Coordinate Geometry - Distance Formula
$begingroup$
If the points $A(4,3)$ and $B(x,5)$ are on the circle with center $O(2,3)$ find the value of $x$.
Since $AO=BO$
Using distance formula,
we get $x=2$
However my approach was why can't section formula be used where $(2,3)$ can be considered as the midpoint so that
$M = frac{x_1 + x_2}{2}$
I can't figure out the reason why this method doesn't work.
Any help will be appreciated.
geometry
edited Dec 20 '18 at 12:16
Matko
864
asked Dec 20 '18 at 12:00
Naimisha GirishNaimisha Girish
14
$endgroup$
add a comment |
$begingroup$
If the points $A(4,3)$ and $B(x,5)$ are on the circle with center $O(2,3)$ find the value of $x$.
Since $AO=BO$
Using distance formula,
we get $x=2$
However my approach was why can't section formula be used where $(2,3)$ can be considered as the midpoint so that
$M = frac{x_1 + x_2}{2}$
I can't figure out the reason why this method doesn't work.
Any help will be appreciated.
geometry
edited Dec 20 '18 at 12:16
Matko
864
asked Dec 20 '18 at 12:00
Naimisha GirishNaimisha Girish
14
$endgroup$
add a comment |
$begingroup$
If the points $A(4,3)$ and $B(x,5)$ are on the circle with center $O(2,3)$ find the value of $x$.
Since $AO=BO$
Using distance formula,
we get $x=2$
However my approach was why can't section formula be used where $(2,3)$ can be considered as the midpoint so that
$M = frac{x_1 + x_2}{2}$
I can't figure out the reason why this method doesn't work.
Any help will be appreciated.
geometry
edited Dec 20 '18 at 12:16
Matko
864
asked Dec 20 '18 at 12:00
Naimisha GirishNaimisha Girish
14
$endgroup$
If the points $A(4,3)$ and $B(x,5)$ are on the circle with center $O(2,3)$ find the value of $x$.
Since $AO=BO$
Using distance formula,
we get $x=2$
However my approach was why can't section formula be used where $(2,3)$ can be considered as the midpoint so that
$M = frac{x_1 + x_2}{2}$
I can't figure out the reason why this method doesn't work.
Any help will be appreciated.
geometry
geometry
edited Dec 20 '18 at 12:16
Matko
864
asked Dec 20 '18 at 12:00
Naimisha GirishNaimisha Girish
14
edited Dec 20 '18 at 12:16
Matko
864
asked Dec 20 '18 at 12:00
Naimisha GirishNaimisha Girish
14
edited Dec 20 '18 at 12:16
Matko
864
edited Dec 20 '18 at 12:16
Matko
864
edited Dec 20 '18 at 12:16
Matko
864
864
asked Dec 20 '18 at 12:00
Naimisha GirishNaimisha Girish
14
asked Dec 20 '18 at 12:00
Naimisha GirishNaimisha Girish
14
asked Dec 20 '18 at 12:00
Naimisha GirishNaimisha Girish
14
14
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
That would require the line segment $AB$ to be the diameter for $M$ to be equal to $O$ which is not necessarily true.
It is possible that $AB$ doesn't pass through $O$.
answered Dec 20 '18 at 12:12
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
add a comment |
$begingroup$
You should sketch it on the cooredinate plane.
$hspace{3cm}$
If $A(4,3)$ lies on the circle with center $O(2,3)$, then $AO=2$ is the radius.
The point $B(x,5)$ lies somewhere on the blue line $y=5$. In order for $B$ to lie on the circle too, it must be a common (touching or intersection) point of the circle and the blue line. And we see it is the touching point $B(2,5)$.
When you consider $M(2,3)$ as the middle point of $AB$ and calculate $frac{x_1+x_2}{2}=frac{4+x}{2}=2 Rightarrow x=0$, you will find the point $C(0,3)$ which is on the circle, but it is not on the blue line. Also, $frac{y_1+y_2}{2}=frac{3+5}{2}=4ne 3$.
answered Dec 20 '18 at 13:31
farruhotafarruhota
20.3k2739
$endgroup$
add a comment |
$begingroup$
Note : $AB$ is a chord.
