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 $\displaystyle y=x^2-4x+1$ 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 $\displaystyle 2kx=x^2-4x+1$
and this worked down to $\displaystyle x=2+k \pm \sqrt[2]{k^2+4k+3}$
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 $\displaystyle y=x^2-4x+1$ 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 $\displaystyle 2kx=x^2-4x+1$
and this worked down to $\displaystyle x=2+k \pm \sqrt[2]{k^2+4k+3}$
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.

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

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

5. Solve the equation $\displaystyle x^2-4x+1=2kx$ for $\displaystyle x$ in terms of $\displaystyle k$.
That will give the $\displaystyle x$-coordinates of $\displaystyle a~\&~b$ in terms of $\displaystyle k$.
Use the equation $\displaystyle y=2kx$ to find the $\displaystyle y$-coordinates of $\displaystyle a~\&~b$ in terms of $\displaystyle k$.
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 $\displaystyle y=x^2-4x+1$ 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 $\displaystyle 2kx=x^2-4x+1$
and this worked down to $\displaystyle x=2+k \pm \sqrt[2]{k^2+4k+3}$
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,
$\displaystyle 2kx = x^2 - 4x +1$
so
$\displaystyle x^2 - (4+2k)x + 1 = 0$

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

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