Locus question.

Apr 2010
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0
how abot this question:
The locus of points the sum of whose distances from the points (1,0) and (-1,0) equals 2.

What is the locus of points and how is the distance?​
 
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mr fantastic

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how abot this question:
The locus of points the sum of whose distances from the points (1,0) and (-1,0) equals 2.

What is the locus of points and how is the distance?
Do the obvious thing and start by plotting the points (1, 0) and (-1, 0). Now think some more about the given condition.
 
Apr 2010
43
0
Do the obvious thing and start by plotting the points (1, 0) and (-1, 0). Now think some more about the given condition.
i did put them on the plot
and we get a line on the x axis from -1 to 1 so the distance is 2 .
but what do we do next?
 
Jun 2009
806
275
i did put them on the plot
and we get a line on the x axis from -1 to 1 so the distance is 2 .
but what do we do next?
No. No. Not that way.

Let P (x, y) be the point. Find the distance between (x, y) and (1, 0), and (x, y) and (-1, 0). Then add the distances and equate it to the given value. Then simplify the equation to get the locus.
 

mr fantastic

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i did put them on the plot
and we get a line on the x axis from -1 to 1 so the distance is 2 .
but what do we do next?
You're menat to note that any point on the line segment joining (-1, 0) and (1, 0) satisfies the given condition. Therefore ....
 
Dec 2009
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Hi sandy,

you can also ask if it is possible for the point "d" located anywhere off the line segment from (-1,0) to (1,0) to be a total distance of 2 away from both points.
 

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Apr 2010
43
0
No. No. Not that way.

Let P (x, y) be the point. Find the distance between (x, y) and (1, 0), and (x, y) and (-1, 0). Then add the distances and equate it to the given value. Then simplify the equation to get the locus.
i done that it will be
sqrt[ (x-1)^2 + y^2 ] + sqrt[ (x+1)^2 + y^2 ] =2

bt how do we solve this?
 
Dec 2009
3,120
1,342
i done that it will be
sqrt[ (x-1)^2 + y^2 ] + sqrt[ (x+1)^2 + y^2 ] =2

bt how do we solve this?
Hi sandy,

\(\displaystyle \sqrt{(x-1)^2+y^2}+\sqrt{(x+1)^2+y^2}=2\)

\(\displaystyle \sqrt{(1+x)^2+y^2}=2-\sqrt{(1-x)^2+y^2}\)

Square both sides

\(\displaystyle \sqrt{(1+x)^2+y^2}\sqrt{(1+x)^2+y^2}=\left(2-\sqrt{(1-x)^2+y^2}\right)\left(2-\sqrt{(1-x)^2+y^2}\right)\)

\(\displaystyle (1+x)^2+y^2=4-4\sqrt{(1-x)^2+y^2}+(1-x)^2+y^2\)

\(\displaystyle y^2\) is common

\(\displaystyle 1+2x+x^2=4-4\sqrt{(1-x)^2+y^2}+1-2x+x^2\)

\(\displaystyle 1+x^2\) is common

\(\displaystyle 2x=4-4\sqrt{(1-x)^2+y^2}-2x\)

\(\displaystyle 4x=4\left(1-\sqrt{(1-x)^2+y^2}\right)\)

\(\displaystyle x=1-\sqrt{(1-x)^2+y^2}\)

\(\displaystyle 1-x=\sqrt{(1-x)^2+y^2}\)

Square both sides

\(\displaystyle (1-x)^2=(1-x)^2+y^2\)

The only way these can be equal is if y=0.

You should note that you can see the solution geometrically,
by just following mr fantastic's hint.

The distance from (-1,0) to (1,0) is 1-(-1)=2.

The shortest distance between 2 points is a straight line.
Hence any point not between (-1,0) and (1,0) must be a combined distance >2 from those two points.

Any point between (-1,0) and (1,0) is a combined distance of 2 from the two points.

Visually, placing any point above the line segment, you can see that the
combined hypotenuses of the two right-angled triangles formed
have sum of lengths > the combined lengths of the bases.

Hence, any point not on the line segment is more than 2 combined units from those points.
 
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