Lines in the Plane

The first graphs and functions you encountered in algebra were straight lines and their equations.  These lines were easy to graph, and the equations were easy to evaluate and to solve.  They described a variety of physical, biological and business phenomena such as  $d = rt$ relating the distance d traveled to the rate r and time t of travel, and  $C = \dfrac{5}{9} ( F – 32)$ for converting the temperature in Fahrenheit degrees (F) to Celsius (C).
 
The first part of calculus, differential calculus, will deal with the ideas and techniques and applications of tangent lines to the graphs of functions, so it is important that you understand the graphs and properties and
equations of straight lines.

Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-1.2-Lines-in-the-Plane.pdf
This work is licensed under a Creative Commons Attribution 3.0 License.

The real numbers (consisting of all integers, fractions, rational and irrational numbers) can be represented as a line, called the real number line (Fig. 1). Once we have selected a starting location, called the origin, a positive direction (usually up or to the right), and unit of length, then every number can be located as a point on the number line. If we move from a point $x = a$ to point $x = b$ on the line (Fig. 2), then we will have moved an increment of $b – a$. This increment is denoted by the symbol $∆x$ ( read "delta x" ).

The Greek capital letter delta, ∆, will appear often in the future and will represent the "change" in something. If b is larger than a, then we will have moved in the positive direction, and $∆x = b – a$ will be positive. If b is smaller than a, then $∆x = b – a$ will be negative and we will have moved in the negative direction. Finally, if $∆x = b – a$ is zero, then $a=b$ and we did not move at all.

We can also use the ∆ notation and absolute values to write the distance that we have moved. On the number line, the distance from $x = a$ to $x = b$ is

dist(a,b) =  $\left \{ \begin {array} {lI} b-a \text { if } b \geq a \\ a-b \text { if } b < a \end {array} \right.$

or simply, dist(a,b) = $| b – a | = | ∆x | = \sqrt{(∆x)^2}$ .

The midpoint of the segment from $x = a$ to $x = b$ is the point $M = \dfrac{a + b}{2}$ on the number line.


Example 1: Find the length and midpoint of the interval from $x = –3$ to $x = 6$.

Solution: $Dist \; (–3,6) = | 6 – (–3) | = | 9 | = 9$. The midpoint is at $\dfrac{(–3) + (6)}{2} = 3/2$.

Practice 1: Find the length and midpoint of the interval from $x = –7$ to $x = –2$.

(Note: Solutions to Practice Problems are given at the end of each section, after the Problems).

A real number plane (Fig. 3) is determined by two perpendicular number lines, called the coordinate axes, which intersect at a point, called the origin of the plane or simply the origin. Each point $P$ in the plane can be described by an ordered pair $(x,y)$ of numbers which specify how far, and in which directions, we must move from the origin to reach the point $P$. The point $P = (x,y)$ can then be located in the plane by starting at the origin and moving $x$ units horizontally and then y units vertically. Similarly, each point in the plane can be labeled with the ordered pair $(x,y)$ which directs us how to reach that point from the origin. In this book, a point in the plane will be labeled either with a name, say $P$, or with an ordered pair $(x,y)$, or with both $P = (x,y)$. This coordinate system is called the rectangular coordinate system or the Cartesian coordinate system after Rene Descartes, and the resulting plane is called the Cartesian Plane.

The coordinate axes divide the plane into four quadrants which are labeled quadrants I, II, III and IV as in Fig. 4 We will often call the horizontal axes the x-axis and the vertical axis the y-axis and then refer to the plane as the xy-plane. This choice of $x$ and y as labels for the axes is simply a common choice, and we will sometimes prefer to use different labels and even different units of measure on the two axes.

If we move from a point $P = (x_1,y_1)$ to a point $Q = (x_2,y_2)$ in the plane, then we will have two increments or changes to consider. The increment in the $x$ or horizontal direction is $x_2 – x_1$ which is denoted by $∆x = x_2 – x_1$. The increment in the $y$ or vertical direction is $∆y = y_2 – y_1$. These increments are shown in Fig. 5 . $∆x$ does not represent $∆$ times $x$, it represents the difference in the $x$ coordinates: $∆x = x_2 – x_1$.

