Before we begin, let's assign names to the given lines for easier reference.

ℓ : k : y = \frac{1}{4} x + 2 4 y - x = - 60

To find the distance between

ℓ

and

k,

we will follow a three-step process.

Pick a point on line $ℓ$ and construct the perpendicular line $p$ through it.
Find the intersection point between lines $k$ and $p .$
Find the distance between the point chosen in the first step and the point found in the second step.

Finding the Equation of a Perpendicular Line

The slope of

ℓ

is

\frac{1}{4} .

Because they are parallel, this is also the slope of

k .

This implies that the slope of the perpendicular line

p

must be

m = - 4 .

As our point of intersection with line

ℓ,

we will use the

y -

intercept,

(0, 2) .

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 - 2 = - 4 (x - 0)

▼

Simplify right-hand side

Multiply

y - 2 = - 4 x

AddEqn

$LHS + 2 = RHS + 2$

y = - 4 x + 2

This new equation,

y = - 4 x + 2,

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.

{4 y - x = - 60 y = - 4 x + 2 (I) (II)

Let's find the solution using the Substitution Method.

{4 y - x = - 60 y = - 4 x + 2 (I) (II)

▼

Solve by substitution

Substitute

$(I):$ $y = - 4 x + 2$

{4 (- 4 x + 2) - x = - 60 y = - 4 x + 2

Distr

$(I):$ Distribute $4$

{- 16 x + 8 - x = - 60 y = - 4 x + 2

SubTerms

$(I):$ Subtract terms

{- 17 x + 8 = - 60 y = - 4 x + 2

SubEqn

$(I):$ $LHS - 8 = RHS - 8$

{- 17 x = - 68 y = - 4 x + 2

DivEqn

$(I):$ $LHS / (- 17) = RHS / (- 17)$

{x = 4 y = - 4 x + 2

To find the

y -

coordinate, we will substitute

x = 4

into the second equation.

{x = 4 y = - 4 x + 2

▼

Solve by substitution

Substitute

$(II):$ $x = 4$

{x = 4 y = - 4 (4) + 2

Multiply

$(II):$ Multiply

{x = 4 y = - 16 + 2

AddTerms

$(II):$ Add terms

{x = 4 y = - 14

The point of intersection is

(4, - 14)

which we will call point

P .

Finding the Distance between the Two Points

Finally, to find the distance between

ℓ

and

k,

we must find the distance between the point on

ℓ,

(0, 2),

and the point on

k,

P (4, - 14) .

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(0, 2)$ & $(4, - 14)$

d = (0 - 4)^{2} + (2 - (- 14))^{2}

▼

Simplify right-hand side

SubNeg

$a - (- b) = a + b$

d = (0 - 4)^{2} + (2 + 14)^{2}

AddSubTerms

Add and subtract terms

d = (- 4)^{2} + (16)^{2}

CalcPow

Calculate power

d = 16 + 256

AddTerms

Add terms

d = 272

Rewrite

Rewrite $272$ as $16 \cdot 17$

d = 16 \cdot 17

RootProd

$a \cdot b = a \cdot b$

d = 16 \cdot 17

CalcRoot

Calculate root

d = 417

The distance between the lines is

417 .

Finding the Equation of a Perpendicular Line

Finding the Intersection Point between Lines k and p

Finding the Distance between the Two Points

Finding the Intersection Point between Lines $k$ and $p$

Exercise

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