Find the coordinate of the vertex of a trapezoid
Question: In the trapezoid $ABCD, A(4,5), B(1,1), C(-1,4), overleftrightarrow{AB}paralleloverleftrightarrow{CD},overline{CD}=2$, find the coordinate of $D$.
My attempt:
because $overleftrightarrow{AB}paralleloverleftrightarrow{CD}$, therefore they have the same slope, we get $$m_{overleftrightarrow{CD}}=frac{4}{3}$$
By using the point-slope formula, we get the linear equation of $overleftrightarrow{CD}$ is $$L_{overleftrightarrow{CD}}:y-4=frac{4}{3}(x+1)$$
which is $$L_{overleftrightarrow{CD}}:4x-3y=-16$$
because $D$ is on $L_{overleftrightarrow{CD}}$, therefore we can assume the coordinate of $D$ is $$D(t, frac{4t+16}{3})$$
by using $overline{CD}=2$, sqare both sides we get $$(t+1)^2+(frac{4t+16}{3}-4)^2=4$$
after simplifying and factoring we get
$$(5t-1)(5t+11)=0$$
we get $t=frac{1}{5}$ or $-frac{11}{5}$, therefore the coordinate of $D$ is
$$(frac{1}{5}, frac{28}{5})text{or}(-frac{11}{5}, frac{12}{5})$$
But there's only one answer $(frac{1}{5}, frac{28}{5})$, why? $D$ can be above $C$ or below it, can't it?
geometry
asked Dec 5 '18 at 12:45
DavidDavid
344
add a comment |
Question: In the trapezoid $ABCD, A(4,5), B(1,1), C(-1,4), overleftrightarrow{AB}paralleloverleftrightarrow{CD},overline{CD}=2$, find the coordinate of $D$.
My attempt:
because $overleftrightarrow{AB}paralleloverleftrightarrow{CD}$, therefore they have the same slope, we get $$m_{overleftrightarrow{CD}}=frac{4}{3}$$
By using the point-slope formula, we get the linear equation of $overleftrightarrow{CD}$ is $$L_{overleftrightarrow{CD}}:y-4=frac{4}{3}(x+1)$$
which is $$L_{overleftrightarrow{CD}}:4x-3y=-16$$
because $D$ is on $L_{overleftrightarrow{CD}}$, therefore we can assume the coordinate of $D$ is $$D(t, frac{4t+16}{3})$$
by using $overline{CD}=2$, sqare both sides we get $$(t+1)^2+(frac{4t+16}{3}-4)^2=4$$
after simplifying and factoring we get
$$(5t-1)(5t+11)=0$$
we get $t=frac{1}{5}$ or $-frac{11}{5}$, therefore the coordinate of $D$ is
$$(frac{1}{5}, frac{28}{5})text{or}(-frac{11}{5}, frac{12}{5})$$
But there's only one answer $(frac{1}{5}, frac{28}{5})$, why? $D$ can be above $C$ or below it, can't it?
geometry
asked Dec 5 '18 at 12:45
DavidDavid
344
add a comment |
Question: In the trapezoid $ABCD, A(4,5), B(1,1), C(-1,4), overleftrightarrow{AB}paralleloverleftrightarrow{CD},overline{CD}=2$, find the coordinate of $D$.
My attempt:
because $overleftrightarrow{AB}paralleloverleftrightarrow{CD}$, therefore they have the same slope, we get $$m_{overleftrightarrow{CD}}=frac{4}{3}$$
By using the point-slope formula, we get the linear equation of $overleftrightarrow{CD}$ is $$L_{overleftrightarrow{CD}}:y-4=frac{4}{3}(x+1)$$
which is $$L_{overleftrightarrow{CD}}:4x-3y=-16$$
because $D$ is on $L_{overleftrightarrow{CD}}$, therefore we can assume the coordinate of $D$ is $$D(t, frac{4t+16}{3})$$
by using $overline{CD}=2$, sqare both sides we get $$(t+1)^2+(frac{4t+16}{3}-4)^2=4$$
after simplifying and factoring we get
$$(5t-1)(5t+11)=0$$
we get $t=frac{1}{5}$ or $-frac{11}{5}$, therefore the coordinate of $D$ is
$$(frac{1}{5}, frac{28}{5})text{or}(-frac{11}{5}, frac{12}{5})$$
But there's only one answer $(frac{1}{5}, frac{28}{5})$, why? $D$ can be above $C$ or below it, can't it?
