Chapter 9 ยท Straight Lines

9.1 Slope of a Line and the Angle between Two Lines

The slope of a line, $m=\tan\theta=\dfrac{y_2-y_1}{x_2-x_1}$, measures its steepness. Parallel lines share a slope, perpendicular slopes multiply to $-1$, and $\tan\theta=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|$ gives the angle between two lines.

A school is building a wheelchair ramp up to a doorway $0.5\ \text{m}$ above the ground, and the guideline says the ramp must be no steeper than "$1$ in $12$". How long must the ramp be along the ground, and what angle will it make with the ground? To answer, we need a single number that measures steepness โ€” how much a line rises for each unit it runs horizontally. That number is the slope. A $1$ in $12$ ramp has slope $\tfrac1{12}$, so it needs at least $12\times0.5=6\ \text{m}$ of horizontal run, and it makes an angle of only about $4.8^\circ$ with the ground.

In this section we define the angle a line makes with the $x$-axis (its inclination) and its slope, and prove that the slope can be read straight from the coordinates of any two points on the line. Slopes then answer three geometric questions with simple algebra: are two lines parallel? are they perpendicular? and, if neither, at what angle do they cross? We start with a quick recall of the two Class 10 formulae that the whole chapter uses.

A Quick Recall: Distance and Section Formulae

Points of the plane are described by ordered pairs $(x,y)$ (Section 2.1). From Class 10 Mathematics (Chapter 7, Coordinate Geometry) you know two formulae for points $P(x_1,y_1)$ and $Q(x_2,y_2)$:

๐Ÿ“ Recall โ€” Distance, section and mid-point formulae

Distance. $$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$$ It is Pythagoras' theorem in the right triangle whose legs are the horizontal and vertical gaps $|x_2-x_1|$ and $|y_2-y_1|$.

Section formula. The point $R$ that divides $PQ$ internally in the ratio $p:q$ (so $PR:RQ=p:q$, with $p,q\gt0$) is $$R\left(\dfrac{px_2+qx_1}{p+q},\ \dfrac{py_2+qy_1}{p+q}\right).$$ (Class 10 often writes the ratio as $m_1:m_2$; we use $p:q$ because in this chapter $m$ stands for a slope.)

Mid-point. Taking $p=q$: the mid-point of $PQ$ is $\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$.

โœ๏ธ Example 1 โ€” Warming up with the recall formulae
For $A(-2,3)$ and $B(4,-5)$ find (a) the length $AB$ (b) the mid-point $M$ of $AB$ (c) the point $R$ dividing $AB$ internally in the ratio $1:3$.
(a) $AB=\sqrt{(4-(-2))^2+(-5-3)^2}=\sqrt{36+64}=\sqrt{100}=10$ units.
(b) $M=\left(\dfrac{-2+4}{2},\dfrac{3+(-5)}{2}\right)=(1,-1)$.
(c) With $p=1$, $q=3$: $R=\left(\dfrac{1\cdot4+3\cdot(-2)}{4},\dfrac{1\cdot(-5)+3\cdot3}{4}\right)=\left(-\tfrac12,\ 1\right)$.
Check Ratio $1:3$ means $R$ is $\tfrac14$ of the way from $A$ to $B$. The step from $A$ to $B$ is $(6,-8)$, and $A+\tfrac14(6,-8)=(-2+1.5,\ 3-2)=(-0.5,\ 1)$ โœ“.

Inclination and Slope of a Line

๐Ÿ“˜ Definition 9.1 โ€” Inclination of a line

The inclination of a line is the angle $\theta$ that the line makes with the positive direction of the $x$-axis, measured anticlockwise (from the $x$-axis round to the line). It always satisfies $$0^\circ\le\theta\lt180^\circ.$$ Lines parallel to the $x$-axis (including the $x$-axis itself) have inclination $0^\circ$; lines parallel to the $y$-axis have inclination $90^\circ$.

