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by thephoenix » Sun Apr 11, 2010 2:21 am
there are two ways of solving it
Ist
applying two pnt formula (dist b/n x1,y1 and x2,y2 is sqrt[(x1-x2)^2+(y1-y2)^2]............1
and pythagoras theorem i.e a^2+b^2=htpotenuous^2..........2

from1 we get s^2+t^2=4...........3
and from 2 we get t=s Sqrt(3).........4

putting 4 in 3 we get s=1

another way is the fig is symmetrical about y axis i.e mirror image so the values get interchanged
i.e x,y will become y,-x
so -rut3,1 will become 1,rut 3
hence s=1
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by eaakbari » Sun Apr 11, 2010 3:47 am
there are two ways of solving it
Ist
applying two pnt formula (dist b/n x1,y1 and x2,y2 is sqrt[(x1-x2)^2+(y1-y2)^2]............1
and pythagoras theorem i.e a^2+b^2=htpotenuous^2..........2

from1 we get s^2+t^2=4...........3
and from 2 we get t=s Sqrt(3).........4

putting 4 in 3 we get s=1

another way is the fig is symmetrical about y axis i.e mirror image so the values get interchanged
i.e x,y will become y,-x
so -rut3,1 will become 1,rut 3
hence s=1
Though your answer is correct, there is some mistake in your calculation as t is not equal to s. And I feel we cannot use the second method you suggested as we do not know if the axis of symmetry coincides with the y axis. The answer seems to suppose that but we cannot jump to that.

Solution

From distance formula we can find distance between (root3, 1) and (0,0) which turns out to be 2 which is the radius

Hence s^2 + t^2 = 4 --1
Now using distance formula between points (root3,1) and (s,t) which will be equal to the hypotenuse of the triangle which is (1,1,root2) triplet hence hyp = 2root2

On equating distance formula and hyp, we obtain
s = t/root3
Substitute in 1 and we get
s= 1, t=root3
which is the answer, s= 1
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by neerajkumar1_1 » Sun Apr 11, 2010 4:32 am
eaakbari wrote:
there are two ways of solving it
Ist
applying two pnt formula (dist b/n x1,y1 and x2,y2 is sqrt[(x1-x2)^2+(y1-y2)^2]............1
and pythagoras theorem i.e a^2+b^2=htpotenuous^2..........2

from1 we get s^2+t^2=4...........3
and from 2 we get t=s Sqrt(3).........4

putting 4 in 3 we get s=1

another way is the fig is symmetrical about y axis i.e mirror image so the values get interchanged
i.e x,y will become y,-x
so -rut3,1 will become 1,rut 3
hence s=1
Though your answer is correct, there is some mistake in your calculation as t is not equal to s. And I feel we cannot use the second method you suggested as we do not know if the axis of symmetry coincides with the y axis. The answer seems to suppose that but we cannot jump to that.

Solution

From distance formula we can find distance between (root3, 1) and (0,0) which turns out to be 2 which is the radius

Hence s^2 + t^2 = 4 --1
Now using distance formula between points (root3,1) and (s,t) which will be equal to the hypotenuse of the triangle which is (1,1,root2) triplet hence hyp = 2root2

On equating distance formula and hyp, we obtain
s = t/root3
Substitute in 1 and we get
s= 1, t=root3
which is the answer, s= 1

For either of the solution i cant seem to understand how u get s=t/root3....
if we use the distance formula between point p and q.... we get sqrt((s + root3)^2 + (t-1)^2).... which ends up getting very complex....
please explain....
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by thephoenix » Sun Apr 11, 2010 5:14 am
hypotenuse=sqrt((s + root3)^2 + (t-1)^2)=sqrt[s^2+3+2s*rut3+t^2+1-2t]=sqrt[s^2+t^2+4-2t+2*s*sqrt3]
since radius is 2 ; s^2+t^2=2^2=4
since H^2=A^2+B^2{where A=B=2]----->[sqrt[s^2+t^2+4-2t+2*s*sqrt3]]^2=4+4--->s^2+t^2+4-2t+2*s*sqrt3=8
since s^2+t^2=4--->8-2t+2s (sqrt 3)=8 or 2t=2srut3 or t= s* sqrt 3
inserting value of t in the eqn s^2+t^2=2^2=4 we get s=1
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by outreach » Sun Apr 11, 2010 9:04 am
thephoenix wrote:there are two ways of solving it
Ist
applying two pnt formula (dist b/n x1,y1 and x2,y2 is sqrt[(x1-x2)^2+(y1-y2)^2]............1
and pythagoras theorem i.e a^2+b^2=htpotenuous^2..........2

from1 we get s^2+t^2=4...........3
and from 2 we get t=s Sqrt(3).........4

putting 4 in 3 we get s=1

another way is the fig is symmetrical about y axis i.e mirror image so the values get interchanged
i.e x,y will become y,-x
so -rut3,1 will become 1,rut 3
hence s=1
can we assume that O is at (0,0)
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by vipulbhatt2003 » Sun Apr 11, 2010 9:18 am
This is the way i solved:

