Mixing
How to solve this problem? I did not understand how to draw it's diagram. please help.
$begingroup$
At the midpoint of line segment $AB$ which is $p$ units long, a perpendicular $MR$ is erected with length $q$ units. An arc is described from $R$ with a radius equal to $frac{1}{2}AB$, meeting $AB$ at $T$. Then $AT$ and $TB$ are the roots of:
$textbf{(A)} x^2+px+q^2=0\ textbf{(B)} x^2-px+q^2=0\ textbf{(C)} x^2+px-q^2=0\ textbf{(D)} x^2-px-q^2=0\ textbf{(E)} x^2-px+q=0$
algebra-precalculus geometry
edited Jan 23 at 5:43
André 3000
12.7k22243
asked Jan 23 at 5:34
AbhiAbhi
11
$endgroup$
add a comment |
$begingroup$
At the midpoint of line segment $AB$ which is $p$ units long, a perpendicular $MR$ is erected with length $q$ units. An arc is described from $R$ with a radius equal to $frac{1}{2}AB$, meeting $AB$ at $T$. Then $AT$ and $TB$ are the roots of:
$textbf{(A)} x^2+px+q^2=0\ textbf{(B)} x^2-px+q^2=0\ textbf{(C)} x^2+px-q^2=0\ textbf{(D)} x^2-px-q^2=0\ textbf{(E)} x^2-px+q=0$
algebra-precalculus geometry
edited Jan 23 at 5:43
André 3000
12.7k22243
asked Jan 23 at 5:34
AbhiAbhi
11
$endgroup$
add a comment |
$begingroup$
At the midpoint of line segment $AB$ which is $p$ units long, a perpendicular $MR$ is erected with length $q$ units. An arc is described from $R$ with a radius equal to $frac{1}{2}AB$, meeting $AB$ at $T$. Then $AT$ and $TB$ are the roots of:
$textbf{(A)} x^2+px+q^2=0\ textbf{(B)} x^2-px+q^2=0\ textbf{(C)} x^2+px-q^2=0\ textbf{(D)} x^2-px-q^2=0\ textbf{(E)} x^2-px+q=0$
algebra-precalculus geometry
edited Jan 23 at 5:43
André 3000
12.7k22243
asked Jan 23 at 5:34
AbhiAbhi
11
$endgroup$
At the midpoint of line segment $AB$ which is $p$ units long, a perpendicular $MR$ is erected with length $q$ units. An arc is described from $R$ with a radius equal to $frac{1}{2}AB$, meeting $AB$ at $T$. Then $AT$ and $TB$ are the roots of:
$textbf{(A)} x^2+px+q^2=0\ textbf{(B)} x^2-px+q^2=0\ textbf{(C)} x^2+px-q^2=0\ textbf{(D)} x^2-px-q^2=0\ textbf{(E)} x^2-px+q=0$
algebra-precalculus geometry
algebra-precalculus geometry
edited Jan 23 at 5:43
André 3000
12.7k22243
asked Jan 23 at 5:34
AbhiAbhi
11
edited Jan 23 at 5:43
André 3000
12.7k22243
asked Jan 23 at 5:34
AbhiAbhi
11
edited Jan 23 at 5:43
André 3000
12.7k22243
edited Jan 23 at 5:43
André 3000
12.7k22243
edited Jan 23 at 5:43
André 3000
12.7k22243
12.7k22243
asked Jan 23 at 5:34
AbhiAbhi
11
asked Jan 23 at 5:34
AbhiAbhi
11
asked Jan 23 at 5:34
AbhiAbhi
11
11
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In such questions whenever you feel you are not able to visualize, and
the problem involves both the concepts of geometry as well as
algebraic equations always go with coordinate geometry or vectors.
For the diagram you can consider points $A$ and $B$ as having the
coordinates $(0,0)$ and $(p,0)$ since the line segment $AB$ is of
length $p$ units.
Now similarly calculate the midpoint coordinates $M$ then erect a
perpendicular and find the coordinates of $R$. When you get the
coordinates of $R$ just take it as a center and form the equation of
circle using it.
The equation of circle centered at a point $(h,k)$ is given by
$$(x-h)^{2}+(y-k)^{2} = r^2$$
Here the $(h,k)$ will be the coordinates of $R$ and the radius is
given in the problem.
