Exercise 1 Page 529

The shortest distance between a point and a line is the length of the line segment that is perpendicular to the given line. We will need to find the perpendicular line and then we can find the intersection point. Finally, we can calculate the distance from the given point to the point of intersection.

Finding the Perpendicular Line

Perpendicular lines have negative reciprocal slopes. This means the product of their slopes is equal to -1. m_1 * m_2=- 1 The given line has a slope of 1. y=x+4 ⇔ y= 1x+4By substituting this value into the equation above for m_1, we can find the slope of the perpendicular line.

m_1 * m_2=- 1

Substitute

m_1= 1

1 * m_2=- 1

OneMult

1* a=a

m_2=- 1

Now, to find the equation of the perpendicular line, we can substitute the known point in the slope-intercept form, using that the slope is - 1.

y=-1 x+b

SubstituteII

x= 6, y= 4

4=-1( 6)+b

▼

Solve for b

MultNegPos

(- a)b = - ab

4=-6+b

AddEqn

LHS+6=RHS+6

10=b

RearrangeEqn

Rearrange equation

b= 10

We can add this value of b, along with the known slope, into the slope-intercept form to have a complete equation for the perpendicular line. y= -1x+ 10 ⇔ y=- x+10

Finding the Point of Intersection

To find the 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=x+4 y=- x+10 Since both equations have y isolated, it is most convenient to use the Substitution Method.

y=x+4 & (I) y=- x+10 & (II)

Substitute

(II): y= x+4

y=x+4 x+4=- x+10

▼

(II): Solve for x

AddEqn

(II): LHS+x=RHS+x

y=x+4 2x+4=10

SubEqn

(II): LHS-4=RHS-4

y=x+4 2x=6

DivEqn

(II): .LHS /2.=.RHS /2.

y=x+4 x=3

Having solved the second equation for x, we can substitute this value into the first equation to find the value of y.

y=x+4 x=3

Substitute

(I): x= 3

y= 3+4 x=3

AddTerms

(I): Add terms

y=7 x=3

The lines intersect at (3,7).

Finding the Distance

Now that we know the two endpoints of the segment, we can use the Distance Formula to calculate the length of the segment.

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

SubstitutePoints

Substitute ( 6,4) & ( 3,7)

d=sqrt(( 3- 6)^2+( 7- 4)^2)

▼

Simplify right-hand side

SubTerms

Subtract terms

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

CalcPow

Calculate power

d=sqrt(9+9)

SumToProdTwoTerms

a+a=2a

d=sqrt(2* 9)

SqrtProd

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

d=sqrt(2)*sqrt(9)

CalcRoot

Calculate root

d=sqrt(2)* 3

Multiply

Multiply

d=3sqrt(2)

The distance from the point (6,4) to the line y=x+4 is 3sqrt(2).