Line and Definition of Slope

The slope of a line is a number that shows how steep the line is and whether it goes up or down as we move from left to right. With the slope we can write the equation of a line, decide whether two lines are parallel or perpendicular, and describe how fast one quantity changes when another one changes.

On this page: Rise and run · Definition · Sign of the slope · Angle of inclination · Parallel and perpendicular · Equations of a line · Slope in real life · Common mistakes · Practice

Slope of a Line (Steepness)

Imagine a point that moves along a non-vertical line from the point $P_1(x_1, y_1)$ to the point $P_2(x_2, y_2).$ Its $y$-coordinate changes by ${y_2-y_1}$ and its $x$-coordinate changes by ${x_2-x_1}.$ The vertical change ${y_2-y_1}$ is called the rise, and the horizontal change ${x_2-x_1}$ is called the run.

The steeper the line, the larger the rise for the same run. That is why steepness is measured by the ratio of the rise to the run. For example, if a line climbs $3$ units while we move $6$ units to the right, the ratio is ${\frac{3}{6}=\frac{1}{2}}\text{:}$ the line climbs half a unit for every unit we move to the right.

Definition

In words: the slope is “rise over run”, the change in $y$ divided by the change in $x.$ The formula needs ${x_1\ne x_2}.$ On a vertical line all points have the same $x$-coordinate, so the run is $0,$ and division by zero is impossible. That is why the slope of a vertical line is undefined. (Some books say that a vertical line has no slope. This does not mean that the slope is $0.$)

The order of the points

It does not matter which point is called $P_1$ and which is called $P_2.$ If we swap them, both the numerator and the denominator change sign, and the fraction stays the same:

$$\frac{y_1-y_2}{x_1-x_2}=\frac{-(y_2-y_1)}{-(x_2-x_1)}=\frac{y_2-y_1}{x_2-x_1}.$$

The order must be the same above and below. If the numerator starts with $y_2,$ the denominator must start with $x_2.$

Any two points give the same slope

The slope belongs to the whole line, not to a pair of points. In the picture the points $A,$ $B,$ $C$ and $D$ lie on the same line. Each of the two shaded right triangles has a horizontal leg and a vertical leg, and their hypotenuses $AB$ and $CD$ lie on this line. The horizontal legs are parallel, so the angles at $A$ and $C$ are equal (they are corresponding angles). Both triangles also have a right angle, so they are similar. In similar triangles the ratios of corresponding sides are equal. For the small triangle the rise is $1$ and the run is $2;$ for the large one the rise is $2$ and the run is $4\text{:}$

$$\frac{1}{2}=\frac{2}{4}.$$

Whichever two points of this line we take, the slope is $\frac{1}{2}.$

Slope as a rate of change

The changes in the coordinates are often written with the Greek letter $\Delta$ (delta): ${\Delta x=x_2-x_1}$ and ${\Delta y=y_2-y_1}.$ Then

$$m=\frac{\Delta y}{\Delta x},\qquad \Delta y=m\cdot\Delta x.$$

If $x$ increases by $1,$ then $y$ changes by $m.$ So the slope is the rate of change of $y$ with respect to $x\text{:}$ it tells us by how many units $y$ changes when $x$ increases by one unit. For a line this rate is the same everywhere.

Positive, Negative, Zero and Undefined Slope

We look at a line from left to right, in the direction in which $x$ increases.

Slope	The line	Reason
Positive, ${m\gt0}$	rises from left to right	when $x$ increases, $y$ increases: the rise is positive
Negative, ${m\lt0}$	falls from left to right	when $x$ increases, $y$ decreases: the rise is negative (it is sometimes called a fall)
Zero, ${m=0}$	is horizontal	all points have the same $y$-coordinate: the rise is $0$
Undefined	is vertical	all points have the same $x$-coordinate: the run is $0$

The absolute value of the slope shows the steepness: the larger $|m|$ is, the steeper the line. A line with slope $4$ is steeper than a line with slope $\frac{1}{2},$ and a line with slope $-2$ is steeper than a line with slope $1,$ although ${-2\lt1}.$

Example 1. Find the slope of the line through the points: a) $(6, 2)$ and $(9, 8);$ b) $(2, 9)$ and $(4, 3);$ c) $(-2, 7)$ and $(5, 7);$ d) $(3, 1)$ and $(3, 5).$

Solution. We use the formula ${m=\frac{y_2-y_1}{x_2-x_1}}.$

a) $m=\frac{8-2}{9-6}=\frac{6}{3}=2.$ The line rises $2$ units for every unit to the right.

b) $m=\frac{3-9}{4-2}=\frac{-6}{2}=-3.$ The line falls $3$ units for every unit to the right.

c) $m=\frac{7-7}{5-(-2)}=\frac{0}{7}=0.$ The line is horizontal.

d) Both points have ${x=3},$ so the run is ${3-3=0}$ and the slope is undefined. The line is vertical.

