# the line y=2x-1 cuts the curve y=x^2-7x-11 at two points calculate the distance between them giving your answer in the form a√b, where a>1

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y=2x-1

y=x^2-7x-11

First we will determine the point where the line and the curve intersect:

x^2-7x-11 = 2x-1

x^2 -9x -10=0

(x-10)(x+1) =0

x1 = 10 and x2= -1

==> y1= 2(10)-1 = 19

==> y2= 2(-1) -1 = -3

The the points of intersection are:

(10,19) and (-1, -3)

Now to calculate the distance between both points ;

d= sqrt[(x2-x1)^2 + (y2-y1)^2]

= sqrt(11^2+22^2) = sqrt(11+ 2(11)^2)

= sqrt(11^2(1+2) = sqrt(11^2) * sqrt(3)

= 11*sqrt(3)

We shall solve the x coordinates of the intersection by substituting y= 2x-1 in the second equation and solve fot x.

2x-1 = x^2-7x-11 Or

0 = x^2-7x-11 - (2x-1) Or

x^2-9x+10 = 0

(x-10)(x+1) = 0.

So x=10 Or x= -1are the x coordinates of the inter section of the curves. The corresponding y coordinates are got by putting x=10 and x=1 in y =2x-1.

For x=10, y = 2*10-1 = 19.

For x = -1, y = 2*(-1)-1 = -3.

So the (x , y) coordinates of intersection points are ((10,19) and (-1, -3).

Therefore the distance between the points of intersection is

sqrt{(-1-10)^2+(-3-19)^2} = sqrt{11^2+22^2}

= 11 sqrt(1+2) = 11 (11)^(1/2).