Exercise 71 Page 244

Before we begin, let's assign names to the given lines for easier reference. l :& y= 2x + 3 k :& y= 2x - 7 To find the distance between l and k, we will follow a three-step process.

1. Pick an arbitrary point on line l and construct the perpendicular line p through it.
2. Find the intersection point between lines k and p.
3. Find the distance between the point chosen in the first step and the point found in the second step.

   
   

   Finding the Equation of the Perpendicular Line

   The slope of the given lines is 2, which implies that the slope of the perpendicular line must be m= - 12. As our point of intersection with line l, we will use the y-intercept, ( 0, 3). We can substitute these values into the point-slope form to write the equation of the line.
   y-y_1=m(x-x_1)

   SubstituteValues

   Substitute values

   y- 3 = - 1/2(x- 0)

   ▼

   Solve for y

   SubTerm

   Subtract term

   y-3=- 1/2x

   AddEqn

   LHS+3=RHS+3

   y=- 1/2x+3
   This new equation, y=- 12x+3, is the equation of line p.

   Finding the Intersection Point Between Lines k and p

   To find the intersection point between lines k and p, we can create a system of equations. y=2x-7 & (I) y=- 12x+3 & (II) Since the y-variable is already isolated, we will use the Substitution Method.
   y=2x-7 & (I) y=- 12x+3 & (II)

   ▼

   Solve by substitution

   Substitute

   (I): y= - 12x+3

   - 12x+3=2x-7 y=- 12x+3

   MultEqn

   (I): LHS * 2=RHS* 2

   - x+6=4x-14 y=- 12x+3

   AddEqn

   (I): LHS+x=RHS+x

   6=5x-14 y=- 12x+3

   AddEqn

   (I): LHS+14=RHS+14

   20=5x y=- 12x+3

   DivEqn

   (I): .LHS /5.=.RHS /5.

   4=x y=- 12x+3

   RearrangeEqn

   (I): Rearrange equation

   x=4 y=- 12x+3
   To find the y-coordinate, we will substitute x=4 into the second equation.
   x=4 y=- 12x+3

   ▼

   Solve by substitution

   Substitute

   (II): x= 4

   x=4 y=- 12( 4)+3

   MoveRightFacToNumOne

   (II): 1/b* a = a/b

   x=4 y=- 42+3

   CalcQuot

   (II): Calculate quotient

   x=4 y=- 2+3

   AddTerms

   (II): Add terms

   x=4 y=1
   The point of intersection of lines k and p is (4,1).

   Finding the Distance Between the Two Points

   Finally, to find the distance between lines l and k, we must find the distance between the point on l, (0,3), and the point on k, (4,1). To do so, we will substitute them into the Distance Formula.
   d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)

   SubstitutePoints

   Substitute ( 0,3) & ( 4,1)

   d = sqrt(( 0- 4)^2 + ( 3- 1)^2)

   ▼

   Simplify right-hand side

   SubTerms

   Subtract terms

   d=sqrt((- 4)^2+2^2)

   NegBaseToPosBase

   (- a)^2=a^2

   d=sqrt(4^2+2^2)

   CalcPow

   Calculate power

   d=sqrt(16+4)

   AddTerms

   Add terms

   d=sqrt(20)

   SplitIntoFactors

   Split into factors

   d=sqrt(4* 5)

   SqrtProd

   sqrt(a* b)=sqrt(a)*sqrt(b)

   d=sqrt(4)* sqrt(5)

   CalcRoot

   Calculate root

   d=2sqrt(5)
   The distance between the lines is 2sqrt(5).