Example 2. The line through the points $(1, 2)$ and $(4, y)$ has slope $3.$ Find $y.$

Solution. By the slope formula, ${\frac{y-2}{4-1}=3}.$ So ${y-2=9}$ and ${y=11}.$

Example 3. Do the points $A(-1, 1),$ $B(1, 5)$ and $C(4, 11)$ lie on the same line?

Solution. The slope of $AB$ is ${\frac{5-1}{1-(-1)}=\frac{4}{2}=2},$ and the slope of $BC$ is ${\frac{11-5}{4-1}=\frac{6}{3}=2}.$ The lines $AB$ and $BC$ have the same slope and the common point $B.$ Through a given point there is only one line with a given slope, so $AB$ and $BC$ are one and the same line. The three points lie on the same line (they are collinear).

Angle of Inclination

The angle of inclination of a line that crosses the $x$-axis is the smallest angle $\varphi$ measured counterclockwise from the positive direction of the $x$-axis to the line. For a horizontal line we take ${\varphi=0^\circ}.$

So ${0^\circ\le\varphi\lt180^\circ}.$ If the line rises, the angle $\varphi$ is acute; if the line falls, $\varphi$ is obtuse; for a vertical line ${\varphi=90^\circ}.$

Theorem. For a non-vertical line the slope $m$ and the angle of inclination $\varphi$ are related by ${m=\tan\varphi}.$

For an obtuse angle the tangent is negative: ${\tan\varphi=-\tan(180^\circ-\varphi)}.$ For example, ${\tan135^\circ=-\tan45^\circ=-1}.$

Proof

Take two points $P_1$ and $P_2$ on the line, with $P_2$ to the right of $P_1,$ and draw the right triangle whose legs are the run and the rise.

If the line rises, the angle of the triangle at $P_1$ is equal to $\varphi\text{:}$ the horizontal leg is parallel to the $x$-axis, so these are corresponding angles. In a right triangle the tangent of an acute angle is the opposite leg divided by the adjacent leg, so ${\tan\varphi=\frac{y_2-y_1}{x_2-x_1}=m}.$

If the line falls, the angle $\varphi$ is obtuse, and the acute angle between the line and the $x$-axis is ${180^\circ-\varphi}.$ The angle of the triangle at $P_1$ is equal to it (again these are corresponding angles). Its tangent is the length of the vertical leg divided by the length of the horizontal leg: ${\tan(180^\circ-\varphi)}={\frac{y_1-y_2}{x_2-x_1}}={-m}.$ Since ${\tan(180^\circ-\varphi)=-\tan\varphi},$ we get ${-\tan\varphi=-m},$ so again ${m=\tan\varphi}.$

For a horizontal line ${m=0}$ and ${\tan0^\circ=0}.$

Angle $\varphi$	Slope ${m=\tan\varphi}$	The line
$0^\circ$	$0$	is horizontal
$30^\circ$	${\frac{\sqrt{3}}{3}\approx0.58}$	rises
$45^\circ$	$1$	rises
$60^\circ$	${\sqrt{3}\approx1.73}$	rises
$90^\circ$	undefined	is vertical
$120^\circ$	${-\sqrt{3}\approx-1.73}$	falls
$135^\circ$	$-1$	falls
$150^\circ$	${-\frac{\sqrt{3}}{3}\approx-0.58}$	falls

The tangent of $90^\circ$ is not defined, and this agrees with the fact that the slope of a vertical line is undefined.

Example 4. Find the angle of inclination of a line with slope a) ${m=1};$ b) ${m=-1}.$

Solution.

a) ${\tan\varphi=1},$ so ${\varphi=45^\circ}$ (in radians $\frac{\pi}{4}$).

b) ${\tan\varphi=-1}.$ Since ${0^\circ\le\varphi\lt180^\circ},$ the angle is ${\varphi=135^\circ}$ (in radians $\frac{3\pi}{4}$).

