Thread: Theoretical Coordinate Geometry

1. Theoretical Coordinate Geometry

Okay, I came across this question in an old mock paper. The problem is that I couldn't do it and neither could my teacher. Any help would be appreciated.

The line y=2kx cuts the curve $y=x^2-4x+1$ at the points a and b. Without finding the co-ordinates of a and b, find, in terms of k, the midpoint of [ab].

Now I tried putting $2kx=x^2-4x+1$
and this worked down to $x=2+k \pm \sqrt[2]{k^2+4k+3}$
But I feel that would be finding the co-ordinates of a and b.
Again, any help would be appreciated.

2. Opinion

Without knowing what a and b are, how can you determine what the midpoint is?

3. That's what I was wondering. Maybe there's some relationship between the midpoint and the axis of the curve...

4. Originally Posted by FreeT
Okay, I came across this question in an old mock paper. The problem is that I couldn't do it and neither could my teacher. Any help would be appreciated.

The line y=2kx cuts the curve $y=x^2-4x+1$ at the points a and b. Without finding the co-ordinates of a and b, find, in terms of k, the midpoint of [ab].

Now I tried putting $2kx=x^2-4x+1$
and this worked down to $x=2+k \pm \sqrt[2]{k^2+4k+3}$
But I feel that would be finding the co-ordinates of a and b.
Again, any help would be appreciated.
Hi FreeT,

It would seem that you are on the right track.

Once you have found the coordinates of a and b in terms of k,
you can then determine the midpoint of ab.

$a(x_1, y_1)$ and $b(x_2, y_2)$

Midpoint of ab = $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

5. Solve the equation $x^2-4x+1=2kx$ for $x$ in terms of $k$.
That will give the $x$-coordinates of $a~\&~b$ in terms of $k$.
Use the equation $y=2kx$ to find the $y$-coordinates of $a~\&~b$ in terms of $k$.
The midpoints are just the averages.

6. Originally Posted by FreeT
Okay, I came across this question in an old mock paper. The problem is that I couldn't do it and neither could my teacher. Any help would be appreciated.

The line y=2kx cuts the curve $y=x^2-4x+1$ at the points a and b. Without finding the co-ordinates of a and b, find, in terms of k, the midpoint of [ab].

Now I tried putting $2kx=x^2-4x+1$
and this worked down to $x=2+k \pm \sqrt[2]{k^2+4k+3}$
But I feel that would be finding the co-ordinates of a and b.
Again, any help would be appreciated.
The tricky point is avoiding actually solving the equation. Here is one way.

By substitution,
$2kx = x^2 - 4x +1$
so
$x^2 - (4+2k)x + 1 = 0$

If a and b are the roots of this equation, then their sum must be the negative of the coefficient of x, i.e.
$a+b = 4+2k$.

From here you can find $\frac{a+b}{2}$ and the associated value of y.