Exercise 42 Page 162

Let's begin by writing the line in slope-intercept form. The distance from a point to a line is defined as the shortest distance. This is going to be the segment that runs perpendicular to the line.

Finding the Perpendicular Line

Lines that run perpendicular to each other have slopes that are opposite reciprocals. This means the product of their slopes will equal -1. m_1* m_2=- 1 Examining the slope-intercept form of the line, we know that it has a slope of - ab. By substituting this value into the formula, we can find the slope of the perpendicular line. The slope of the perpendicular line is ba. To find the equation of the line, we substitute the known point in the slope-intercept form, setting the slope to ba, and then solve for the y-intercept.

y=b/ax+c

SubstituteII

y= y_0, x= x_0

y_0=b/a x_0+c

▼

Solve for c

SubEqn

LHS-b/ax_0=RHS-b/ax_0

y_0-b/ax_0=c

RearrangeEqn

Rearrange equation

c=y_0-b/ax_0

MoveRightFacToNum

a/c* b = a* b/c

c=y_0-bx_0/a

NumberToFrac

a = a* a/a

c=ay_0/a-bx_0/a

SubFrac

Subtract fractions

c=ay_0-bx_0/a

Now we can write the equation of the perpendicular line. y=b/ax+ay_0-bx_0/a

Finding the Point of Intersection

To find our distance, we also need to know where the given line and the perpendicular line intersect. By setting up a system of equations, we can find the point of intersection. y=- abx & (I) y= bax+ ay_0-bx_0a & (II) Since both equations have y isolated, it's convenient to use the Substitution Method. We will substitute the value of y from Equation (I) into Equation (II).

y=b/ax+ay_0-bx_0/a

Substitute

y= - a/bx

- a/bx=b/ax+ay_0-bx_0/a

SubEqn

LHS-b/ax=RHS-b/ax

- a/bx-b/ax=ay_0-bx_0/a

FactorOut

Factor out x

x(- a/b-b/a)=ay_0-bx_0/a

DivEqn

.LHS /(- a/b-b/a).=.RHS /(- a/b-b/a).

x=(ay_0-bx_0/a)/(- a/b-b/a)

From here, we subtract the fractions in the denominator before multiplying the numerator and the denominator by ab. x=(ay_0-bx_0/a)* ab/(- a/b-b/a) * ab ⇔ x=b(ay_0-bx_0)/- a^2-b^2 Now, we simplify a little more by multiplying the numerator and denominator by -1. x=b(ay_0-bx_0)* (-1)/(- a^2-b^2)* (-1) ⇔ x=b(bx_0-ay_0)/a^2+b^2 Having solved the second equation for x, we can substitute this value into the first equation to find the value of y.

y=- a/bx

Substitute

x= b(bx_0-ay_0)/a^2+b^2

y=- a/b* b(bx_0-ay_0)/a^2+b^2

▼

Simplify right-hand side

MultFrac

Multiply fractions

y=- ab(bx_0-ay_0)/b(a^2+b^2)

ReduceFrac

a/b=.a /b./.b /b.

y=- a(bx_0-ay_0)/a^2+b^2

MoveNegFracToNum

Put minus sign in numerator

y=- a(bx_0-ay_0)/a^2+b^2

DistrNegSignSwap

-(b-a)=a-b

y=a(ay_0-bx_0)/a^2+b^2

Finding the Distance

When we know the coordinates of the two points making up the segment, we can use the Distance Formula to calculate it.

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

SubstitutePoints

Substitute ( b(bx_0-ay_0)/a^2+b^2,a(ay_0-bx_0)/a^2+b^2) & ( x_0,y_0)

d=sqrt(( b(bx_0-ay_0)/a^2+b^2- x_0)^2+( a(ay_0-bx_0)/a^2+b^2- y_0)^2)

▼

Simplify right-hand side

NumberToFrac

a = (a^2+b^2)* a/(a^2+b^2)

d=sqrt((b(bx_0-ay_0)/a^2+b^2-x_0(a^2+b^2)/a^2+b^2)^2+(a(ay_0-bx_0)/a^2+b^2-y_0(a^2+b^2)/a^2+b^2)^2)

Distr

Distribute coefficients

d=sqrt((b^2x_0-aby_0/a^2+b^2-a^2x_0+b^2x_0/a^2+b^2)^2+(a^2y_0-abx_0/a^2+b^2-a^2y_0+b^2y_0/a^2+b^2)^2)