For the final part you need to know the value of $T$. Now
geometrically, when two shapes intersect like a circle and a line(in
this case) there are intersection points. So just solve the equations
of both circle and the line $AB$ ( which is nothing but the equation
of $X$ axis). Simultaneously solving both the equations means that you
want to find the intersection points of the circle and the line.
Finally you will get the equation in terms of $x$,$p$ and $q$. Since
it is a quadratic equation, there will be two roots which will
correspond to $AT$ and $TB$.
Hope this helps.......
answered Jan 23 at 6:08
SNEHIL SANYALSNEHIL SANYAL
616110
$endgroup$
add a comment |
$begingroup$
Construct the image step-bt-step,
collecting and analyzing all the given information.
Segment $AB$: $|AB|=p$.
The point $M$ is the midpoint of $AB$,
hence $|AM|=|MB|=tfrac p2$.From the point $M$ erect the perpendicular $MRperp AB$,
of length $q$, that is, $|MR|=q$.
Note that the radius $r$ of the arc from $R$
is stated to be $tfrac12|AB|$, and the arc touches $AB$ at the point $T$.
Than can only mean that
the point $T$ is the midpoint of $AB$, hence $T=M$,
$|AT|=|TB|=tfrac p2$ and also,
$q=|MR|=r=tfrac12|AB|=tfrac p2$.
Next, the algebraic part. We know that $|AT|$ and $|TB|$
must be the roots of quadratic equation
begin{align}
(x-|AT|)(x-|BT|)&=
x^2-(|AT|+|TB|)x+|AT|cdot|TB|=0
tag{1}label{1}
.
end{align}
Using all the previously collected pieces of information,
namely that $|AT|+|TB|=p$, $|AT|=|TB|=tfrac12p=q$
we can express eqref{1} as
begin{align}
x^2-px+q^2&=0
tag{2}label{2}
,
end{align}
which has a match with one of the given options.
answered Jan 23 at 7:25
g.kovg.kov
6,3071818
$endgroup$
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
$begingroup$
In such questions whenever you feel you are not able to visualize, and
the problem involves both the concepts of geometry as well as
algebraic equations always go with coordinate geometry or vectors.
For the diagram you can consider points $A$ and $B$ as having the
coordinates $(0,0)$ and $(p,0)$ since the line segment $AB$ is of
length $p$ units.
Now similarly calculate the midpoint coordinates $M$ then erect a
perpendicular and find the coordinates of $R$. When you get the
coordinates of $R$ just take it as a center and form the equation of
circle using it.
The equation of circle centered at a point $(h,k)$ is given by
$$(x-h)^{2}+(y-k)^{2} = r^2$$
Here the $(h,k)$ will be the coordinates of $R$ and the radius is
given in the problem.
For the final part you need to know the value of $T$. Now
geometrically, when two shapes intersect like a circle and a line(in
this case) there are intersection points. So just solve the equations
of both circle and the line $AB$ ( which is nothing but the equation
of $X$ axis). Simultaneously solving both the equations means that you
want to find the intersection points of the circle and the line.
Finally you will get the equation in terms of $x$,$p$ and $q$. Since
it is a quadratic equation, there will be two roots which will
correspond to $AT$ and $TB$.
Hope this helps.......
answered Jan 23 at 6:08
SNEHIL SANYALSNEHIL SANYAL
616110
$endgroup$
add a comment |
$begingroup$
In such questions whenever you feel you are not able to visualize, and
the problem involves both the concepts of geometry as well as
algebraic equations always go with coordinate geometry or vectors.
For the diagram you can consider points $A$ and $B$ as having the
coordinates $(0,0)$ and $(p,0)$ since the line segment $AB$ is of
length $p$ units.
Now similarly calculate the midpoint coordinates $M$ then erect a
perpendicular and find the coordinates of $R$. When you get the
coordinates of $R$ just take it as a center and form the equation of
circle using it.
The equation of circle centered at a point $(h,k)$ is given by
$$(x-h)^{2}+(y-k)^{2} = r^2$$
Here the $(h,k)$ will be the coordinates of $R$ and the radius is
given in the problem.
For the final part you need to know the value of $T$. Now
geometrically, when two shapes intersect like a circle and a line(in
this case) there are intersection points. So just solve the equations
of both circle and the line $AB$ ( which is nothing but the equation
of $X$ axis). Simultaneously solving both the equations means that you
want to find the intersection points of the circle and the line.
Finally you will get the equation in terms of $x$,$p$ and $q$. Since
it is a quadratic equation, there will be two roots which will
correspond to $AT$ and $TB$.