Example 5. Find, to the nearest degree, the angle of inclination of the line through the points a) $(1, 1)$ and $(4, 3);$ b) $(0, 5)$ and $(2, 1).$

Solution.

a) ${m=\frac{3-1}{4-1}=\frac{2}{3}}.$ The inverse tangent key $\tan^{-1}$ of a calculator (in degree mode) gives ${\varphi=\tan^{-1}\frac{2}{3}\approx34^\circ}.$

b) ${m=\frac{1-5}{2-0}=\frac{-4}{2}=-2}.$ A calculator gives ${\tan^{-1}(-2)\approx-63^\circ},$ but an angle of inclination cannot be negative. The tangent does not change when $180^\circ$ is added to the angle, so we add $180^\circ$ and get ${\varphi\approx-63^\circ+180^\circ=117^\circ}.$

For a negative slope a calculator returns a negative angle. Add $180^\circ$ to it to get the angle of inclination.

Parallel and Perpendicular Lines

The second condition can be written as ${m_2=-\frac{1}{m_1}}\text{:}$ the slopes of perpendicular lines are negative reciprocals of each other. For example, lines with slopes $2$ and $-\frac{1}{2}$ are perpendicular, and so are lines with slopes $-\frac{3}{4}$ and $\frac{4}{3}.$

The slope of a vertical line is undefined, so the theorem says nothing about vertical lines. Any two different vertical lines are parallel. A vertical line is perpendicular to every horizontal line and to no other line.

Proof for parallel lines

Suppose first that neither line is horizontal. Then each of them crosses the $x$-axis; let $\varphi_1$ and $\varphi_2$ be their angles of inclination.

If the lines are parallel, they cross the $x$-axis at two different points, and $\varphi_1$ and $\varphi_2$ are corresponding angles: the $x$-axis is the transversal. Corresponding angles at parallel lines are equal, ${\varphi_1=\varphi_2},$ so ${m_1=\tan\varphi_1=\tan\varphi_2=m_2}.$

Conversely, let ${m_1=m_2},$ that is, ${\tan\varphi_1=\tan\varphi_2}.$ Two different angles between $0^\circ$ and $180^\circ$ cannot have the same tangent, so ${\varphi_1=\varphi_2}.$ The lines cannot cross the $x$-axis at the same point: two lines through the same point of the axis that have the same angle of inclination coincide, and our lines are different. So they cross the axis at two different points, the corresponding angles $\varphi_1$ and $\varphi_2$ are equal, and therefore the lines are parallel.

If one of the lines is horizontal, its slope is $0.$ The other line is parallel to it exactly when it is horizontal too, that is, when its slope is also $0.$

Proof for perpendicular lines

In both parts of the proof the lines meet. Perpendicular lines always meet. And if ${m_1m_2=-1},$ then ${m_1\ne m_2}$ (equal slopes would give ${m_1^2=-1},$ which is impossible), so by the first part of the theorem the lines are not parallel. Let $A$ be the point where the lines meet. If we move both lines up by the same distance, their slopes and the angle between them do not change, so we may assume that $A$ lies above the $x$-axis.

If the lines are perpendicular, then ${m_1m_2=-1}.$ Neither line is vertical, so neither is horizontal (a line perpendicular to a horizontal line is vertical). So both lines cross the $x$-axis: $L_1$ at a point $B$ and $L_2$ at a point $C.$ We name the lines so that $B$ is to the left of $C.$ The triangle $ABC$ has a right angle at $A,$ so its angles at $B$ and $C$ are acute. Therefore $L_1$ rises, $L_2$ falls, and the foot $H$ of the altitude $AH$ lies between $B$ and $C.$ Let ${AH=h}$ and ${HC=c}.$

The angle of the triangle at $B$ is $\varphi_1.$ The angle $HAC$ is also equal to $\varphi_1,$ because each of them adds up to $90^\circ$ with the angle at $C$ (in the right triangles $ABC$ and $AHC$). From the right triangle $AHC$ we get ${\tan\varphi_1=\frac{c}{h}},$ and the tangent of the angle at $C$ is $\frac{h}{c}.$ The angle of inclination of $L_2$ is $\varphi_2,$ and the angle at $C$ is ${180^\circ-\varphi_2},$ so ${\tan\varphi_2=-\frac{h}{c}}.$ Therefore

