A line has equation y = kx + 6 and a curve has equation y = x^2 + 3x + 2k, where k is a constant.

(i) For the case where k = 2, the line and the curve intersect at points A and B. Find the distance AB and the coordinates of the mid-point of AB.

### 1 Answer | Add Yours

`y = kx+6`

`y = x^2+3x+2k`

when k= 2

`y = 2x+6`

`y= x^2+3x+4`

At intersecting points;

`2x+6 = x^2+3x+4`

`0=x^2+x-2`

`0=x^2+2x-x-2`

`0=x(x+2)-1(x+2)`

`0=(x+2)(x-1)`

`x=-2` and `x=1`

When x= -2 then y = 2

When x = 1 then y = 8

`A=(-2,2)`

`B=(1,8)`

Distance AB `= sqrt((-2-1)^2+(2-8)^2) = sqrt(3^2+6^2) = 3 sqrt(5)`

Mid point of AB `= (1/2)(-2+1),(1/2)(2+8) = (-1/2 , 5)`

*So the distance AB is `3sqrt(5)` units and mid point of AB is `(-1/2,5)` *

**Sources:**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes