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{{ printedBook.courseTrack.name }} {{ printedBook.name }} In this lesson, the relationship between the slopes of parallel and perpendicular lines will be explored.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

The lines $ℓ_{1}$ and $ℓ_{2}$ are positioned as shown in the graph. Move the point $C$ vertically.

Compare the slope triangles. What conclusion can be made about the triangles and the lines?

In the previous graph, the slope triangles $△ABF$ and $△CDE$ can be mapped onto each other by a translation. Since translations are rigid motions, it can be concluded that $△ABF$ and $△CDE$ are congruent triangles. Because the slope triangles are congruent, the slopes of the lines are equal. This means that the lines are parallel.

In a coordinate plane, two distinct non-vertical lines are parallel if and only if their slopes are equal.

Let $ℓ_{1}$ and $ℓ_{2}$ be parallel lines, and $m_{1}$ and $m_{2}$ be their respective slopes. Then, the following statement is true.

$ℓ_{1}∥ℓ_{2}⇔m_{1}=m_{2}$

The slope of a vertical line is not defined. Therefore, this theorem only applies to non-vertical lines. Any two distinct vertical lines are parallel.

Since the theorem consists of a biconditional statement, the proof will consists in two parts.

- If two distinct non-vertical lines are parallel, then their slopes are equal.
- If the slopes of two distinct non-vertical lines are equal, then the lines are parallel.

${y=m_{1}x+b_{1}y=m_{2}x+b_{2} (I)(II) $

Substitute

$(I):$ $y=m_{2}x+b_{2}$

${m_{2}x+b_{2}=m_{1}x+b_{1}y=m_{2}x+b_{2} $

$(I):$ Solve for $x$

${x=m_{2}−m_{1}b_{1}−b_{2} y=m_{2}x+b_{2} $

${x=m_{2}−m_{1}b_{1}−b_{2} y=m_{2}x+b_{2} $

Substitute

$(II):$ $x=m_{2}−m_{1}b_{1}−b_{2} $

${x=m_{2}−m_{1}b_{1}−b_{2} y=m_{2}(m_{2}−m_{1}b_{1}−b_{2} )+b_{2} $

$ℓ_{1}∥ℓ_{2}⇒m_{1}=m_{2}$

${y=mx+b_{1}y=mx+b_{2} (I)(II) $

Substitute

$(I):$ $y=mx+b_{2}$

${mx+b_{2}=mx+b_{1}y=mx+b_{2} $

SubEqn

$(I):$ $LHS−mx=RHS−mx$

${b_{2}=b_{1}y=mx+b_{2} $

$m_{1}=m_{2}⇒ℓ_{1}∥ℓ_{2}$

Both directions of the biconditional statement have been proved.

$ℓ_{1}∥ℓ_{2}⇔m_{1}=m_{2}$

If the equation of a linear function is written in slope-intercept form, its slope can be identified.

By rewriting the given equation in slope-intercept form, find the slope of a parallel line to the line whose equation is shown. If necessary, round your answer to $2$ decimal places.

Consider the equation of a line written in slope-intercept form.
$y=-2x−5 $
Which of the following is the equation of the line that passes through the point $(2,-3)$ and is parallel to the given line. ### Hint

### Solution

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class=\"mord\">7<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.625em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">y<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.72777em;vertical-align:-0.08333em;\"><\/span><span class=\"mord text\"><span class=\"mord\">-<\/span><\/span><span class=\"mord\">2<\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">\u2212<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.625em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">y<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:1.190108em;vertical-align:-0.345em;\"><\/span><span class=\"mord\"><span class=\"mopen nulldelimiter\"><\/span><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.845108em;\"><span style=\"top:-2.6550000000000002em;\"><span class=\"pstrut\" style=\"height:3em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span 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style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">4<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.625em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">y<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.72777em;vertical-align:-0.08333em;\"><\/span><span class=\"mord text\"><span class=\"mord\">-<\/span><\/span><span class=\"mord\">2<\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><\/span><\/span><\/span>"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":3}

Start by identifying the slope of the given line. Then, use the Slopes of Parallel Lines Theorem.

