Exercise 70 Page 244

We are given two horizontal lines. &y=- 6 &y=1 Therefore, the line perpendicular to both lines could be any vertical line. Let's consider the line x=3. It intersects the lines at P(3,- 6) and Q(3,1), respectively.

The distance between these two lines is the distance between P and Q. Let's use the Distance Formula to find it!

d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)

SubstitutePoints

Substitute ( 3,- 6) & ( 3,1)

d = sqrt(( 3- 3)^2 + ( - 6- 1)^2)

▼

Simplify right-hand side

SubTerms

Subtract terms

d=sqrt(0^2+(- 7)^2)

NegBaseToPosBase

(- a)^2=a^2

d=sqrt(0^2+7^2)

CalcPow

Calculate power

d=sqrt(0+49)

AddTerms

Add terms

d=sqrt(49)

CalcRoot

Calculate root

d=7

Therefore, the distance between the lines is 7 units.