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Big Ideas Math Geometry, 2014

BI

Big Ideas Math Geometry, 2014 View details

5. Equations of Parallel and Perpendicular Lines

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Exercise 23 Page 160

The shortest distance between a point and a line is the length of the line segment that is perpendicular to the given line.

sqrt(29) units or about 5.4 units

Practice makes perfect

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 To help us identify the slope of the given line, let's first convert it into slope-intercept form.

5x+2y=4

▼

Write in slope-intercept form

LHS-5x=RHS-5x

2y=- 5x+4

.LHS /2.=.RHS /2.

y=- 5x+4/2

Write as a sum of fractions

y=- 5x/2+4/2

Calculate quotient

y=- 5x/2+2

MoveNegNumToFrac

Put minus sign in front of fraction

y=-5x/2+2

MovePartNumRight

a* b/c=a/c* b

y=-5/2x+2

The given line has a slope of - 52. y= -5/2x+2 By substituting this value into the equation above for m_1, we can find the slope of the perpendicular line.

m_1 * m_2=- 1

m_1= -5/2

-5/2 * m_2=- 1

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Solve for m_2

LHS * 2=RHS* 2

- 5* m_2=- 2

.LHS /(- 5).=.RHS /(- 5).

m_2=- 2/- 5

- a/- b=a/b

m_2=2/5

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 25.

y=2/5 x+b

x= 15, y= - 21

- 21=2/5( 15)+b

▼

Solve for b

MoveRightFacToNum

a/c* b = a* b/c

- 21=2* 15/5+b

a/b=.a /5./.b /5.

- 21=2* 3/1+b

a/1=a

- 21=2* 3+b

Multiply

- 21=6+b

LHS-6=RHS-6

- 27=b

Rearrange equation

b= - 27

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= 2/5x+( - 27) ⇔ y=2/5x-27

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. Note that we will use the slope-intercept form of the given equation. y=- 52x+2 y= 25x-27 Since both equations have y isolated, it is most convenient to use the Substitution Method.

y=- 52x+2 & (I) y= 25x-27 & (II)

(II): y= - 52x+2

y=- 52x+2 - 52x+2= 25x-27

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(II): Solve for x

(II): LHS * 10=RHS* 10

y=- 52x+2 - 25x+20=4x-270

(II): LHS-4x=RHS-4x

y=- 52x+2 - 29x+20=- 270

(II): LHS-20=RHS-20

y=- 52x+2 - 29x=- 290

(II): .LHS /(- 29).=.RHS /(- 29).

y=- 52x+2 x=10

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

y=- 52x+2 x=10

(I): x= 10

y=- 52( 10)+2 x=10

▼

(I): Solve for y

MoveRightFacToNum

(I): a/c* b = a* b/c

y=- 502+2 x=10

(I): Calculate quotient

y=- 25+2 x=10

(I): Add terms

y=- 23 x=10

The lines intersect at (10,- 23).

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 ( 15,- 21) & ( 10,- 23)

d=sqrt(( 10- 15)^2+( - 23-( - 21))^2)

▼

Simplify right-hand side

a-(- b)=a+b

d=sqrt((10-15)^2+(- 23+21)^2)

Add and subtract terms

d=sqrt((- 5)^2+(- 2)^2)

Calculate power

d=sqrt(25+4)

Add terms

d=sqrt(29)

Use a calculator

d=5.38516...

Round to 1 decimal place(s)

d≈ 5.4

The distance from the point (15,- 21) to the line 5x+2y=4 is about 5.4 units.

Subchapter links

Communicate Your Answer p.155

Monitoring Progress p.156-159

Exercises p.160-162