# Math Help - Line intersecting implicit function

1. ## Line intersecting implicit function

Hi, I have to find the co-ordinates where the line y=-x+1 intersects f(x,y) = x^2 + y^2 -xy - y -1

I can do this with explicit functions by making them equal, but it doesn't work with this.

The actual question asks for the co-ordinates where the tangent to f(x,y), is at 45 deg to the horizontal. So I found the gradient function which = 1 (that got me the linear equation), now i'm stuck.

Not sure if this should be in calc. section, so sorry if I got it wrong.
Thanks, Rich.

2. Hello hciR
Originally Posted by hciR
Hi, I have to find the co-ordinates where the line y=-x+1 intersects f(x,y) = x^2 + y^2 -xy - y -1

I can do this with explicit functions by making them equal, but it doesn't work with this.

The actual question asks for the co-ordinates where the tangent to f(x,y), is at 45 deg to the horizontal. So I found the gradient function which = 1 (that got me the linear equation), now i'm stuck.

Not sure if this should be in calc. section, so sorry if I got it wrong.
Thanks, Rich.
I assume that the curve is $f(x,y)=0$, in which case I agree that $y = 1-x$ when $\frac{dy}{dx}=1$.

Just substitute $y = 1-x$ into $f(x,y)=0$ to eliminate $y$, and solve the resulting quadratic in $x$. I make the answer
$x=-\tfrac13, \;1$
Then find the corresponding values of $y$.

3. The line y = -x + 1 lies in the x-y plane, or where f(x,y) = 0.

So sub that in:
0 = x^2 + y^2 - xy -y -1,
0 = x^2 + (-x+1)^2 -x(-x+1) -(-x+1) -1

Now solve for x.

EDIT - I got:

$
x = \frac {2 \pm \sqrt {16} } 6 = 1, -\frac 1 3
$

4. Oh ok, i was going about it the wrong way thanks.

As an aside:- I managed to get those co-ordinates for x earlier in the question, where it asked for the stationary points. Is it just the nature of the shape that the x co-ordinates are the same, when the gradient is 0 and 45 deg?

Also, when working out the S.P I got four possible Y values, and only knew which ones to use by plotting the graph, is there another mathematical way of doing it?
Thanks.