Exercise 14 Page 194

Practice makes perfect

First, we can trace along the given line until we find two points that lie perfectly on grid lines.

The slopes of parallel lines are the same. Therefore, to find the slope of a parallel line, we need to know the slope of the given line. Let's calculate it by substituting the points that lie on the given line into the Slope Formula.

m=y_2-y_1/x_2-x_1

SubstitutePoints

Substitute ( -3,3) & ( -2,-1)

m=-1- 3/-2-( -3)

SubNeg

a-(- b)=a+b

m=-1-3/-2+3

AddSubTerms

Add and subtract terms

m=- 4/1

DivByOne

a/1=a

m=- 4

The slope of the given line and, consequently, of the parallel line is - 4. We can write a partial equation of our line recalling the slope-intercept form. y= mx+b ↓ y= - 4x+b Next, by substituting the given point ( -4, 0) in the above equation, we can find the y-intercept b.

y=mx+b

SubstituteValues

Substitute values

0= - 4( -4)+b

▼

Solve for b

MultNegNegOnePar

- a(- b)=a* b

0=16+b

SubEqn

LHS-16=RHS-16

-16=b

RearrangeEqn

Rearrange equation

b=-16

Now that we know that b= 19, we can write the equation of our line. y=- 4x+( -16) ⇒ y = -4 x - 16

Now, we want to find the equation of a perpendicular line that passes through the given point. When two lines are perpendicular, their slopes are negative reciprocals. This means that their product must be -1. m_1* m_2=- 1 From Part A, we know that the slope of the given line is - 4. We can substitute - 4 for m_1 into the above equation to find m_2, the slope of the perpendicular line.

m_1* m_2=- 1

Substitute

m_1= - 4

( - 4)* m_2=- 1

DivEqn

.LHS /(- 4).=.RHS /(- 4).

m_2=- 1/- 4

DivNegNeg

- a/- b=a/b

m_2=1/4

The slope of the perpendicular line is 14. We can write its partial equation recalling the slope-intercept form. y= mx+b ↓ y= 1/4x+b Next, by substituting the given point ( -4, 0) in the above equation, we can find the y-intercept b.

y=mx+b

SubstituteValues

Substitute values

0= 1/4( - 4)+b

▼

Solve for b

MultPosNeg

a(- b)=- a * b

0=-1+b

AddEqn

LHS+1=RHS+1

1=b