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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- May 15th 2010, 05:35 AMsandyLocus question.
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? - May 15th 2010, 03:06 PMmr fantastic
- May 15th 2010, 09:01 PMsandy
- May 15th 2010, 09:11 PMsa-ri-ga-ma
- May 15th 2010, 10:46 PMmr fantastic
- May 16th 2010, 04:19 AMArchie Meade
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. - May 16th 2010, 09:36 AMsandy
- May 16th 2010, 11:26 AMArchie Meade
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.