Consider both the general form of the slope-intercept form of a line and the given line. $Slope-Intercept Form:Given Line: y=mx+by=-2x−5 $
In the slope-intercept form, $m$ represents the slope and $b$ the $y-$intercept. Since the given line is in this form, the slope of the line is $-2.$ By the Slopes of Parallel Lines Theorem, a parallel line to this line also has a slope of $-2.$ $Given Line:Parallel Line: y=-2x−5y=-2x+b $
It is given that this parallel line passes through the point $(2,-3).$ By substituting this point into the equation of the parallel line, its $y-$intercept can be found.
The equation of the parallel line to $y=-2x−5$ through the point $(2,-3)$ is $y=-2x+1.$
$Given Line:Parallel Line: y=-2x−5y=-2x+1 $

Up to now, it has been discussed how to find the equation of a parallel line to a line whose equation is given. What about finding the equation of a parallel line to a line whose *graph* is given?

The first transatlantic telegraph cable was laid between Valentia in western Ireland and Trinity Bay Newfoundland in the $1850s.$ With the invention of fiber optic cables, the number of transatlantic cables has increased. The following map shows some of these cables.

Kevin wants to write the equation of a line parallel to the first transatlantic cable and passes through point $V(-4,-3),$ a specific location in Virginia beach. Kevin draws a coordinate plane, then line $ℓ_{1},$ on which the cable lies, and the point $V$ as shown. a If Kevin wants to use transformations to find the equation of the parallel line, what would be his next step? Use transformations to write the equation of the parallel line. Write the equation in slope-intercept form.

b Use the Slopes of Parallel Lines Theorem to find the equation of the line. Write the equation in slope-intercept form.

a **Example Next Step: ** The next step would be determining the number of units needed to vertically translate the given line so that it passes through the given point.

$y=41 x−2$

$y=41 x−2$

b **Equation: ** $y=41 x−2$

a By which transformation can a line be mapped onto a parallel line?

b What is the slope of the given line?

a Parallel lines can be mapped onto each other by a translation. Therefore, the next step will be to determine the number of units needed to translate the line so that it passes through $V.$
### Step $1$

### Step $2$

As can be seen in the graph, if the given line is translated $5$ units down, a parallel line through $V(-4,-3)$ is obtained. As a result, the $y-$values of the parallel line will be $5$ less than the corresponding $y-$values of the given line. With this knowledge, the equation of the parallel line can be found by following two steps.

- Find the equation of the given line.
- Translate the given line $5$ units down.

By observing the graph, it can be seen that the line $ℓ_{1}$ passes through the points $(0,3)$ and $(4,4).$ When a slope triangle is constructed between these points, the slope of the line is calculated as $41 .$

It can be seen that the line intercepts the $y-$axis at $(0,3).$ Therefore, the $y-$intercept is $3.$ Knowing the slope and the $y-$intercept of line is enough to write its equation in slope-intercept form. $ℓ_{1}:y=41 x+3 $

Recall that the $y-$values of the parallel line are $5$ less than the corresponding $y-$values of the given line. Therefore, to obtain the equation of the desired line, $5$ units must be subtracted from the obtained equation. $Given Line y=f(x)⇓y=41 x+3 Parallel Line y=f(x)−5⇓y=41 x+3−5y=41 x−2 $

The equation of the parallel line to $ℓ_{1}$ through $V(-4,-3)$ is $y=41 x−2.$ b Using the points $(0,3)$ and $(4,4),$ the slope of the line is calculated as $41 .$
Therefore, by the Slopes of Parallel Lines Theorem, all parallel lines to $ℓ_{1}$ have a slope of $41 .$ Then, the equation of the parallel line passing through $P$ can be written as follows, where $b$ is the $y-$intercept. $y=41 x+b $
Since $V(-4,-3)$ should lie on the line, the value of $b$ can be found by substituting its coordinates into the above equation.
The equation of the parallel line to $ℓ_{1}$ that passes through $V(-4,-3)$ is $y=41 x−2.$ The equation is the same as the equation obtained in Part A, as expected.

$y=41 x+b$

SubstituteII

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

$-3=41 (-4)+b$

Solve for $b$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$-3=4-4 +b$

MoveNegNumToFrac

Put minus sign in front of fraction

$-3=-44 +b$

QuotOne

$aa =1$

$-3=-1+b$

AddEqn

$LHS+1=RHS+1$

$-2=b$

RearrangeEqn

Rearrange equation

$b=-2$

Recall that a system of linear equations can have infinitely many solutions, one solution or no solution. Under which conditions does a system of linear equations have infinitely many solutions or no solution?