SubFrac

Subtract fractions

d=sqrt((- aby_0-a^2x_0/a^2+b^2)^2+(- abx_0-b^2y_0/a^2+b^2)^2)

To simplify this further, there are two things we need to do. First, remember that the square of a negative is a positive number. Then, we can use the properties of perfect square trinomials.

d=sqrt((- aby_0-a^2x_0/a^2+b^2)^2+(- abx_0-b^2y_0/a^2+b^2)^2)

▼

(- a)^2 = a^2

FactorOut

Factor out - 1

d=sqrt((- (aby_0+a^2x_0)/a^2+b^2)^2+(- (abx_0+b^2y_0)/a^2+b^2)^2)

MoveNegNumToFrac

Put minus sign in front of fraction

d=sqrt((- aby_0+a^2x_0/a^2+b^2)^2+(-abx_0+b^2y_0/a^2+b^2)^2)

NegBaseToPosPow

(- a)^2 = a^2

d=sqrt((aby_0+a^2x_0/a^2+b^2)^2+(abx_0+b^2y_0/a^2+b^2)^2)

PowQuot

(a/b)^m=a^m/b^m

d=sqrt((aby_0+a^2x_0)^2/(a^2+b^2)^2+(abx_0+b^2y_0)^2/(a^2+b^2)^2)

AddFrac

Add fractions

d=sqrt((aby_0+a^2x_0)^2+(abx_0+b^2y_0)^2/(a^2+b^2)^2)

ExpandPosPerfectSquare

(a+b)^2=a^2+2ab+b^2

d=sqrt(a^2b^2y_0^2 + 2a^3x_0by_0 + a^4x_0^2+a^2x_0^2b^2+2ax_0b^3y_0+b^4y_0^2/(a^2+b^2)^2)

Finally, we can remove common factors from the numerator and the denominator before fully simplifying the distance.

d=sqrt(a^2b^2y_0^2 + 2a^3x_0by_0 + a^4x_0^2+a^2x_0^2b^2+2ax_0b^3y_0+b^4y_0^2/(a^2+b^2)^2)

▼

Factor

FactorOut

Factor out a^2

d=sqrt(a^2( b^2 y_0^2 + 2ax_0b y_0 + a^2 x_0^2)+a^2x_0^2b^2+2ax_0b^3y_0+b^4y_0^2/(a^2+b^2)^2)

FactorOut

Factor out b^2

d=sqrt(a^2( b^2 y_0^2 + 2ax_0by_0 + a^2 x_0^2)+b^2(a^2x_0^2+2ax_0by_0+b^2 y_0^2)/(a^2+b^2)^2)

CommutativePropAdd

Commutative Property of Addition

d=sqrt(a^2(a^2 x_0^2 + 2ax_0by_0 + b^2 y_0^2)+b^2(a^2x_0^2+2ax_0by_0+b^2 y_0^2)/(a^2+b^2)^2)

FactorOut

Factor out (a^2+b^2)

d=sqrt((a^2+b^2)(a^2 x_0^2 + 2ax_0by_0 + b^2 y_0^2)/(a^2+b^2)^2)

ReduceFrac

a/b=.a /(a^2+b^2)./.b /(a^2+b^2).

d=sqrt(a^2 x_0^2 + 2ax_0by_0 + b^2 y_0^2/a^2+b^2)

FacPosPerfectSquare

a^2+2ab+b^2=(a+b)^2

d=sqrt((ax_0 + by_0)^2/a^2+b^2)

SqrtQuot

sqrt(a/b)=sqrt(a)/sqrt(b)

d=sqrt((ax_0 + by_0)^2)/sqrt(a^2+b^2)

SqrtToAbs

sqrt(a^2)=|a|

d=|ax_0 + by_0|/sqrt(a^2+b^2)

Having found the formula, we can try it using the point (1,1) and the line 2x+3y=0. This means a= 2, b= 3, x_0= 1, and y_0=1

d=|ax_0 + by_0|/sqrt(a^2+b^2)

SubstituteValues

Substitute values

d=| 2( 1) + 3(1)|/sqrt(2^2+ 3^2)

▼

Simplify right-hand side

Multiply

Multiply

d=|2 + 3|/sqrt(2^2+ 3^2)

CalcPow

Calculate power

d=|2 + 3|/sqrt(4+9)

AddTerms

Add terms

d=|5|/sqrt(13)

AbsPos

|5|=5

d=5/sqrt(13)

UseCalc

Use a calculator

d≈ 1.39

The distance is approximately 1.39 units.