๐Ÿ“˜ Definition 9.2 โ€” Slope of a line

If a line has inclination $\theta\ne90^\circ$, its slope (or gradient) is $$m=\tan\theta.$$ A vertical line ($\theta=90^\circ$) has no slope: $\tan90^\circ$ is not defined. A horizontal line has slope $\tan0^\circ=0$.

The tangent of an obtuse angle is negative ($\tan(180^\circ-\alpha)=-\tan\alpha$, Section 3.2), so a line with an obtuse inclination has a negative slope. Because $\tan$ takes every real value exactly once on $0^\circ\le\theta\lt180^\circ$, $\theta\ne90^\circ$, each real number is the slope of lines of exactly one inclination.

๐Ÿ“ Theorem 9.1 โ€” Slope from two points

Let a line pass through two points $P(x_1,y_1)$ and $Q(x_2,y_2)$ with $x_1\ne x_2$. Then its slope is $$m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\text{rise}}{\text{run}}.$$ If $x_1=x_2$ (and $P\ne Q$), the line is vertical and has no slope.

Proof. The expression does not change if $P$ and $Q$ are swapped (numerator and denominator both change sign), so we may label the points with $x_1\lt x_2$. Let $R(x_2,y_1)$ be the point level with $P$ and directly below or above $Q$; then $PR=x_2-x_1\gt0$ and $PR$ is parallel to the $x$-axis.

Case 1: $y_2\gt y_1$ (the line rises to the right). The line meets the $x$-axis and $PR$ at equal corresponding angles, so $\angle QPR=\theta$, which is acute. In the right triangle $PRQ$, $\tan\theta=\dfrac{RQ}{PR}=\dfrac{y_2-y_1}{x_2-x_1}$.

Case 2: $y_2\lt y_1$ (the line falls to the right). Now $\theta$ is obtuse and $\angle QPR=180^\circ-\theta$, so $\tan(180^\circ-\theta)=\dfrac{RQ}{PR}=\dfrac{y_1-y_2}{x_2-x_1}$. Since $\tan\theta=-\tan(180^\circ-\theta)$, again $\tan\theta=\dfrac{y_2-y_1}{x_2-x_1}$.

Case 3: $y_2=y_1$. The line is horizontal, $\theta=0^\circ$, and both sides are $0$.

Finally, if $x_1=x_2$ both points lie on one vertical line, whose inclination is $90^\circ$. $\blacksquare$

So the slope is the same whichever two points of the line you use โ€” it is the rise per unit run, everywhere along the line. Its sign and size tell you how the line looks:

slopeinclinationthe lineโ€ฆ
$m\gt0$acute, $0^\circ\lt\theta\lt90^\circ$rises from left to right
$m\lt0$obtuse, $90^\circ\lt\theta\lt180^\circ$falls from left to right
$m=0$$\theta=0^\circ$is horizontal
not defined$\theta=90^\circ$is vertical

The larger $|m|$ is, the steeper the line: slope $5$ climbs five units for every unit across, slope $\tfrac15$ only one unit for every five across.