Midpoint: $M=(x/2+2,,4)$.
1) Slope $AB$:
$m_1:= dfrac{5-3}{x-4}=dfrac{2}{x-4}$.
2) Slope $MO$:
$m_2:= dfrac {4-3}{x/2+2 -2}=2/x.$
$m_1$ is perpendicular to $m_2$ (why?):
$m_1m_2=-1.$
Hence $x/2-2=- x/2;$
$x=2$
answered Dec 20 '18 at 13:00
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
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
$begingroup$
That would require the line segment $AB$ to be the diameter for $M$ to be equal to $O$ which is not necessarily true.
It is possible that $AB$ doesn't pass through $O$.
answered Dec 20 '18 at 12:12
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
add a comment |
$begingroup$
That would require the line segment $AB$ to be the diameter for $M$ to be equal to $O$ which is not necessarily true.
It is possible that $AB$ doesn't pass through $O$.
answered Dec 20 '18 at 12:12
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
add a comment |
$begingroup$
That would require the line segment $AB$ to be the diameter for $M$ to be equal to $O$ which is not necessarily true.
It is possible that $AB$ doesn't pass through $O$.
answered Dec 20 '18 at 12:12
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
That would require the line segment $AB$ to be the diameter for $M$ to be equal to $O$ which is not necessarily true.
It is possible that $AB$ doesn't pass through $O$.
answered Dec 20 '18 at 12:12
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 20 '18 at 12:12
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 20 '18 at 12:12
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 20 '18 at 12:12
Siong Thye GohSiong Thye Goh
101k1466118
101k1466118
add a comment |
add a comment |
$begingroup$
You should sketch it on the cooredinate plane.
$hspace{3cm}$
If $A(4,3)$ lies on the circle with center $O(2,3)$, then $AO=2$ is the radius.
The point $B(x,5)$ lies somewhere on the blue line $y=5$. In order for $B$ to lie on the circle too, it must be a common (touching or intersection) point of the circle and the blue line. And we see it is the touching point $B(2,5)$.
When you consider $M(2,3)$ as the middle point of $AB$ and calculate $frac{x_1+x_2}{2}=frac{4+x}{2}=2 Rightarrow x=0$, you will find the point $C(0,3)$ which is on the circle, but it is not on the blue line. Also, $frac{y_1+y_2}{2}=frac{3+5}{2}=4ne 3$.
answered Dec 20 '18 at 13:31
farruhotafarruhota
20.3k2739
$endgroup$
add a comment |
$begingroup$
You should sketch it on the cooredinate plane.
$hspace{3cm}$
If $A(4,3)$ lies on the circle with center $O(2,3)$, then $AO=2$ is the radius.
The point $B(x,5)$ lies somewhere on the blue line $y=5$. In order for $B$ to lie on the circle too, it must be a common (touching or intersection) point of the circle and the blue line. And we see it is the touching point $B(2,5)$.
When you consider $M(2,3)$ as the middle point of $AB$ and calculate $frac{x_1+x_2}{2}=frac{4+x}{2}=2 Rightarrow x=0$, you will find the point $C(0,3)$ which is on the circle, but it is not on the blue line. Also, $frac{y_1+y_2}{2}=frac{3+5}{2}=4ne 3$.
answered Dec 20 '18 at 13:31
farruhotafarruhota
20.3k2739
$endgroup$
add a comment |
$begingroup$
You should sketch it on the cooredinate plane.
$hspace{3cm}$
If $A(4,3)$ lies on the circle with center $O(2,3)$, then $AO=2$ is the radius.
The point $B(x,5)$ lies somewhere on the blue line $y=5$. In order for $B$ to lie on the circle too, it must be a common (touching or intersection) point of the circle and the blue line. And we see it is the touching point $B(2,5)$.