Consider the following systems of equations. $System I:System II: {y=3x−32y=6x−6 (I)(II) {y=-5x−3y=-5x−1 (I)(II) $ Determine the number of solutions to each system without finding the actual solutions, if exist.**System I: ** Infinitely many solutions

**System II: ** No solution

Determine whether the equations in the system represent parallel lines. What does this say about the number of solutions?

Start by examining the first system of equations.
$System I: {y=3x−32y=6x−6 (I)(II) $
Dividing both sides of the second equation by $2$ will result in the first equation. This result means that the equations in this system represent the same line. ${y=3x−32y=6x−6 (II)÷2 {y=3x−3y=3x−3 $
Therefore, the lines are coincidental, and they have infinitely many points of intersection. Consequently, System (I) has **infinitely many solutions**. The equations in the System (II), however, differ in their $y-$intercepts. $System II: {y=-5x−3y=-5x−1 (I)(II) $
Since the equations are written in slope-intercept form, their slopes can be identified as $-5.$

Equation | Slope | $y-$Intercept |
---|---|---|

$y=-5x+(-3)$ | $-5$ | $-3$ |

$y=-5x+(-1)$ | $-5$ | $-1$ |

By the Slopes of Parallel Lines Theorem, these lines are parallel and do **not** intersect. Therefore, System (II) has **no solution**.

From the previous example, the following conclusions about systems of linear equations can be made.

Condition | Conclusion | Example |
---|---|---|

The lines of the system have the same slope and the same $y-$intercept. | The lines are coincidental. This means that there are infinitely many points of intersection. Therefore, the system has infinitely many solutions. | ${y=3x+1y=3x+1 $ |

The lines of the system have the same slope but different $y-$intercept. | The lines are parallel. This means that there is not a point of intersection. Therefore, the system has no solution. | ${y=3x+1y=3x+5 $ |

The lines of the system have different slopes. | The lines are neither parallel nor coincidental. This means that there is one point of intersection. Therefore, the system has one solution. | ${y=3x+1y=-2x+5 $ |

These conclusions can be seen in the following diagram.

In the graph below, the slope triangles *seem* to be congruent. The congruence of these triangles can be shown by a rotation. Move the slider to rotate line $ℓ_{1}$ counterclockwise around point $A.$

Think about these questions.

- Is it possible to find a relationship between the slopes of the lines before rotating $ℓ_{1}?$
- How does the rotation change the slope of line $ℓ_{1}?$

In the previous graph, it can be seen that the initial angle between the lines measures $90_{∘}.$ Therefore, the lines are perpendicular.

In a coordinate plane, two non-vertical lines are perpendicular if and only if their slopes are negative reciprocals.

If $ℓ_{1}$ and $ℓ_{2}$ are two perpendicular lines with slopes $m_{1}$ and $m_{2},$ respectively, the following relation holds true.

$ℓ_{1}⊥ℓ_{2}⇔m_{1}⋅m_{2}=-1$

This theorem does not apply to vertical lines because their slope is undefined. However, vertical lines are always perpendicular to horizontal lines.

Since the theorem is a biconditional statement, it will be proven in two parts.

- If two non-vertical lines are perpendicular, then the product of their slopes is $-1.$
- If the product of the slopes of two non-vertical lines is $-1,$ then the lines are perpendicular.

Let $m_{1}$ and $m_{2}$ be the slopes of the lines $ℓ_{1}$ and $ℓ_{2},$ respectively. Next, consider the vertical line $x=1.$ This line intersects both $ℓ_{1}$ and $ℓ_{2}.$

Since $ℓ_{1}$ and $ℓ_{2}$ are assumed to be perpendicular, $△AOC$ is a right triangle. Using the Distance Formula, the lengths of the sides of this triangle can be found.

Side | Points | $Distance Formula((x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} ) $ | Length |
---|---|---|---|

$AO$ | $A(1,m_{1})$ $&$ $O(0,0)$ | $(0−1)_{2}+(0−m_{1})_{2} $ | $1+m_{1} $ |

$CO$ | $C(1,m_{2})$ $&$ $O(0,0)$ | $(0−0)_{2}+(0−m_{2})_{2} $ | $1+m_{2} $ |

$CA$ | $C(1,m_{2})$ $&$ $A(1,m_{1})$ | $(1−1)_{2}+(m_{1}−m_{2})_{2} $ | $m_{1}−m_{2}$ |

