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# Proving Relationships of Parallel and Perpendicular Lines

Parallel and perpendicular lines can be analyzed in the coordinate plane. Here, their slopes are used to prove their inherent characteristics.
Rule

## Slopes of Parallel Lines Theorem

In a coordinate plane, two nonvertical lines are parallel if and only if their slopes are equal.

This can be proven using linear equations written in slope-intercept form.

### Proof

info
Slopes of Parallel Lines Theorem

Consider two lines in a coordinate plane. In slope-intercept form, their equations can be written as follows. $y=m_1x+b_1 \quad \text{and} \quad y=m_2x+b_2$ 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.
$y=m_1x+b_2$
${\color{#0000FF}{m_2x+b_2}}=m_1x+b_1$
$m_2x=m_1x+b_1-b_2$
$m_2x-m_1x=b_1-b_2$
$x(m_2-m_1)=b_1-b_2$
$x=\dfrac{b_1-b_2}{m_2-m_1}$
The $x$-coordinate of the point of intersection is $x=\frac{b_1-b_2}{m_2-m_1}.$ However, if the lines are parallel, by definition, there will be no point of intersection. Thus, $x=\frac{b_1-b_2}{m_2-m_1}$ must be undefined, which occurs when the denominator is zero. $m_2-m_1=0 \quad \Leftrightarrow \quad m_2=m_1$ Therefore, the slopes of parallel lines are equal.
The above proof can be summarized in a flowchart.
Rule

## Slopes of Perpendicular Lines Theorem

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

This can be proven using right triangles.

### Proof

info
Slopes of Perpendicular Lines Theorem

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. $m_D=\dfrac{c}{a} \quad \text{and} \quad m_E=\dfrac{\text{-} c}{b}$ Multiplying the slopes gives their product. $\dfrac{c}{a} \cdot \dfrac{\text{-} c}{b} = \dfrac{\text{-} c^2}{ab}$ Notice that the product equals $\text{-} 1$ if and only if the following is true. $\dfrac{\text{-} c^2}{ab} = \text{-} 1 \Leftrightarrow c^2=ab \Leftrightarrow c=\sqrt{ab}$ The Right Triangle 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, $c=\sqrt{ab}$ is a true statement. Therefore, the product of the slopes of two perpendicular lines equals $\text{-} 1.$

$m_D \cdot m_E = \text{-} 1$
Exercise

The points $P(0.8,\text{-} 0.6),$ $Q(\text{-} 1.6,4.2),$ $R(1.2,5.6),$ and $S(3.6,0.8)$ are vertices of a quadrilateral. Prove that quadrilateral is a rectangle.

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

• $\overline{PQ} \parallel \overline{RS}$,
• $\overline{PS} \parallel \overline{QR}$, and
• $\overline{QR} \perp \overline{RS}.$
Note that, if the first two criteria is proven true, it is sufficient to show that one pair of adjacent sides is perpendicular. First, we can determine the slopes of each side using the coordinates of the points and the slope formula. Then we can compare the slopes to see how the sides relate. We'll start with $\overline{PQ}.$
$PQ = \dfrac{y_2-y_1}{x_2-x_1}$
$PQ = \dfrac{{\color{#0000FF}{4.2}}-\left({\color{#009600}{\text{-} 0.6}}\right)}{{\color{#0000FF}{\text{-} 1.6}}-{\color{#009600}{0.8}}}$
$PQ=\dfrac{4.2+0.6}{\text{-} 1.6-0.8}$
$PQ=\dfrac{4.8}{\text{-} 2.4}$
$PQ=\text{-} 2$
Thus, $\overline{PQ}$ has the slope $m_{PQ}=\text{-} 2.$ The slopes of the other line segments are calculated in the same way.
Line segment Slope formula Simplified Slope
$\overline{PQ}$ $\dfrac{4.2-(\text{-} 0.6)}{\text{-} 1.6-0.8}$ $\dfrac{4.8}{\text{-} 2.4}$ $m_{PQ}=\text{-} 2$
$\overline{QR}$ $\dfrac{5.6-4.2}{1.2-(\text{-} 1.6)}$ $\dfrac{1.4}{2.8}$ $m_{QR}=0.5$
$\overline{RS}$ $\dfrac{0.8-5.6}{3.6-1.2}$ $\dfrac{\text{-} 4.8}{2.4}$ $m_{RS}=\text{-} 2$
$\overline{SP}$ $\dfrac{\text{-} 0.6-0.8}{0.8-3.6}$ $\dfrac{\text{-} 1.4}{\text{-} 2.8}$ $m_{SP}=0.5$

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

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