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

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saj-94 | (Level 1) Salutatorian

Posted September 13, 2013 at 11:15 AM via web

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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.

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

Posted September 13, 2013 at 11:33 AM (Answer #1)

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`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)`

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