Question #e2be8

1 Answer
May 28, 2017

See below.

Explanation:

Find the intersection of the line and curve:
2x+y=14
y=14-2x

Substitute in the curve:
2x^2-y^2=2xy-6
2x^2-(14-2x)^2=2x(14-2x)-6
2x^2-(4x^2-56x+196)=28x-4x^2-6
-2x^2+56x-196=-4x^2+28x-6
2x^2+28x-190=0
x^2+14x-95=0
(x+19)(x-5)=0

x=-19 Then y=14-2(-19)=52
x=5 Then y=14-2(5)=4

Thus the two points of intersection are:
(-19,52) and (5,4)

Now find the distance:
d=sqrt((x_1-x_2)^2+(y_1-y_2)^2)
d=sqrt((-19-5)^2+(52-4)^2)
d=sqrt(24^2+48^2)
d=sqrt(24^2(1+2^2))
d=24sqrt(1+4)
d=24sqrt(5)

QED