The distance between the points $P = (x_1,y_1)$ and $Q = (x_2,y_2)$ is simply an application of the Pythagorean formula for right triangles, and

dist(P,Q) = $\sqrt{(∆x)^2 + (∆y)^2} = \sqrt{ (x_2–x_1)^2 + (y_2–y_1)^2}$

The midpoint $M$ of the line segment joining $P$ and $Q$ is $M = (\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2})$

Example 2: Find an equation describing the points $P = (x,y)$ which are equidistant from $Q = (2,3)$ and $R = (5,–1)$. (Fig. 6) 

Solution: The points $P=(x,y)$ must satisfy $dist(P,Q) = dist(P,R)$ so

$\sqrt{(x–2)^2+(y–3)^2} =\sqrt {(x–5)^2+(y–(–1))^2}$

By squaring each side we get $(x–2)^2+(y–3)^2 = (x–5)^2+(y+1)^2$

Then $x^2 – 4x + 4 + y^2 – 6y + 9 = x^2 – 10x + 25 + y^2 + 2y + 1$

so $–4x – 6y + 13 = –10x + 2y + 26$ and $y = .75x – 1.625$, a straight line. Every point on the line $y = .75x – 1.625$ is equally distant from $Q$ and $R$.

Practice 2: Find an equation describing all points $P = (x,y)$ equidistant from $Q = (1,–4)$ and $R = (0,–3)$.

A circle with radius $r$ and center at the point $C = (a,b)$ consists of all points $P = (x,y)$ which are at a distance of $r$ from the center $C$: the points $P$ which satisfy $dist(P,C) = r$.

Example 3: Find the equation of a circle with radius $r = 4$ and center $C = (5,–3)$. (Fig. 7)

Solution: A circle is the set of points $P=(x,y)$ which are at a fixed distance $r$ from the center point $C$, so this circle will be the set of points $P=(x,y)$ which are at a distance of 4 units from the point $C = (5,–3)$. P will be on this circle if $dist(P,C) = 4$.
Using the distance formula and simplifying,

$\sqrt{(x–5)^2 + (y+3)^2} = 4$ so $(x–5)^2 + (y+3)^2=16$ or

$x^2 – 10x + 25 + y^2 + 6y + 9 = 16$.

Practice 3: Find the equation of a circle with radius $r = 5$ and center $C = (–2,6)$.

In one dimension on the number line, our only choice was to move in the positive direction (so the x–values were increasing) or in the negative direction. In two dimensions in the plane, we can move in infinitely many directions and a precise means of describing direction is needed. The slope of the line segment joining $P = (x_1,y_1)$ to $Q = (x_2,y_2)$ , is

$m = {slope \; from \; P \; to \; Q } = \dfrac{rise}{run} = \dfrac{y_2–y_1}{x_2–x_1} = \dfrac{∆y}{∆x}$

In Fig. 8, the slope of a line measures how fast we rise or fall as we move from left to right along the line. It measures the rate of change of the y-coordinate with respect to changes in the x-coordinate. Most of our work will occur in 2 dimensions, and slope will be a very useful concept which will appear often.

If $P$ and $Q$ have the same $x$ coordinate, then $x_1 = x_2$ and $∆x = 0$. The line from $P$ to $Q$ is vertical and the slope $m = ∆y/∆x$ is undefined because $∆x = 0$. If $P$ and $Q$ have the same y coordinate, then $y_1 = y_2$ and $∆y = 0$, so the line is horizontal and the slope is $m = ∆y/∆x = 0/∆x = 0$ (assuming $∆x ≠ 0$).

Practice 4: For $P = (–3,2)$ and $Q = (5,–14)$, find $∆x, ∆y$, and the slope of the line segment from $P$ to $Q$.

If the coordinates of $P$ or $Q$ contain variables, then the slope $m$ is still given by $∆y/∆x$ , but we will need to use algebra to evaluate and simplify $m$.

Example 4: Find the slope of the line segment from $P = (1,3)$ to $Q = (1+h, 3 + 2h)$. (Fig. 9)

Solution: $y_1 = 3$ and $y_2 = 3 + 2h$ so $∆y = (3 + 2h) – (3) = 2h . x_1 = 1$ and $x_2 = 1 + h$ so $∆x = (1 + h) – (1) = h$, and the slope is $m = \dfrac{∆y}{∆x} = \dfrac{2h}{h} = 2$.

In this example, the value of $m$ is the constant 2 and does not depend on the value of $h$.

Practice 5: Find the slope and midpoint of the line segment from $P = (2,–3)$ to $Q = (2 + h, –3 + 5h)$.

Example 5: Find the slope between the points $P = (x, x^2 + x )$ and $Q = (a, a^2 + a )$ for $a ≠ x$.

