Question

Question #0dcca

Answer 1

`17sqrt(5)`

Explanation:

First, let's find the two points at which the two equations intersect.

From the first equation, we have `2x-y = 7 => y = 2x-7`.

If we substitute this into the quadratic, we get

`(2x-7)^2-x(x+2x-7)=11`

`=> 4x^2-28x+49-3x^2+7x=11`

`=> x^2-21x+38 = 0`

`=> (x-2)(x-19) = 0`

`=> x = 2 or x = 19`

We now have our two `x`-coordinates of the intersections as `x_1 = 2` and `x_2 = 19`. We can substitute these into the first equation to get the points:

`y_1 = 2(2)-7 = -3`

`y_2 = 2(19)-7 = 31`

So, we have the two points `A(2, -3)` and `B(19, 31)`. We can now find the distance between them using the formula

`"distance"_(AB) = sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`=sqrt((19-2)^2+(31-(-3))^2)`

`=sqrt(1445)`

`=17sqrt(5)`