$$m_1m_2=\frac{c}{h}\cdot\left(-\frac{h}{c}\right)=-1.$$

If ${m_1m_2=-1},$ then the lines are perpendicular. The slopes have opposite signs; let ${m_1\gt0}$ and ${m_2\lt0}.$ Then $L_1$ rises and $L_2$ falls. Going down from $A$ along $L_1$ we move to the left, and along $L_2$ to the right, so $L_1$ crosses the $x$-axis at a point $B$ to the left of $A,$ and $L_2$ crosses it at a point $C$ to the right of $A.$ So the foot $H$ of the altitude $AH$ of the triangle $ABC$ lies between $B$ and $C.$ Let ${AH=h},$ ${BH=d}$ and ${HC=c}.$ From the right triangles $ABH$ and $AHC$ we get ${m_1=\frac{h}{d}}$ and ${m_2=-\frac{h}{c}}.$ The equality ${m_1m_2=-1}$ becomes ${\frac{h}{d}\cdot\frac{h}{c}=1},$ that is, ${\frac{h}{d}=\frac{c}{h}}.$

Here $\frac{h}{d}$ is the tangent of the angle $ABH$ and $\frac{c}{h}$ is the tangent of the angle $HAC.$ Two acute angles with equal tangents are equal, so ${\angle HAC=\angle ABH}.$ In the right triangle $ABH$ we have ${\angle BAH=90^\circ-\angle ABH}.$ Therefore ${\angle BAC}={\angle BAH+\angle HAC}={90^\circ\text{:}}$ the lines are perpendicular.

Equations of a Line

An equation of a line is an equation in $x$ and $y$ that is true for the coordinates of every point of the line and false for every other point. Using the slope, such equations are easy to write.

Vertical and horizontal lines

Theorem. The vertical line through the point $(a, 0)$ has the equation ${x=a}.$ The horizontal line through the point $(0, b)$ has the equation ${y=b}.$

On the vertical line every point has the $x$-coordinate $a,$ whatever its $y$-coordinate is. On the horizontal line every point has the $y$-coordinate $b.$ For example, ${x=3}$ is the vertical line through $(3, 0),$ and ${y=-2}$ is the horizontal line through $(0, -2).$ The $x$-axis is the line ${y=0},$ and the $y$-axis is the line ${x=0}.$

Point-slope form

Theorem. The line that passes through the point $P_1(x_1, y_1)$ and has slope $m$ is given by the equation ${y-y_1=m(x-x_1)}.$ This is called the point-slope form of the equation of the line.

Proof

Let $P(x, y)$ be a point different from $P_1.$

If ${x\ne x_1},$ the point $P$ lies on the line exactly when the slope of $P_1P$ is equal to $m\text{:}$

$$\frac{y-y_1}{x-x_1}=m.$$

Multiplying both sides by ${x-x_1}$ gives the equivalent equation ${y-y_1=m(x-x_1)}.$

If ${x=x_1},$ then ${y\ne y_1}.$ Such a point is not on the line, because the line is not vertical, and it does not satisfy the equation: the left side ${y-y_1}$ is not $0,$ and the right side is $0.$

The coordinates of $P_1$ itself satisfy the equation: both sides become $0.$

Example 6. Write the equation of the line through the point $(2, 3)$ with slope $4.$

Solution. ${y-3=4(x-2)}.$ We remove the parentheses: ${y-3=4x-8},$ so ${y=4x-5}.$

Example 7. Write the equation of the line through the points $(1, 5)$ and $(3, 1).$

Solution. First we find the slope: ${m=\frac{1-5}{3-1}=\frac{-4}{2}=-2}.$ Then we use the point-slope form with the point $(1, 5)\text{:}$ ${y-5=-2(x-1)},$ so ${y=-2x+7}.$ Check with the second point: ${-2\cdot3+7=1}.$

Slope-intercept form

The $y$-intercept of a line is the $y$-coordinate of the point where the line crosses the $y$-axis. If the $y$-intercept is $b,$ the line passes through the point $(0, b).$

Theorem. The line with slope $m$ and $y$-intercept $b$ is given by the equation ${y=mx+b}.$ This is called the slope-intercept form of the equation of the line.

