First, let's find the two points at which the two equations intersect.
From the first equation, we have 2x-y = 7 => y = 2x-72x−y=7⇒y=2x−7.
If we substitute this into the quadratic, we get
(2x-7)^2-x(x+2x-7)=11(2x−7)2−x(x+2x−7)=11
=> 4x^2-28x+49-3x^2+7x=11⇒4x2−28x+49−3x2+7x=11
=> x^2-21x+38 = 0⇒x2−21x+38=0
=> (x-2)(x-19) = 0⇒(x−2)(x−19)=0
=> x = 2 or x = 19⇒x=2orx=19
We now have our two xx-coordinates of the intersections as x_1 = 2x1=2 and x_2 = 19x2=19. We can substitute these into the first equation to get the points:
y_1 = 2(2)-7 = -3y1=2(2)−7=−3
y_2 = 2(19)-7 = 31y2=2(19)−7=31
So, we have the two points A(2, -3)A(2,−3) and B(19, 31)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)distanceAB=√(x2−x1)2+(y2−y1)2
=sqrt((19-2)^2+(31-(-3))^2)=√(19−2)2+(31−(−3))2
=sqrt(1445)=√1445
=17sqrt(5)=17√5