โœ๏ธ Example 2 โ€” Slopes and inclinations
(a) Find the slope of the line through $(3,-2)$ and $(-1,4)$. Does it rise or fall?
(b) Find the slope of the line through $(-2,5)$ and $(-2,-1)$.
(c) Find the slope of a line with inclination $120^\circ$, and the inclination of a line with slope $\tfrac{1}{\sqrt3}$.
(d) A ramp rises $0.5\ \text{m}$ over a horizontal run of $6\ \text{m}$. Find its slope and the angle it makes with the ground.
(a) $m=\dfrac{4-(-2)}{-1-3}=\dfrac{6}{-4}=-\dfrac32$. Negative, so the line falls from left to right.
(b) The $x$-coordinates are equal ($-2$), so the run is $0$: the line is vertical and has no slope. (Writing $\tfrac{-6}{0}$ is meaningless, not "infinity".)
(c) $m=\tan120^\circ=-\tan60^\circ=-\sqrt3$. For $\tan\theta=\tfrac1{\sqrt3}$ with $0^\circ\le\theta\lt180^\circ$: $\theta=30^\circ$ (the only angle in that range; $210^\circ$ is not allowed).
(d) $m=\dfrac{0.5}{6}=\dfrac1{12}\approx0.0833$. The angle with the ground is the inclination: $\theta=\tan^{-1}\tfrac1{12}\approx4.76^\circ$. (Here $\tan^{-1}$ is just the calculator key that undoes $\tan$.)
โœ๏ธ Example 3 โ€” Finding an unknown coordinate
(a) The line through $(2,k)$ and $(5,7)$ has slope $2$. Find $k$.
(b) For which $k$ does the line through $(k,-1)$ and $(3,k)$ have inclination $45^\circ$? Is there a $k$ giving inclination $135^\circ$?
(a) $\dfrac{7-k}{5-2}=2$, so $7-k=6$ and $k=1$. Check: $(2,1)$ to $(5,7)$ rises $6$ over a run of $3$: slope $2$ โœ“.
(b) 45ยฐ Slope $\tan45^\circ=1$: $\dfrac{k-(-1)}{3-k}=1$ (we need $k\ne3$), so $k+1=3-k$ and $k=1$. Check: $(1,-1)$ and $(3,1)$: slope $\tfrac22=1$ โœ“.
(b) 135ยฐ Slope $\tan135^\circ=-1$: $\dfrac{k+1}{3-k}=-1$ gives $k+1=k-3$, i.e. $1=-3$ โ€” impossible.
Why The run is $3-k$ and the rise is $k+1$; their sum is always $4$. Slope $-1$ needs rise $=-$run, i.e. a sum of $0$. So no line of this family has inclination $135^\circ$.
Interactive 1 โ€” The Slope Explorer

Drag the points $P$ and $Q$ (they snap to whole-number coordinates), or press Play to send $Q$ round a square about $P$. The orange run and red rise make the slope triangle, and the purple arc is the inclination $\theta$. Watch the sign of $m$ change as the line passes through vertical. Switch on the partner lines through $K(5,-3)$: the parallel partner has the same slope, the perpendicular partner has slope $-\tfrac1m$, so the product of the two slopes is always $-1$.

rise / runโ€”
slope $m$โ€”
inclination $\theta$โ€”
perpendicular slopeโ€”
product of slopesโ€”

Parallel and Perpendicular Lines

Two lines are parallel exactly when they make equal corresponding angles with the $x$-axis, that is, when they have equal inclinations. Since equal inclinations (other than $90^\circ$) mean equal tangents, slopes give an instant test.

๐Ÿ“ Theorem 9.2 โ€” Slopes of parallel lines

Two non-vertical lines with slopes $m_1$ and $m_2$ are parallel (or are the same line) if and only if $$m_1=m_2.$$ Any two vertical lines are parallel.

Proof. Let the inclinations be $\theta_1,\theta_2$, both in $[0^\circ,180^\circ)$ and not $90^\circ$. The lines are parallel exactly when $\theta_1=\theta_2$ (corresponding angles with the $x$-axis). On this range $\tan$ never takes the same value twice, so $\theta_1=\theta_2$ exactly when $\tan\theta_1=\tan\theta_2$, i.e. $m_1=m_2$. $\blacksquare$

๐Ÿ“ Theorem 9.3 โ€” Slopes of perpendicular lines

Two non-vertical lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $$m_1m_2=-1,\quad\text{that is,}\quad m_2=-\dfrac{1}{m_1}.$$ (A horizontal line, slope $0$, is perpendicular to every vertical line, which has no slope; the formula does not cover that pair.)

Proof. Neither line is vertical, and if they are perpendicular neither can be horizontal (its partner would be vertical). So the inclinations are not $0^\circ$ or $90^\circ$. Label the lines so that $\theta_1\lt\theta_2$.