Hope this helps.......
answered Jan 23 at 6:08
SNEHIL SANYALSNEHIL SANYAL
616110
$endgroup$
add a comment |
$begingroup$
In such questions whenever you feel you are not able to visualize, and
the problem involves both the concepts of geometry as well as
algebraic equations always go with coordinate geometry or vectors.
For the diagram you can consider points $A$ and $B$ as having the
coordinates $(0,0)$ and $(p,0)$ since the line segment $AB$ is of
length $p$ units.
Now similarly calculate the midpoint coordinates $M$ then erect a
perpendicular and find the coordinates of $R$. When you get the
coordinates of $R$ just take it as a center and form the equation of
circle using it.
The equation of circle centered at a point $(h,k)$ is given by
$$(x-h)^{2}+(y-k)^{2} = r^2$$
Here the $(h,k)$ will be the coordinates of $R$ and the radius is
given in the problem.
For the final part you need to know the value of $T$. Now
geometrically, when two shapes intersect like a circle and a line(in
this case) there are intersection points. So just solve the equations
of both circle and the line $AB$ ( which is nothing but the equation
of $X$ axis). Simultaneously solving both the equations means that you
want to find the intersection points of the circle and the line.
Finally you will get the equation in terms of $x$,$p$ and $q$. Since
it is a quadratic equation, there will be two roots which will
correspond to $AT$ and $TB$.
Hope this helps.......
answered Jan 23 at 6:08
SNEHIL SANYALSNEHIL SANYAL
616110
$endgroup$
In such questions whenever you feel you are not able to visualize, and
the problem involves both the concepts of geometry as well as
algebraic equations always go with coordinate geometry or vectors.
For the diagram you can consider points $A$ and $B$ as having the
coordinates $(0,0)$ and $(p,0)$ since the line segment $AB$ is of
length $p$ units.
Now similarly calculate the midpoint coordinates $M$ then erect a
perpendicular and find the coordinates of $R$. When you get the
coordinates of $R$ just take it as a center and form the equation of
circle using it.
The equation of circle centered at a point $(h,k)$ is given by
$$(x-h)^{2}+(y-k)^{2} = r^2$$
Here the $(h,k)$ will be the coordinates of $R$ and the radius is
given in the problem.
For the final part you need to know the value of $T$. Now
geometrically, when two shapes intersect like a circle and a line(in
this case) there are intersection points. So just solve the equations
of both circle and the line $AB$ ( which is nothing but the equation
of $X$ axis). Simultaneously solving both the equations means that you
want to find the intersection points of the circle and the line.
Finally you will get the equation in terms of $x$,$p$ and $q$. Since
it is a quadratic equation, there will be two roots which will
correspond to $AT$ and $TB$.
Hope this helps.......
answered Jan 23 at 6:08
SNEHIL SANYALSNEHIL SANYAL
616110
answered Jan 23 at 6:08
SNEHIL SANYALSNEHIL SANYAL
616110
answered Jan 23 at 6:08
SNEHIL SANYALSNEHIL SANYAL
616110
answered Jan 23 at 6:08
SNEHIL SANYALSNEHIL SANYAL
616110
616110
add a comment |
add a comment |
$begingroup$
Construct the image step-bt-step,
collecting and analyzing all the given information.
Segment $AB$: $|AB|=p$.
The point $M$ is the midpoint of $AB$,
hence $|AM|=|MB|=tfrac p2$.From the point $M$ erect the perpendicular $MRperp AB$,
of length $q$, that is, $|MR|=q$.
Note that the radius $r$ of the arc from $R$
is stated to be $tfrac12|AB|$, and the arc touches $AB$ at the point $T$.
Than can only mean that
the point $T$ is the midpoint of $AB$, hence $T=M$,
$|AT|=|TB|=tfrac p2$ and also,
$q=|MR|=r=tfrac12|AB|=tfrac p2$.
Next, the algebraic part. We know that $|AT|$ and $|TB|$
must be the roots of quadratic equation
begin{align}
(x-|AT|)(x-|BT|)&=
x^2-(|AT|+|TB|)x+|AT|cdot|TB|=0
tag{1}label{1}
.
end{align}
Using all the previously collected pieces of information,
namely that $|AT|+|TB|=p$, $|AT|=|TB|=tfrac12p=q$
we can express eqref{1} as
begin{align}
x^2-px+q^2&=0
tag{2}label{2}
,
end{align}
which has a match with one of the given options.
answered Jan 23 at 7:25
g.kovg.kov
6,3071818
$endgroup$
add a comment |
$begingroup$
Construct the image step-bt-step,
collecting and analyzing all the given information.