geometry
asked Dec 5 '18 at 12:45
DavidDavid
344
Question: In the trapezoid $ABCD, A(4,5), B(1,1), C(-1,4), overleftrightarrow{AB}paralleloverleftrightarrow{CD},overline{CD}=2$, find the coordinate of $D$.
My attempt:
because $overleftrightarrow{AB}paralleloverleftrightarrow{CD}$, therefore they have the same slope, we get $$m_{overleftrightarrow{CD}}=frac{4}{3}$$
By using the point-slope formula, we get the linear equation of $overleftrightarrow{CD}$ is $$L_{overleftrightarrow{CD}}:y-4=frac{4}{3}(x+1)$$
which is $$L_{overleftrightarrow{CD}}:4x-3y=-16$$
because $D$ is on $L_{overleftrightarrow{CD}}$, therefore we can assume the coordinate of $D$ is $$D(t, frac{4t+16}{3})$$
by using $overline{CD}=2$, sqare both sides we get $$(t+1)^2+(frac{4t+16}{3}-4)^2=4$$
after simplifying and factoring we get
$$(5t-1)(5t+11)=0$$
we get $t=frac{1}{5}$ or $-frac{11}{5}$, therefore the coordinate of $D$ is
$$(frac{1}{5}, frac{28}{5})text{or}(-frac{11}{5}, frac{12}{5})$$
But there's only one answer $(frac{1}{5}, frac{28}{5})$, why? $D$ can be above $C$ or below it, can't it?
geometry
geometry
asked Dec 5 '18 at 12:45
DavidDavid
344
asked Dec 5 '18 at 12:45
DavidDavid
344
asked Dec 5 '18 at 12:45
DavidDavid
344
asked Dec 5 '18 at 12:45
DavidDavid
344
asked Dec 5 '18 at 12:45
DavidDavid
344
344
add a comment |
add a comment |
1 Answer
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Referring to the graph below, because the other point $D'$ does not make it trapezoid, but a broken line.
$hspace{3cm}$
Note: The ordering is important: $ABCD$ and $ABDC$ are different figures.
answered Dec 5 '18 at 13:23
farruhotafarruhota
19.6k2737
add a comment |
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1 Answer
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1 Answer
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Referring to the graph below, because the other point $D'$ does not make it trapezoid, but a broken line.
$hspace{3cm}$
Note: The ordering is important: $ABCD$ and $ABDC$ are different figures.
answered Dec 5 '18 at 13:23
farruhotafarruhota
19.6k2737
add a comment |
Referring to the graph below, because the other point $D'$ does not make it trapezoid, but a broken line.
$hspace{3cm}$
Note: The ordering is important: $ABCD$ and $ABDC$ are different figures.
answered Dec 5 '18 at 13:23
farruhotafarruhota
19.6k2737
add a comment |
Referring to the graph below, because the other point $D'$ does not make it trapezoid, but a broken line.
$hspace{3cm}$
Note: The ordering is important: $ABCD$ and $ABDC$ are different figures.
answered Dec 5 '18 at 13:23
farruhotafarruhota
19.6k2737
Referring to the graph below, because the other point $D'$ does not make it trapezoid, but a broken line.
$hspace{3cm}$
Note: The ordering is important: $ABCD$ and $ABDC$ are different figures.
answered Dec 5 '18 at 13:23
farruhotafarruhota
19.6k2737
answered Dec 5 '18 at 13:23
farruhotafarruhota
19.6k2737
answered Dec 5 '18 at 13:23
farruhotafarruhota
19.6k2737
answered Dec 5 '18 at 13:23
farruhotafarruhota
19.6k2737
19.6k2737
add a comment |
add a comment |
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