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...
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
jeew-m | College Teacher | (Level 1) Educator Emeritus
`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)`