Question #0dcca

1 Answer
sente
Oct 13, 2016

#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)#

Related questions
See all questions in Vertex Form of a Quadratic Equation
Impact of this question
959 views around the world
You can reuse this answer
Creative Commons License