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?
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?
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.