f(3,4)$.
answered Jan 18 at 4:04
Lee MosherLee Mosher
49.3k33685
answered Jan 18 at 4:04
Lee MosherLee Mosher
49.3k33685
answered Jan 18 at 4:04
Lee MosherLee Mosher
49.3k33685
answered Jan 18 at 4:04
Lee MosherLee Mosher
49.3k33685
49.3k33685
add a comment |
add a comment |
$begingroup$
Working in homogeneous coordinates, the image of $(3,4)$ under the affine map that sends $triangle{RPQ}$ to $triangle{LMK}$ is (writing all of the coordinates as column vectors) $$begin{bmatrix}K&L&M \ 1&1&1 end{bmatrix} begin{bmatrix}R&P&Q\1&1&1end{bmatrix}^{-1}begin{bmatrix}3\4\1end{bmatrix}.$$ However, there’s a much simpler solution to this particular problem.
Once you’ve plotted these points as instructed, observe that $(3,4)$ lies on the segment $overline{PR}$ (you can easily verify this algebraically). Since $overline{PR}congoverline{LK}$, you just have to find the point on the latter segment that’s the same distance from $L$ as $(3,4)$ is from $P$. Equivalently, solve for $t$ in $(1-t)P+tR=(3,4)$ and plug that value into $(1-t)L+tK$.
answered Jan 19 at 2:04
amdamd
30k21050
$endgroup$
add a comment |
$begingroup$
Working in homogeneous coordinates, the image of $(3,4)$ under the affine map that sends $triangle{RPQ}$ to $triangle{LMK}$ is (writing all of the coordinates as column vectors) $$begin{bmatrix}K&L&M \ 1&1&1 end{bmatrix} begin{bmatrix}R&P&Q\1&1&1end{bmatrix}^{-1}begin{bmatrix}3\4\1end{bmatrix}.$$ However, there’s a much simpler solution to this particular problem.
Once you’ve plotted these points as instructed, observe that $(3,4)$ lies on the segment $overline{PR}$ (you can easily verify this algebraically). Since $overline{PR}congoverline{LK}$, you just have to find the point on the latter segment that’s the same distance from $L$ as $(3,4)$ is from $P$. Equivalently, solve for $t$ in $(1-t)P+tR=(3,4)$ and plug that value into $(1-t)L+tK$.
answered Jan 19 at 2:04
amdamd
30k21050
$endgroup$
add a comment |
$begingroup$
Working in homogeneous coordinates, the image of $(3,4)$ under the affine map that sends $triangle{RPQ}$ to $triangle{LMK}$ is (writing all of the coordinates as column vectors) $$begin{bmatrix}K&L&M \ 1&1&1 end{bmatrix} begin{bmatrix}R&P&Q\1&1&1end{bmatrix}^{-1}begin{bmatrix}3\4\1end{bmatrix}.$$ However, there’s a much simpler solution to this particular problem.
Once you’ve plotted these points as instructed, observe that $(3,4)$ lies on the segment $overline{PR}$ (you can easily verify this algebraically). Since $overline{PR}congoverline{LK}$, you just have to find the point on the latter segment that’s the same distance from $L$ as $(3,4)$ is from $P$. Equivalently, solve for $t$ in $(1-t)P+tR=(3,4)$ and plug that value into $(1-t)L+tK$.
answered Jan 19 at 2:04
amdamd
30k21050
$endgroup$
Working in homogeneous coordinates, the image of $(3,4)$ under the affine map that sends $triangle{RPQ}$ to $triangle{LMK}$ is (writing all of the coordinates as column vectors) $$begin{bmatrix}K&L&M \ 1&1&1 end{bmatrix} begin{bmatrix}R&P&Q\1&1&1end{bmatrix}^{-1}begin{bmatrix}3\4\1end{bmatrix}.$$ However, there’s a much simpler solution to this particular problem.
Once you’ve plotted these points as instructed, observe that $(3,4)$ lies on the segment $overline{PR}$ (you can easily verify this algebraically). Since $overline{PR}congoverline{LK}$, you just have to find the point on the latter segment that’s the same distance from $L$ as $(3,4)$ is from $P$. Equivalently, solve for $t$ in $(1-t)P+tR=(3,4)$ and plug that value into $(1-t)L+tK$.
answered Jan 19 at 2:04
amdamd
30k21050
answered Jan 19 at 2:04
amdamd
30k21050
answered Jan 19 at 2:04
amdamd
30k21050
answered Jan 19 at 2:04
amdamd
30k21050
30k21050
add a comment |
add a comment |
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