When you consider $M(2,3)$ as the middle point of $AB$ and calculate $frac{x_1+x_2}{2}=frac{4+x}{2}=2 Rightarrow x=0$, you will find the point $C(0,3)$ which is on the circle, but it is not on the blue line. Also, $frac{y_1+y_2}{2}=frac{3+5}{2}=4ne 3$.
answered Dec 20 '18 at 13:31
farruhotafarruhota
20.3k2739
$endgroup$
You should sketch it on the cooredinate plane.
$hspace{3cm}$
If $A(4,3)$ lies on the circle with center $O(2,3)$, then $AO=2$ is the radius.
The point $B(x,5)$ lies somewhere on the blue line $y=5$. In order for $B$ to lie on the circle too, it must be a common (touching or intersection) point of the circle and the blue line. And we see it is the touching point $B(2,5)$.
When you consider $M(2,3)$ as the middle point of $AB$ and calculate $frac{x_1+x_2}{2}=frac{4+x}{2}=2 Rightarrow x=0$, you will find the point $C(0,3)$ which is on the circle, but it is not on the blue line. Also, $frac{y_1+y_2}{2}=frac{3+5}{2}=4ne 3$.
answered Dec 20 '18 at 13:31
farruhotafarruhota
20.3k2739
answered Dec 20 '18 at 13:31
farruhotafarruhota
20.3k2739
answered Dec 20 '18 at 13:31
farruhotafarruhota
20.3k2739
answered Dec 20 '18 at 13:31
farruhotafarruhota
20.3k2739
20.3k2739
add a comment |
add a comment |
$begingroup$
Note : $AB$ is a chord.
Midpoint: $M=(x/2+2,,4)$.
1) Slope $AB$:
$m_1:= dfrac{5-3}{x-4}=dfrac{2}{x-4}$.
2) Slope $MO$:
$m_2:= dfrac {4-3}{x/2+2 -2}=2/x.$
$m_1$ is perpendicular to $m_2$ (why?):
$m_1m_2=-1.$
Hence $x/2-2=- x/2;$
$x=2$
answered Dec 20 '18 at 13:00
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
add a comment |
$begingroup$
Note : $AB$ is a chord.
Midpoint: $M=(x/2+2,,4)$.
1) Slope $AB$:
$m_1:= dfrac{5-3}{x-4}=dfrac{2}{x-4}$.
2) Slope $MO$:
$m_2:= dfrac {4-3}{x/2+2 -2}=2/x.$
$m_1$ is perpendicular to $m_2$ (why?):
$m_1m_2=-1.$
Hence $x/2-2=- x/2;$
$x=2$
answered Dec 20 '18 at 13:00
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
add a comment |
$begingroup$
Note : $AB$ is a chord.
Midpoint: $M=(x/2+2,,4)$.
1) Slope $AB$:
$m_1:= dfrac{5-3}{x-4}=dfrac{2}{x-4}$.
2) Slope $MO$:
$m_2:= dfrac {4-3}{x/2+2 -2}=2/x.$
$m_1$ is perpendicular to $m_2$ (why?):
$m_1m_2=-1.$
Hence $x/2-2=- x/2;$
$x=2$
answered Dec 20 '18 at 13:00
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
Note : $AB$ is a chord.
Midpoint: $M=(x/2+2,,4)$.
1) Slope $AB$:
$m_1:= dfrac{5-3}{x-4}=dfrac{2}{x-4}$.
2) Slope $MO$:
$m_2:= dfrac {4-3}{x/2+2 -2}=2/x.$
$m_1$ is perpendicular to $m_2$ (why?):
$m_1m_2=-1.$
Hence $x/2-2=- x/2;$
$x=2$
answered Dec 20 '18 at 13:00
Peter SzilasPeter Szilas
11.3k2822
edited Dec 20 '18 at 15:05
answered Dec 20 '18 at 13:00
Peter SzilasPeter Szilas
11.3k2822
answered Dec 20 '18 at 13:00
Peter SzilasPeter Szilas
11.3k2822
answered Dec 20 '18 at 13:00
Peter SzilasPeter Szilas
11.3k2822
11.3k2822
add a comment |
add a comment |
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