Solution: $y_1 = x^2+x$ and $y_2 = a^2+a$ so $∆y = (a^2 + a) – (x^2 + x)$. $x_1 = x$ and $x_2 = a$ so $∆x = a–x$ and the slope is $m = \dfrac{∆y}{∆x} = \dfrac{(a^2+a) – (x^2+x)}{a–x} = \dfrac{a^2 – x^2 + a – x}{a – x}$

$= \dfrac{ (a–x) (a+x) + (a–x)}{a – x}$

$= \dfrac{(a–x) . {(a+x) + 1}}{a – x} = (a + x) + 1$.

In this example, the value of $m$ depends on the values of both $a$ and $x$.

Practice 6: Find the slope between $P = (x, 3x^2 + 5x)$ $Q = (a, 3a^2 + 5a)$ for $a ≠ x$.

In application problems it is important to read the information and the questions very carefully. Including the units of measurement of the variables can help you avoid "silly" answers.

Example 6: In 1970 the population of Houston was 1,233,535 and in 1980 it was 1,595,138. Find the slope of the line through the points (1970, 1233535) and (1980, 1595138).

Solution: $m = \frac{∆y}{∆x} =\dfrac{1595138 – 1233535}{1980 – 1970} = \dfrac{361603}{10} = 36,160.3$

But 36,160.3 is just a number which may or may not have any meaning to you. If we include the units of measurement along with the numbers we will get a more meaningful result:

$m = \dfrac{∆y}{∆x} = \dfrac{1595138 \;people – 1233535 \;people}{\;year 1980 – \;year 1970}$

$= \dfrac{361603 \;people}{10 \;years} = 36,160.3 \;people/year$

which says that during the decade from 1970 to 1980 the population of Houston grew at an average rate of 36,160 people per year.

If the x–unit is time in hours and the y-unit is distance in kilometers, then m is $\dfrac{∆y \; kilometers}{∆x \; hours}$, so the units for $m$ are kilometers/hour ("kilometers per hour"), a measure of velocity, the rate of change of distance with respect to time. If the x-unit is the number of employees at a bicycle factory and the y-unit is the number of bicycles manufactured, then $m$ is $\dfrac {∆y \;bicycles}{∆x \;employees}$, and the units for $a$ are bicycles/employee ("bicycles per employee"), a measure of the rate of production per employee.

Every line has the property that the slope of the segment between any two points on the line is the same, and this constant slope property of straight lines leads to ways of finding equations to represent nonvertical lines.

Point–Slope Equation

In calculus, we will usually know a point on the line and the slope of the line so the point–slope form will be the easiest to apply, and the other forms of equations for straight lines can be derived from the point–slope form.

If $L$ is a nonvertical line through a known point $P = (x_1,y_1)$ with a known slope m (Fig. 10), then the equation of the line $L$ is

Point-Slope: $y – y_1 = m(x – x_1)$

Example 7: Find the equation of the line through (2,–3) with slope 5.
Solution: The solution is simply a matter of knowing and using the point–slope formula. $m = 5, y_1 = –3$ and $x_1 = 2$ so $y – (–3) = 5(x – 2)$. This equation simplifies to $y = 5x –13$ (Fig. 11).

The equation of a vertical line through a point $P = (a,b)$ is $x = a$. The only points $Q = (x,y)$ on the vertical line through the point $P$ have the same x–coordinate as $P$.

Two–Point and Slope–Intercept Equations

If two points $P = (x_1,y_1)$ and $Q = (x_2,y_2)$ are on the line $L$, then we can calculate the slope between them and use the first point and the point–slope equation to get the equation of $L$:

Two Points: $y – y_1 = m(x – x_1)$ where $m =\dfrac{y_2 – y_1}{x_2 – x_1}$

Once we have the slope $m$, it does not matter whether we use $P$ or $Q$ as the point. Either choice will give the same simplified equation for the line.

It is common practice to rewrite the equation of the line in the form $y = mx + b$, the slope-intercept form of the line. The line $y = mx + b$ has slope $m$ and crosses the y-axis at the point ( 0, b ).

Practice 7: Use the $∆y/∆x$ definition of slope to calculate the slope of the line $y = mx + b$.

The point-slope and the two-point formulas are usually more useful for finding the equation of a line, but the slope-intercept form is usually the most useful form for an answer because it allows us to easily picture the graph of the line and to quickly calculate y-values.

Angles Between Lines

The angle of inclination of a line with the x-axis is the smallest angle θ which the line makes with the positive x-axis as measured from the x-axis counterclockwise to the line (Fig. 12). Since the slope $m = ∆y/∆x$ and since $tan(θ) =$ opposite/adjacent, we have that $m = tan(θ)$.
The slope of the line is the tangent of the angle of inclination of the line.

