Lines in the Plane

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.