Segment $AB$: $|AB|=p$.
The point $M$ is the midpoint of $AB$,
hence $|AM|=|MB|=tfrac p2$.From the point $M$ erect the perpendicular $MRperp AB$,
of length $q$, that is, $|MR|=q$.
Note that the radius $r$ of the arc from $R$
is stated to be $tfrac12|AB|$, and the arc touches $AB$ at the point $T$.
Than can only mean that
the point $T$ is the midpoint of $AB$, hence $T=M$,
$|AT|=|TB|=tfrac p2$ and also,
$q=|MR|=r=tfrac12|AB|=tfrac p2$.
Next, the algebraic part. We know that $|AT|$ and $|TB|$
must be the roots of quadratic equation
begin{align}
(x-|AT|)(x-|BT|)&=
x^2-(|AT|+|TB|)x+|AT|cdot|TB|=0
tag{1}label{1}
.
end{align}
Using all the previously collected pieces of information,
namely that $|AT|+|TB|=p$, $|AT|=|TB|=tfrac12p=q$
we can express eqref{1} as
begin{align}
x^2-px+q^2&=0
tag{2}label{2}
,
end{align}
which has a match with one of the given options.
answered Jan 23 at 7:25
g.kovg.kov
6,3071818
$endgroup$
add a comment |
$begingroup$
Construct the image step-bt-step,
collecting and analyzing all the given information.
Segment $AB$: $|AB|=p$.
The point $M$ is the midpoint of $AB$,
hence $|AM|=|MB|=tfrac p2$.From the point $M$ erect the perpendicular $MRperp AB$,
of length $q$, that is, $|MR|=q$.
Note that the radius $r$ of the arc from $R$
is stated to be $tfrac12|AB|$, and the arc touches $AB$ at the point $T$.
Than can only mean that
the point $T$ is the midpoint of $AB$, hence $T=M$,
$|AT|=|TB|=tfrac p2$ and also,
$q=|MR|=r=tfrac12|AB|=tfrac p2$.
Next, the algebraic part. We know that $|AT|$ and $|TB|$
must be the roots of quadratic equation
begin{align}
(x-|AT|)(x-|BT|)&=
x^2-(|AT|+|TB|)x+|AT|cdot|TB|=0
tag{1}label{1}
.
end{align}
Using all the previously collected pieces of information,
namely that $|AT|+|TB|=p$, $|AT|=|TB|=tfrac12p=q$
we can express eqref{1} as
begin{align}
x^2-px+q^2&=0
tag{2}label{2}
,
end{align}
which has a match with one of the given options.
answered Jan 23 at 7:25
g.kovg.kov
6,3071818
$endgroup$
Construct the image step-bt-step,
collecting and analyzing all the given information.
Segment $AB$: $|AB|=p$.
The point $M$ is the midpoint of $AB$,
hence $|AM|=|MB|=tfrac p2$.From the point $M$ erect the perpendicular $MRperp AB$,
of length $q$, that is, $|MR|=q$.
Note that the radius $r$ of the arc from $R$
is stated to be $tfrac12|AB|$, and the arc touches $AB$ at the point $T$.
Than can only mean that
the point $T$ is the midpoint of $AB$, hence $T=M$,
$|AT|=|TB|=tfrac p2$ and also,
$q=|MR|=r=tfrac12|AB|=tfrac p2$.
Next, the algebraic part. We know that $|AT|$ and $|TB|$
must be the roots of quadratic equation
begin{align}
(x-|AT|)(x-|BT|)&=
x^2-(|AT|+|TB|)x+|AT|cdot|TB|=0
tag{1}label{1}
.
end{align}
Using all the previously collected pieces of information,
namely that $|AT|+|TB|=p$, $|AT|=|TB|=tfrac12p=q$
we can express eqref{1} as
begin{align}
x^2-px+q^2&=0
tag{2}label{2}
,
end{align}
which has a match with one of the given options.
answered Jan 23 at 7:25
g.kovg.kov
6,3071818
edited Jan 23 at 7:36
answered Jan 23 at 7:25
g.kovg.kov
6,3071818
answered Jan 23 at 7:25
g.kovg.kov
6,3071818
answered Jan 23 at 7:25
g.kovg.kov
6,3071818
6,3071818
add a comment |
add a comment |
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