# Math Help - Locus question.

1. ## Locus 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?

2. Originally Posted by sandy
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.

3. Originally Posted by mr fantastic
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?

4. Originally Posted by sandy
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.

5. Originally Posted by sandy
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 ....

6. 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.

7. Originally Posted by sa-ri-ga-ma
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?

8. Originally Posted by sandy
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,

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

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

Square both sides

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

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

$y^2$ is common

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

$1+x^2$ is common

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

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

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

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

Square both sides

$(1-x)^2=(1-x)^2+y^2$

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.