Parallel and Perpendicular Lines

Two parallel lines $L_1$ and $L_2$ make equal angles with the x-axis so their angles of inclination will be equal (Fig. 13) and so will their slopes. Similarly, if their slopes $m_1$ and $m_2$ are equal, then the equations of the lines will always differ by a constant:

$y_1 – y_2 = {m_1x+b_1} – {m_2x+b_2}$

$= (m_1–m_2)x + (b_1–b_2)$

$= b_1 – b_2$

which is a constant so the lines will be parallel. These two ideas can be combined into a single statement:

Two nonvertical lines $L_1$ and $L_2$ with slopes $m_1$ and $m_2$ are parallel if and only if $m_1 = m_2$.

Practice 8: Find the equation of the line in Fig. 14 which contains the point (–2,3) and is parallel to the line $3x + 5y = 17$.

If two lines are perpendicular and neither line is vertical, the situation is a bit more complicated (Fig. 15).

Assume $L_1$ and $L_2$ are two nonvertical lines that intersect at the origin (for simplicity) and that $P = (x_1,y_1)$ and $Q = (x_2,y_2)$ are points away from the origin on $L_1$ and $L_2$ , respectively. Then the slopes of $L_1$ and $L_2$ will be $m_1 = y_1/x_1$ and $m_2 = y_2/x_2$ . The line connecting $P$ and $Q$ forms the third side of the triangle $OPQ$ , and this will be a right triangle if and only if $L_1$ and $L_2$ are perpendicular. In particular, $L_1$ and $L_2$ are perpendicular if and only if the triangle $OPQ$ satisfies the Pythagorean theorem:

${dist(O,P) }^2 + {dist(O,Q) }^2 = {dist(P,Q) }^2$ or

$( x_1–0)^2 + (y_1–0)^2 + ( x_2–0)^2 + (y_2–0)^2$

$= ( x_1 – x_2)^2 + (y_1 – y_2)^2$.

By squaring and simplifying, this last equation reduces to

$0 = –2x_1x_2 – 2y_1y_2$ so $y_2/x_2 = – x_1/y_1$ and

$m_2 = y_2/x_2 = – x_1/y_1 = –\dfrac{1}{(y_1/x_1)} = –\dfrac{1}{m_1}$.

We have just proved the following result:

Two nonvertical lines $L_1$ and $L_2$ with slopes $m_1$ and $m_2$ are perpendicular if and only if their slopes are negative reciprocals of each other: $m_2 = –\dfrac{1}{m_1}$

Practice 9: Find the line which goes through the point (2,–5) and is perpendicular to the line $3y – 7x = 2$.

Example 8: Find the distance (the shortest distance) from the point (1,8) to the line $L$: $3y – x = 3$.

Solution: This is a sophisticated problem which requires several steps to solve.

First we need a picture of the problem (Fig. 16). We will find the line $L*$ through the point (1,8) and perpendicular to $L$. Then we will find the point $P$ where $L$ and $L*$ intersect, and, finally, we will find the distance from $P$ to (1,8).

(i) $L$ has slope 1/3 so $L*$ has slope $m = –\dfrac{1}{1/3} = –3$ , and $L*$ has the equation $y – 8 = –3(x – 1)$ which simplifies to $y = –3x + 11$.

(ii) We can find the point of intersection of $L$ and $L*$ by replacing the $y$ in the equation for $L$ with the $y$ from $L*$ so $3(–3x + 11) – x = 3$. Then $x = 3$ so $y = –3x + 11 = –3(3) + 11 = 2$ , so $L$ and $L*$ intersect at $P = (3,2)$.

(iii) Finally, the distance from $L$ to (1,8) is just the distance from the point (1,8) to the point $P = (3,2)$ which is $\sqrt{(1 – 3)^2 + (8 – 2)^2} = \sqrt{ 40} ≈ 6.325$.

Angle Formed by Intersecting Lines

If two lines which are not perpendicular intersect at a point and neither line is vertical, then we can use some geometry and trigonometry to determine the angles formed by the intersection of the lines (Fig. 17). Since θ₂ is an exterior angle of the triangle ABC, θ₂ is equal to the sum of the two opposite interior angles so $θ_2 = θ_1 + θ$ and $θ = θ_2 – θ_1$. Then, from trigonometry,

$tan(θ) = tan(θ_2 – θ_1) = \dfrac{tan (θ_2) – tan (θ_1)}{1 + tan (θ_2)tan (θ_1)} = \dfrac{m_2 – m_1}{1 + m_2m_1}$

The inverse tangent of an angle is between $–π/2$ and $π/2$ ( –90^o and 90^o) so $θ = arctan(\dfrac{m_2 – m_1}{1 + m_2m_1})$ always gives the smaller of the angles.