2 Steps:

Step 1: As the 2 lines are perpendicular, the product of the slopes of these two lines would be equal to -1.
Therefore, (1-0)/(-sqrt(3)-0) * (t-0/s-0) = -1
This gives us s= t/ sqrt(3) ---------------------eq 1


Step 2: Using distance formula , Equating the distance of two points on the circle as both the lines are radius:
square (s) + square (t) = sqaure(1) + square ( -sqrt(3)) = 1+3 = 4

square (s) + square (t) = 4 ------------------- eq 2


solving eq 1 and eq 2, you will get s= 1
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by harshavardhanc » Sun Apr 11, 2010 9:37 am
neerajkumar1_1 wrote:need a solution
one more method, hope you find it easy.

radius = 2

drop a perpedicular from P on X axis. let the base be R. . You will find that <POR = 30 degrees.

Therefore, angle on the other side, <QO (something) = 60 degrees.

Hence, S= 2 Cos 60 = 1 .
Regards,
Harsha
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by kstv » Sun Apr 11, 2010 9:41 am
thephoenix wrote:there are two ways of solving it
another way is the fig is symmetrical about y axis i.e mirror image so the values get interchanged
i.e x,y will become y,-x so -rut3,1 will become 1,rut 3 hence s=1
in that case (s,t) should it not be (√3,1)

slope (PO X QO) = -1 (perpendicular to each other)
slope of QO = t/s = √3 ----------- eq 1
PO = QO = 2 radius of the circle
s²+t²= 4----------------eq 2
s²+(s√3)²=4 or s = =-1
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by eaakbari » Sun Apr 11, 2010 10:09 am
vipulbhatt2003 wrote:This is the way i solved:

2 Steps:

Step 1: As the 2 lines are perpendicular, the product of the slopes of these two lines would be equal to -1.
Therefore, (1-0)/(-sqrt(3)-0) * (t-0/s-0) = -1
This gives us s= t/ sqrt(3) ---------------------eq 1


Step 2: Using distance formula , Equating the distance of two points on the circle as both the lines are radius:
square (s) + square (t) = sqaure(1) + square ( -sqrt(3)) = 1+3 = 4

square (s) + square (t) = 4 ------------------- eq 2


solving eq 1 and eq 2, you will get s= 1
Really like your method, had not thought of that. Good thinking

Thanks
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by kstv » Sun Apr 11, 2010 10:02 pm
@ vipulbhatt2003
I did not read your solution before posting. So mine is a fascimile of your solution.
Great men think alike or Fools seldom differ.
for my sake I vote for you as a potentially Great man.

this idea of thephoenix is still nagging

there are two ways of solving it
another way is the fig is symmetrical about y axis i.e mirror image so the values get interchanged
i.e x,y will become y,-x so -rut3,1 will become 1,rut 3 hence s=1

instead of y axis if it was the line x = y
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by harshavardhanc » Sun Apr 11, 2010 11:18 pm
kstv wrote:
this idea of thephoenix is still nagging

there are two ways of solving it
another way is the fig is symmetrical about y axis i.e mirror image so the values get interchanged
i.e x,y will become y,-x so -rut3,1 will become 1,rut 3 hence s=1
kstv,

the figure is not symmetrical about the Y axis. See my previous post. Y axis doesn't bisect the right angle. Hence, the other point would not be the mirror image.

https://www.beatthegmat.com/post243314.html#243314

Anyway, here are some concept-refreshers :

the mirror image of a point (x,y) about the Y axis would be (-x,y).

the mirror image of a point (x,y) about the X axis would be (x,-y).

It is only the about the line X=Y or X=-Y, where the coordinates are interchanged during reflection.

In the case of reflection about X=Y, point (x,y) becomes (y,x)

and reflection about the line X= - Y, point (x,y) becomes (-y,-x).

So the second method, which is making you confused, need not be considered. Don't worry! :)
Regards,
Harsha
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