$AO_{2}+CO_{2}=CA_{2}$

SubstituteExpressions

Substitute expressions

$(1+m_{1} )_{2}+(1+m_{2} )_{2}=(m_{1}−m_{2})_{2}$

Simplify

PowSqrt

$(a )_{2}=a$

$1+m_{1}+1+m_{2}=(m_{1}−m_{2})_{2}$

AddTerms

Add terms

$2+m_{1}+m_{2}=(m_{1}−m_{2})_{2}$

ExpandNegPerfectSquare

$(a−b)_{2}=a_{2}−2ab+b_{2}$

$2+m_{1}+m_{2}=m_{1}−2m_{1}m_{2}+m_{2}$

SubEqn

$LHS−m_{1}=RHS−m_{1}$

$2+m_{2}=-2m_{1}m_{2}+m_{2}$

SubEqn

$LHS−m_{2}=RHS−m_{2}$

$2=-2m_{1}m_{2}$

DivEqn

$LHS/(-2)=RHS/(-2)$

$-22 =m_{1}m_{2}$

MoveNegDenomToFrac

Put minus sign in front of fraction

$-22 =m_{1}m_{2}$

QuotOne

$aa =1$

$-1=m_{1}m_{2}$

RearrangeEqn

Rearrange equation

$m_{1}⋅m_{2}=-1$

$ℓ_{1}⊥ℓ_{2}⇒m_{1}⋅m_{2}=-1$

Here it is assumed that the slopes of two lines $ℓ_{1}$ and $ℓ_{2},$ are opposite reciprocals. $m_{1}⋅m_{2}=-1 $ Consider the steps taken in Part $1.$ This time, it should be shown that $△AOC$ is a right triangle.

If the lengths of the sides of $△AOC$ satisfy the Pythagorean Theorem, then the triangle is a right triangle. $AO_{2}+CO_{2}=?CA_{2} $ The side lengths, which were previously solved for in Part 1, can be substituted in the above equation.$AO_{2}+CO_{2}=?CA_{2}$

SubstituteExpressions

Substitute expressions

$(1+m_{1} )_{2}+(1+m_{2} )_{2}=?(m_{1}−m_{2})_{2}$

Simplify

PowSqrt

$(a )_{2}=a$

$1+m_{1}+1+m_{2}=?(m_{1}−m_{2})_{2}$

AddTerms

Add terms

$2+m_{1}+m_{2}=?(m_{1}−m_{2})_{2}$

ExpandNegPerfectSquare

$(a−b)_{2}=a_{2}−2ab+b_{2}$

$2+m_{1}+m_{2}=?m_{1}−2m_{1}m_{2}+m_{2}$

SubEqn

$LHS−m_{1}=RHS−m_{1}$

$2+m_{2}=?-2m_{1}m_{2}+m_{2}$

SubEqn

$LHS−m_{2}=RHS−m_{2}$

$2=?-2m_{1}m_{2}$

Substitute

$m_{1}m_{2}=-1$

$2=?-2(-1)$

MultNegNegOnePar

$-a(-b)=a⋅b$

$2=2✓$

$m_{1}⋅m_{2}=-1⇒ℓ_{1}⊥ℓ_{2}$

The biconditional statement has been proven.

$ℓ_{1}⊥ℓ_{2}⇔m_{1}⋅m_{2}=-1$

If the equation of a linear function is written in slope-intercept form, its slope can be easily identified.

Find the slope of the line perpendicular to the given equation's line. This can be done by rewriting the given equation in slope-intercept form. If necessary, round the answer to two decimal places.

Just like with parallel lines, there is an infinite number of perpendicular lines to a given line. Two pieces of information will help in writing the equation of a perpendicular line. These are the slope — which is calculated using the slope of the given line — and a point that lies on the line.

The equation of a line is given in standard form. $2x+y=-3 $ Determine the equation of a perpendicular line to the given line that passes through the point $(-2,-1).${"type":"choice","form":{"alts":["<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.625em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">y<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.72777em;vertical-align:-0.08333em;\"><\/span><span class=\"mord\">0<\/span><span class=\"mord\">.<\/span><span class=\"mord\">5<\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">\u2212<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.625em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">y<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.72777em;vertical-align:-0.08333em;\"><\/span><span class=\"mord\">2<\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">3<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.625em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">y<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">0<\/span><span class=\"mord\">.<\/span><span class=\"mord\">5<\/span><span class=\"mord mathdefault\">x<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.625em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">y<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.66666em;vertical-align:-0.08333em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><\/span><\/span><\/span>"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":2}

Start by identifying the slope of the given line. Then, use the Slopes of Perpendicular Lines Theorem.

