Exercise 4 Page 232

Let's put the three points on the coordinate plane and draw a line $k$ through the two given points. Next we draw a line through the third point that is perpendicular to line $k .$ The distance of the point from the line can be measured along this perpendicular line.

Here is the plan we will follow to find the distance.

We first find the equation of line $k .$
Next we find the equation of the perpendicular line.
Using the two equations, we can find the point of intersection.
The distance of point $(- 4, 0)$ and this intersection point is the distance of point $(- 4, 0)$ from line $k .$

Let's now see the details.

Equation of Line $k$

To find the equation of line

k,

we first use the Slope Formula to find its slope.

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

We can substitute the coordinates of the given points

(4, 1)

and

(- 5, - 5)

, and simplify the expression to find the slope.

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

SubstitutePoints

Substitute $(4, 1)$ & $(- 5, - 5)$

m = \frac{- 5 - 1}{- 5 - 4}

▼

Simplify

SubTerm

Subtract term

m = \frac{- 6}{- 9}

ReduceFrac

$\frac{a}{b} = \frac{a / - 3}{b / - 3}$

m = \frac{2}{3}

We look for the equation in slope-intercept form.

y = m x + b

We already know the slope,

m = \frac{2}{3} .

We can use

(4, 1),

a point on the line, to find the

y -

intercept

b .

y = m x + b

SubstituteII

$m = \frac{2}{3}$ , $(x, y) = (4, 1)$

1 = (\frac{2}{3}) (4) + b

▼

Solve for

b

MoveRightFacToNum

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

1 = \frac{8}{3} + b

SubEqn

$LHS - \frac{8}{3} = RHS - \frac{8}{3}$

- \frac{5}{3} = b

RearrangeEqn

Rearrange equation

b = - \frac{5}{3}

We now have both the slope and the intercept to write the equation of the line through the points

(4, 1)

and

(- 5, - 5) .

y = \frac{2}{3} x - \frac{5}{3}

Equation of the Perpendicular Line

We again look for the equation in slope-intercept form. Since this line is perpendicular to line

k

their slopes multiply to

- 1 .

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

We can use the slope of line

k,

which we found in the previous step, to find the slope of the perpendicular line.

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

Substitute

$m_{1} = \frac{2}{3}$

(\frac{2}{3}) m_{2} = - 1

▼

Simplify

DivEqn

$LHS / \frac{2}{3} = RHS / \frac{2}{3}$

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

MoveNegNumToFrac

Put minus sign in front of fraction

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

DivOneByFracD

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

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

Knowing this slope we can use

(- 4, 0),

a point on the line, to find the

y -

intercept

b .

y = m x + b

SubstituteII

$m = - \frac{3}{2}$ , $(x, y) = (- 4, 0)$

0 = (- \frac{3}{2}) (- 4) + b

▼

Solve for

b

Multiply

0 = 6 + b

SubEqn

$LHS - 6 = RHS - 6$

- 6 = b

RearrangeEqn

Rearrange equation

b = - 6

We now have both the slope and the intercept to write the equation of the line through the point

(- 4, 0)

that is perpendicular to

k .

y = - \frac{3}{2} x - 6

Intersection Point of the Lines

The diagram below shows the two lines and their equations.

To find the coordinates of the intersection point we solve the system of equations.

{y = \frac{2}{3} x - \frac{5}{3} y = - \frac{3}{2} x - 6 (I) (II)

Substitute

$(II):$ $y = \frac{2}{3} x - \frac{5}{3}$

{y = - \frac{3}{2} x - 6 \frac{2}{3} x - \frac{5}{3} = - \frac{3}{2} x - 6

▼

(II):

Solve for

x

MultEqn

$(II):$ $LHS \cdot 6 = RHS \cdot 6$

{y = - \frac{3}{2} x - 6 6 (\frac{2}{3} x - \frac{5}{3}) = 6 (- \frac{3}{2} x - 6)

Distr

$(II):$ Distribute $6$

{y = - \frac{3}{2} x - 6 4 x - 10 = - 9 x - 36

AddEqn

$(II):$ $LHS + 9 x = RHS + 9 x$

{y = - \frac{3}{2} x - 6 13 x - 10 = - 36

AddEqn

$(II):$ $LHS + 10 = RHS + 10$

{y = - \frac{3}{2} x - 6 13 x = - 26

DivEqn

$(II):$ $LHS / 13 = RHS / 13$

{y = - \frac{3}{2} x - 6 x = - 2

We can use this value of

x

and substitute it in the first equation to get the value of

y .

{y = - \frac{3}{2} x - 6 x = - 2 (I) (II)

Substitute

$(I):$ $x = - 2$

{y = - \frac{3}{2} (- 2) - 6 x = - 2

▼

(I):

Solve for

y

Multiply

$(I):$ Multiply

{y = 3 - 6 x = - 2

SubTerm

$(I):$ Subtract term

{y = - 3 x = - 2

The point of intersection is

(- 2, - 3) .

Distance of the Points

We need the distance of this point from

(- 4, 0) .

We can use the Distance Formula.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(- 2, - 3)$ & $(- 4, 0)$

d = (- 4 - (- 2))^{2} + (0 - (- 3))^{2}

▼

Simplify

SubNeg

$a - (- b) = a + b$

d = (- 4 + 2)^{2} + (0 + 3)^{2}

AddTerms

Add terms

d = (- 2)^{2} + 3^{2}

CalcPow

Calculate power

d = 4 + 9

AddTerms

Add terms

d = 13

RoundDec

Round to $1$ decimal place(s)

d \approx 3.6

Answer

Since the two lines are perpendicular, the distance of $(- 4, 0)$ from the point of intersection is the distance we are looking for.

The distance of $(- 4, 0)$ from line $k$ is $13,$ or about $3.6$ units. The correct answer choice is G.