Indeed, the point-slope form with the point $(0, b)$ is ${y-b=m(x-0)},$ that is, ${y=mx+b}.$ In this form both numbers can be read at once: the coefficient of $x$ is the slope, and the constant term is the $y$-intercept. For example, the line ${y=3x-2}$ has slope $3$ and crosses the $y$-axis at the point $(0, -2).$

Example 8. Find the slope and the $y$-intercept of the line ${3x+2y-6=0}.$

Solution. We solve the equation for $y\text{:}$ ${2y=-3x+6},$ so ${y=-\frac{3}{2}x+3}.$ The slope is $-\frac{3}{2}$ and the $y$-intercept is $3.$

In general, if ${B\ne0},$ the line ${Ax+By+C=0}$ has slope ${m=-\frac{A}{B}}$ and $y$-intercept $-\frac{C}{B}.$ If ${B=0}$ (then ${A\ne0}$), the equation becomes ${x=-\frac{C}{A}}\text{:}$ the line is vertical. The equation ${Ax+By+C=0}$ is called the general form of the equation of a line; many textbooks write it as ${Ax+By=C}$ and call it the standard form.

Two equations can describe the same line: ${y=2x+1}$ and ${2y=4x+2}$ have the same slope $2$ and the same $y$-intercept $1.$ Two lines with the same slope are parallel only if their $y$-intercepts are different.

Example 9. Graph the line ${y=-\frac{2}{3}x+4}.$

Solution. The $y$-intercept is $4,$ so the line passes through the point $(0, 4).$ The slope is ${-\frac{2}{3}=\frac{-2}{3}}\text{:}$ for every $3$ units to the right the line goes $2$ units down. From $(0, 4)$ we move $3$ units to the right and $2$ units down and get the point $(3, 2);$ one more such step gives $(6, 0).$ We draw the line through these points.

Example 10. Write the equation of the line through the point $(4, 1)$ that is a) parallel to the line ${y=2x-3};$ b) perpendicular to it.

Solution. The given line has slope $2.$

a) A parallel line has the same slope $2\text{:}$ ${y-1=2(x-4)},$ so ${y=2x-7}.$

b) A perpendicular line has slope $-\frac{1}{2},$ because ${2\cdot\left(-\frac{1}{2}\right)=-1}.$ Its equation is ${y-1=-\frac{1}{2}(x-4)},$ so ${y=-\frac{1}{2}x+3}.$

Slope in Real Life

Roads. A road sign with $8\%$ on it means that the road rises (or drops) $8$ feet for every $100$ feet of horizontal distance. The grade of a road is its slope written as a percent: ${\frac{8}{100}=0.08=8\%}.$ The angle of inclination of such a road is small: ${\tan^{-1}0.08\approx4.6^\circ}.$

A grade of $100\%$ is not a vertical wall: it is the slope $1,$ which is an angle of $45^\circ.$

Speed. If we plot the distance traveled against time, the slope of the graph is the speed. A car that moves at a constant speed and covers $150$ miles in $3$ hours gives a line through the points $(0, 0)$ and $(3, 150).$ Its slope is ${\frac{150-0}{3-0}=50}\text{:}$ the speed is $50$ miles per hour.

Common Mistakes

Mistake	Wrong	Right
Run over rise	${m=\frac{x_2-x_1}{y_2-y_1}}$	${m=\frac{y_2-y_1}{x_2-x_1}}\text{:}$ the change in $y$ is on top
Mixed order of the points	For $(6, 2)$ and $(9, 8)\text{:}$ ${m=\frac{8-2}{6-9}=-2}$	${m=\frac{8-2}{9-6}=2}\text{:}$ the same point comes first above and below
Subtracting a negative number	For $(-2, 1)$ and $(5, 4)$ the run is ${5-2=3}$	The run is ${5-(-2)=7}$
Zero slope and undefined slope	The slope of a horizontal line is undefined, a vertical line has slope $0$	A horizontal line has slope $0,$ the slope of a vertical line is undefined
Slope of a perpendicular line	A line perpendicular to ${y=2x+1}$ has slope $-2$	It has slope $-\frac{1}{2}\text{:}$ change the sign and take the reciprocal
Reading the slope too early	The line ${2y=6x+4}$ has slope $6$	First solve for $y\text{:}$ ${y=3x+2},$ so the slope is $3$
A grade of $100\%$	A road with a $100\%$ grade is vertical	A $100\%$ grade is the slope $1,$ an angle of $45^\circ$

Practice Problems

1. Find the slope of the line through the points: a) $(1, 2)$ and $(5, 10);$ b) $(-3, 4)$ and $(1, -4);$ c) $(0, -5)$ and $(6, -5);$ d) $(2, -1)$ and $(2, 7).$

