{{ item.displayTitle }}

No history yet!

Student

Teacher

{{ item.displayTitle }}

{{ item.subject.displayTitle }}

{{ searchError }}

{{ courseTrack.displayTitle }} {{ statistics.percent }}% Sign in to view progress

{{ printedBook.courseTrack.name }} {{ printedBook.name }}
Parallel and perpendicular lines can be analyzed in the coordinate plane. Here, their slopes are used to prove their inherent characteristics.

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) $

${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} $

${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) $

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

${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.

$ℓ_{1}∥ℓ_{2}⇔m_{1}=m_{2}$

Consider two lines in a coordinate plane. In slope-intercept form, their equations can be written as follows. $y=m_{1}x+b_{1}andy=m_{2}x+b_{2}$ The intersection of these lines is the $x$-$y$ point that satisfies both equations. Setting the equations equal to each other makes it possible to solve for the $x$-coordinate of the intersection point.

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

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

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

SubEqn$LHS−b_{2}=RHS−b_{2}$

$m_{2}x=m_{1}x+b_{1}−b_{2}$

SubEqn$LHS−m_{1}x=RHS−m_{1}x$

$m_{2}x−m_{1}x=b_{1}−b_{2}$

FactorOutFactor out $x$

$x(m_{2}−m_{1})=b_{1}−b_{2}$

DivEqn$LHS/(m_{2}−m_{1})=RHS/(m_{2}−m_{1})$

$x=m_{2}−m_{1}b_{1}−b_{2} $

The above proof can be summarized in a flowchart.

In a coordinate plane two nonvertical lines are perpendicular if and only if the product of their slopes is $-1.$

Consider the perpendicular lines $D$ and $E$ in the coordinate plane, that form a right triangle with the $x$-axis.

Using the points on each line, the slopes of $D$ and $E$ can be written as follows. $m_{D}=ac andm_{E}=b-c $ Multiplying the slopes gives their product. $ac ⋅b-c =ab-c_{2} $ Notice that the product equals $-1$ if and only if the following is true. $ab-c_{2} =-1⇔c_{2}=ab⇔c=ab $ The Geometric Mean Altitude Theorem states that the altidtude of a right triangle is equal to the geometric mean of the segments the altitude creates with the hypotenuse. Notice that $c$ is the altitude of the right triangle in the diagram. Also, $c$ divides the hypotenuse into the segments $a$ and $b.$ Thus, $c=ab $ is a true statement. Therefore, the product of the slopes of two perpendicular lines equals $-1.$

$m_{D}⋅m_{E}=-1$The points $P(0.8,-0.6),$ $Q(-1.6,4.2),$ $R(1.2,5.6),$ and $S(3.6,0.8)$ are vertices of a quadrilateral. Prove that quadrilateral is a rectangle.

Show Solution

To begin, we can graph the quadrilateral in a coordinate plane.

If quadrilateral $PQRS$ is a rectangle, both pairs of opposite sides will be parallel and all pairs of adjacent sides will be perpendicular. Thus, we must show that

- $PQ ∥RS$,
- $PS∥QR $, and
- $QR ⊥RS.$

$PQ=x_{2}−x_{1}y_{2}−y_{1} $

$PQ=-1.6−0.84.2−(-0.6) $

SubNeg$a−(-b)=a+b$

$PQ=-1.6−0.84.2+0.6 $

AddSubTermsAdd and subtract terms

$PQ=-2.44.8 $

CalcQuotCalculate quotient

$PQ=-2$

Line segment | Slope formula | Simplified | Slope |
---|---|---|---|

$PQ $ | $-1.6−0.84.2−(-0.6) $ | $-2.44.8 $ | $m_{PQ}=-2$ |

$QR $ | $1.2−(-1.6)5.6−4.2 $ | $2.81.4 $ | $m_{QR}=0.5$ |

$RS$ | $3.6−1.20.8−5.6 $ | $2.4-4.8 $ | $m_{RS}=-2$ |

$SP$ | $0.8−3.6-0.6−0.8 $ | $-2.8-1.4 $ | $m_{SP}=0.5$ |

From the slopes above, we can see that $PQ $ and $RS$ have equal slopes. Thus, they are parallel. The same can be said for $QR $ and $SP.$ To determine if $QR ⊥RS,$ we can see if the product of their slopes equals $-1.$ $m_{Q}R⋅m_{R}S=0.5(-2)=-1$ Thus, $QR ⊥RS.$ It follows that all pairs of adjacent sides are perpendicular. Therefore, the quadrialteral is a rectangle.

{{ 'mldesktop-placeholder-grade' | message }} {{ article.displayTitle }}!

{{ exercise.headTitle }}

{{ 'ml-heading-exercise' | message }} {{ focusmode.exercise.exerciseName }}