# Theoretical Coordinate Geometry

#### 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 $$\displaystyle x=2+k \pm \sqrt{k^2+4k+3}$$
But I feel that would be finding the co-ordinates of a and b.
Again, any help would be appreciated.

#### wonderboy1953

Opinion

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

#### FreeT

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

#### masters

MHF Helper
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 $$\displaystyle x=2+k \pm \sqrt{k^2+4k+3}$$
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 = $$\displaystyle \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$$

#### Plato

MHF Helper
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.

#### awkward

MHF Hall of Honor
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 $$\displaystyle x=2+k \pm \sqrt{k^2+4k+3}$$
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.

Last edited:
• FreeT