For any equation written in slope-intercept form $y=mx+b,$ the value of $m$ is its slope. Since the given equation is not written in this form, it will be rewritten so that the slope can be identified.
$2x+y=-3⇔y=-2x−3 $
The given equation's slope is $-2.$ Now, by the Slopes of Perpendicular Lines Theorem, the product of the slope of the given line and the slope of a line perpendicular must be $-1.$
$m_{1}⋅m_{2}=-1 $
By substituting $-2$ for $m_{1}$ in the above equation, the slope of a perpendicular line $m_{2}$ can be found.
This means that any perpendicular line to the given line will have a slope of $0.5.$ Therefore, a general equation in slope-intercept form for all the lines perpendicular to the given line can be written. $y=0.5x+b $
It is given that this perpendicular line passes through the point $(-2,-1).$ By substituting this point into the above equation, the value of $b$ will be found.
Therefore, the equation of the perpendicular line to the line with equation $2x+y=-3$ through $(2,-3)$ is $y=0.5x.$
$Given Line:Perpendicular Line: 2x+y=-3y=0.5x $

It has been discussed how to find the equation of a line that is perpendicular to a line whose equation is given. What about finding the equation of a line perpendicular to a line whose *graph* is given?

Mark says that perpendicular lines have the same slope as the given line. Paulina, on the other hand, says that the slopes of a line and a perpendicular line are opposite reciprocals.

a Determine who is correct.

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b Which of the following is the equation of the perpendicular line through $(0,38 )?$

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a What does the Slopes of Perpendicular Lines Theorem state?

b Find the slope of the given line.

a Recall that the Slopes of Perpendicular Lines Theorem states that two nonvertical lines are perpendicular if and only if they have opposite reciprocal slopes. From here, it can be concluded that Paulina is correct and Mark is not.
$Paulina:Mark: ✓× $

b To find the equation of a perpendicular line, start by determining the slope of the given line.

As it can be seen in the graph, the slope of the given line is $23 .$ By the above theorem, the product of the slope of the given line and the slope of a perpendicular line must be $-1.$
$m_{1}⋅m_{2}=-1 $
By substituting $23 $ for $m_{1}$ into the above equation, the value of $m_{2}$ can be found.
The slope of all perpendicular lines to the given line is $-32 .$ Therefore, a general equation in slope-intercept form for all lines perpendicular to the given line can be written as follows. $y=-32 x+b $
Finally, the value of the $y-$intercept $b$ needs to be found. It is known that the perpendicular line passes through the point $(0,38 ).$ Since the $x-$coordinate of this point is $0,$ then the value of $b$ is equal to the $y-$coordinate of the point, which is $38 .$ With this information, the desired line can be written. $Equation of The Perpendicular Line y=-32 x+38 $

The theorems seen in this lesson can be used to identify quadrilaterals and some of their properties.

Determine whether quadrilateral $ABCD$ is a parallelogram. Explain the reasoning.

Yes, see solution.

Recall that a parallelogram is a quadrilateral with two pairs of parallel sides.

From the graph, it appears that $AB$ and $CD$ are parallel and that $BC$ and $DA$ are parallel. To prove this claim, start by finding the slope of each side.

As it can be seen, the slopes of the sides $AB$ and $DC$ are the same, as well as the slopes of $AB$ and $DC.$ Therefore, by the Slopes of Parallel Lines Theorem, $AB$ and $DC$ are parallel and $BC$ and $AD$ are parallel. $AB∥DCandBC∥AD $
Since the given quadrilateral has two pairs of parallel sides, it is a parallelogram.

Yes, see solution.

Start by drawing the diagonals of the rhombus. Then, find the slopes of the diagonals.

Start by drawing the diagonals of the rhombus. Then, find the slopes of the diagonals.

The slope $m_{1}$ of $KL$ is $1$ and the slope $m_{2}$ of $NL$ is $-1.$ Notice that their product is $-1.$
$m_{1}⋅m_{2}=1(-1)=-1 $
By the Slopes of Perpendicular Lines Theorem, it can be concluded that the diagonals are perpendicular to each other.
$KM⊥LN $

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