# Thread: intersection of a line and a curve

1. ## intersection of a line and a curve

Find the coordinates of the points of itersection A and B of the line 2x+y+2=0 and the curve 1/x +2/y= 1/2

i reform the eqn of the line to be y=-2-2x and the curve is y= $\displaystyle \frac{2}{(0.5-1/x)}$

then i sub the two eqn and i 2=(0.5-1/x)(-2-2x)

and so x=-2. but i only got one ans and the qn say there are two point of intersection. How to solve this qn?

2. Originally Posted by helloying
Find the coordinates of the points of itersection A and B of the line 2x+y+2=0 and the curve 1/x +2/y= 1/2

i reform the eqn of the line to be y=-2-2x and the curve is $\displaystyle \frac{2}{(0.5-1/x)}$

then i sub the two eqn and i 2=(0.5-1/x)(-2-2x)

and so x=-2. but i only got one ans and the qn say there are two point of intersection. How to solve this qn?
y=-2-2x, and substitute to get $\displaystyle \frac{1}{x}+\frac{2}{-2-2x}=\frac{1}{2}$
$\displaystyle \frac{1}{x}-\frac{1}{1+x}=\frac{1}{2}$

$\displaystyle \frac{1+x}{x(1+x)}-\frac{x}{x(1+x)}=\frac{1}{2}$

$\displaystyle \frac{1}{x+x^2}=\frac{1}{2}$

$\displaystyle x+x^2=2$
$\displaystyle x^2+x-2=(x+2)(x-1)=0$

x=1 and x=-2 and therefore y=-4 and y=2

so (1,-4) and (-2,2)

3. Originally Posted by helloying
2=(0.5-1/x)(-2-2x)
opening brackets,
$\displaystyle 2=-1-x+\frac{2}{x}+2$

$\displaystyle 2x=-x-x^2+2+2x$

which gives
$\displaystyle x=-2$
and $\displaystyle x=1$