Question #e2be8

1 Answer
Altrigeos
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

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