Exercise 22 Page 246

To write the equation of a line perpendicular to the given equation, we first need to determine its slope.

Calculating the Perpendicular Line's Slope

Two lines are perpendicular when their slopes are negative reciprocals. This means that the product of a given slope and the slope of a line perpendicular to it will be

- 1 .

m_{1} \cdot m_{2} = - 1

For any equation written in slope-intercept form,

y = m x + b,

we can identify its slope as the value of

m .

Looking at the given equation, we can see that its slope is

- \frac{1}{3} .

y = - \frac{1}{3} x - 5

By substituting this value into our negative reciprocal equation for

m_{1},

we can solve for the slope of the perpendicular line,

m_{2} .

m_{1} \cdot m_{2} = - 1

Substitute

$m_{1} = - \frac{1}{3}$

- \frac{1}{3} \cdot m_{2} = - 1

▼

Solve for

m_{2}

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

- \frac{m _{2}}{3} = - 1

MultEqn

$LHS \cdot (- 3) = RHS \cdot (- 3)$

m_{2} = 3

With this, we can identify that any line perpendicular to the given equation will have a slope of

3 .

Writing the Perpendicular Line's Equation

Using the slope

m_{2} = 3,

we can write a general equation in slope-intercept form for all lines perpendicular to the given equation.

y = 3 x + b

By substituting the given point

(3, - 4)

into this equation for

x

and

y,

we can solve for the

y -

intercept

b

of the perpendicular line.

y = 3 x + b

SubstituteII

$x = 3$ , $y = - 4$

- 4 = 3 (3) + b

▼

Solve for

b

Multiply

- 4 = 9 + b

SubEqn

$LHS - 9 = RHS - 9$

- 13 = b

RearrangeEqn

Rearrange equation

b = - 13

Now that we have the

y -

intercept, we can complete the equation.

y = 3 x + (- 13) \Leftrightarrow y = 3 x - 13

The line given by this equation is both perpendicular to

y = - \frac{1}{3} x - 5

and passes through the point

(3, - 4) .