Answer

a) $m=\frac{10-2}{5-1}=\frac{8}{4}=2.$

b) ${m}={\frac{-4-4}{1-(-3)}}={\frac{-8}{4}}={-2.}$

c) $m=\frac{-5-(-5)}{6-0}=\frac{0}{6}=0.$

d) The slope is undefined: this is the vertical line ${x=2}.$

2. A line with slope $\frac{3}{4}$ passes through the point $(2, 1).$ Find the $y$-coordinate of the point on this line with ${x=10}.$

Answer

The run is ${10-2=8},$ so the rise is ${\frac{3}{4}\cdot8=6}$ and ${y=1+6=7}.$

3. Do the three points lie on the same line? a) $(0, -1),$ $(2, 3)$ and $(5, 9);$ b) $(1, 1),$ $(3, 4)$ and $(7, 9).$

Answer

a) Yes: ${\frac{3-(-1)}{2-0}=\frac{4}{2}=2}$ and ${\frac{9-3}{5-2}=\frac{6}{3}=2}.$

b) No: ${\frac{4-1}{3-1}=\frac{3}{2}},$ but ${\frac{9-4}{7-3}=\frac{5}{4}}.$

4. Find the angle of inclination of a line with slope a) $\sqrt{3};$ b) $-\frac{\sqrt{3}}{3};$ c) $0.$

Answer

a) $60^\circ;$ b) $150^\circ;$ c) $0^\circ.$

5. A line passes through the point $(0, 2),$ and its angle of inclination is $135^\circ.$ Write its equation.

Answer

${m=\tan135^\circ=-1}$ and the $y$-intercept is $2,$ so ${y=-x+2}.$

6. Write the equation of the line through the point $(-1, 4)$ with slope $-3$ in slope-intercept form.

Answer

${y-4=-3(x+1)},$ so ${y=-3x+1}.$

7. Write the equation of the line through the points $(2, -1)$ and $(6, 7).$

Answer

${m=\frac{7-(-1)}{6-2}=\frac{8}{4}=2}.$ Then ${y+1=2(x-2)},$ so ${y=2x-5}.$

8. Find the slope and the $y$-intercept of the line ${4x-2y+10=0}.$

Answer

${2y=4x+10},$ so ${y=2x+5}.$ The slope is $2$ and the $y$-intercept is $5.$

9. Are the two lines parallel, perpendicular or neither? a) ${y=3x-1}$ and ${6x-2y+5=0};$ b) ${y=\frac{2}{5}x+1}$ and ${5x+2y=4};$ c) ${y=2x}$ and ${y=-2x+3}.$

Answer

a) Parallel: the second line is ${y=3x+\frac{5}{2}}.$ Both slopes are $3,$ and the $y$-intercepts $-1$ and $\frac{5}{2}$ are different, so these are two different lines.

b) Perpendicular: the second line is ${y=-\frac{5}{2}x+2},$ and ${\frac{2}{5}\cdot\left(-\frac{5}{2}\right)=-1}.$

c) Neither: the slopes $2$ and $-2$ are not equal, and their product is ${2\cdot(-2)=-4}.$

10. Write the equation of the line through the point $(3, -2)$ that is perpendicular to the line ${y=\frac{3}{4}x+1}.$

Answer

The slope is $-\frac{4}{3}.$ Then ${y+2=-\frac{4}{3}(x-3)},$ so ${y=-\frac{4}{3}x+2}.$

11. Show that the triangle with vertices $A(1, 1),$ $B(5, 3)$ and $C(3, 7)$ is a right triangle.

Answer

The slope of $AB$ is ${\frac{3-1}{5-1}=\frac{2}{4}=\frac{1}{2}},$ and the slope of $BC$ is ${\frac{7-3}{3-5}=\frac{4}{-2}=-2}.$ Since ${\frac{1}{2}\cdot(-2)=-1},$ the sides $AB$ and $BC$ are perpendicular: the angle at $B$ is a right angle.

12. A wheelchair ramp rises $1$ foot for every $12$ feet of horizontal distance. Write its slope as a percent and find its angle of inclination. Round both to one decimal place.

Answer

The slope is ${\frac{1}{12}\approx0.083=8.3\%}.$ The angle of inclination is ${\tan^{-1}\frac{1}{12}\approx4.8^\circ}.$

See also: Graphs and coordinates · Analytic geometry formulas · Angle pairs · Trigonometry