The larger angle is $π – θ$ or 180^o – θ^o.

The smaller angle θ formed by two nonperpendicular lines with slopes $m_1$ and $m_2$ is

$θ = arctan(\dfrac{m_2 – m_1}{1 + m_2m_1})$.


Example 9: Find the point of intersection and the angle between $y = x + 3$ and $y = 2x + 1$. (Fig. 18)

Solution: Solving the first equation for y and then substituting into the second equation, $(x + 3) = 2x + 1$ so $x = 2$. Putting this back into either equation, we get $y = 5$. Each of the lines is in the slope–intercept form so it is easy to see that $m_1 = 1$ and $m_2 = 2$ . Then

$tan(θ) = \dfrac{m_2 – m_1}{1 + m_2m_1} = \dfrac{2 – 1}{1 + (2)(1)} = 1/3$ and

$θ = arctan(1/3) = .322 \; radians ≈ 18.435^o$

Practice 1: $Length = Dist( –7, –2 ) = | (–7) – (–2) | = | –5 | = 5$.

The midpoint is at $\dfrac{(–7) + (–2)}{2} =\dfrac{–9}{2} = – 4.5$.

Practice 2: $Dist(P,Q) = Dist(P,r)$ so $\sqrt{(x – 1)^2 + (y + 4)^2} = \sqrt{(x – 0)^2 + (y + 3)^2}$.

Squaring each side and simplifying, we eventually have $y = x – 4$ .

Practice 3: The point $P = ( x , y)$ is on the circle when it is 5 units from the center $C = ( –2, 6)$ so $Dist(P,C) = 5$. Then $Dist( (x,y) , (–2,6) ) = 5$ so

$\sqrt{(x + 2)^2 + (y – 6)^2} = 5$ or $(x + 2)^2 + (y – 6)^2 = 25$.

Practice 4: $∆x = 5 – (–3) = 8$, $∆y = –14 – 2 = –16$ , and $slope = \dfrac{∆y}{∆x} = \dfrac{–16}{8} = – 2$.

Practice 5: $slope =\dfrac{∆y}{∆x} = \dfrac{ (–3 + 5h) – (–3)}{(2 + h) – 2} = \dfrac{5h}{h} = 5$.

The midpoint is at $( \dfrac{ (2) + (2 + h)}{2} , \dfrac{(–3 + 5h) + (–3)}{2} ) = ( 2 +\dfrac{h}{2} , –3 + \dfrac{5h}{2})$.

Practice 6: $slope =\dfrac{∆y}{∆x} = \dfrac{(3a^2 + 5a) – (3x^2 + 5x)}{a – x}$

$= \dfrac{ 3(a^2 – x^2) + 5(a – x)}{a – x} = \dfrac{3(a + x)(a – x) + 5(a – x)}{a – x} = 3(a + x) + 5$.

Practice 7: Let $y_1 = mx_1 + b$ and $y_2 = mx_2 + b$ . Then

$slope =\dfrac{∆y}{∆x} = \dfrac{(mx_2 + b) – (mx_1 + b)}{x_2 – x_1} = \dfrac{m(x_2 – x_1)}{x_2 – x_1 } = m$.

Practice 8: The line $3x + 5y = 17$ has slope $\dfrac{–3}{5}$ so the slope of the parallel line is $m = \dfrac{–3}{5}$.

Using the form $y = \dfrac{–3}{5} x + b$ and the point ( –2, 3) on the line, we have

$3 = \dfrac{–3}{5} (–2) + b$ so $b = \dfrac{9}{5}$ and

$y = \dfrac{–3}{5} x + \dfrac{9}{5}$ or $5y + 3x = 9$ ..

Practice 9: The line $3y – 7x = 2$ has slope $\dfrac{7}{3}$ so the slope of the perpendicular line is $m= \dfrac{–3}{7}$.

Using the form $y = \dfrac{–3}{7} x + b$ and the point ( 2, –5) on the line, we have $–5 =\dfrac{–3}{7} (2) + b$ so $b = \dfrac{–29}{7}$ and $y =\dfrac{–3}{7} x + \dfrac{–29}{7}$ or $7y + 3x = –29$ .