($\Rightarrow$) If the lines are perpendicular, the angle of the triangle they form with the $x$-axis gives $\theta_2=\theta_1+90^\circ$ (an exterior angle equals the sum of the two interior opposite angles, one of which is the right angle). Then $$m_2=\tan(90^\circ+\theta_1)=-\cot\theta_1=-\dfrac1{\tan\theta_1}=-\dfrac1{m_1},$$ using the allied-angle formula from Section 3.3. So $m_1m_2=-1$.

($\Leftarrow$) If $m_1m_2=-1$, the slopes have opposite signs, so one inclination is acute and the other obtuse; with our labelling $0^\circ\lt\theta_1\lt90^\circ\lt\theta_2\lt180^\circ$. Now $\tan\theta_2=-\dfrac1{\tan\theta_1}=\tan(90^\circ+\theta_1)$, and both $\theta_2$ and $90^\circ+\theta_1$ lie between $90^\circ$ and $180^\circ$, where $\tan$ never repeats a value. Hence $\theta_2=\theta_1+90^\circ$ and the lines are perpendicular. $\blacksquare$

In practice: to get a perpendicular slope, flip the fraction and change its sign. Perpendicular to slope $\tfrac23$ is slope $-\tfrac32$; perpendicular to slope $-4$ is slope $\tfrac14$.

โœ๏ธ Example 4 โ€” Testing figures with slopes
(a) Show that $A(1,1)$, $B(5,3)$, $C(4,5)$, $D(0,3)$ are the vertices of a rectangle. Is it a square?
(b) The line through $(-3,5)$ and $(3,7)$ is perpendicular to the line through $(6,10)$ and $(x,22)$. Find $x$.
(c) Find $y$ if the line through $(-1,y)$ and $(2,-4)$ is parallel to the line through $(0,1)$ and $(4,-7)$.
(a) Slopes $m_{AB}=\dfrac{3-1}{5-1}=\dfrac12$, $m_{BC}=\dfrac{5-3}{4-5}=-2$, $m_{CD}=\dfrac{3-5}{0-4}=\dfrac12$, $m_{DA}=\dfrac{1-3}{1-0}=-2$.
(a) Conclude $AB\parallel DC$ and $BC\parallel AD$ (equal slopes), so $ABCD$ is a parallelogram; and $m_{AB}\,m_{BC}=\tfrac12\times(-2)=-1$, so $\angle B=90^\circ$. A parallelogram with a right angle is a rectangle. $AB=\sqrt{16+4}=\sqrt{20}$ but $BC=\sqrt{1+4}=\sqrt5$, so it is not a square.
(b) First slope $\dfrac{7-5}{3-(-3)}=\dfrac13$, so the second must be $-3$: $\dfrac{22-10}{x-6}=-3$, $12=-3x+18$, $x=2$. Check: $(6,10)$ to $(2,22)$: $\tfrac{12}{-4}=-3$ and $\tfrac13\times(-3)=-1$ โœ“.
(c) Second slope $\dfrac{-7-1}{4-0}=-2$. So $\dfrac{-4-y}{2-(-1)}=-2$, $-4-y=-6$, $y=2$. Check: $(-1,2)$ to $(2,-4)$: $\tfrac{-6}{3}=-2$ โœ“.

The Angle between Two Lines

Two lines that cross make two angles at the crossing point, $\phi$ and $180^\circ-\phi$ (vertically opposite angles repeat them). Unless the lines are perpendicular, one angle is acute and the other obtuse, and "the angle between the lines" usually means the acute one. Slopes give it directly.

๐Ÿ“ Theorem 9.4 โ€” Angle between two lines

Let two non-vertical lines have slopes $m_1$ and $m_2$, with $1+m_1m_2\ne0$. Then the acute angle $\theta$ between them satisfies $$\tan\theta=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|,$$ and the obtuse angle between them is $180^\circ-\theta$. If $1+m_1m_2=0$ the lines are perpendicular (Theorem 9.3); if $m_1=m_2$ they are parallel and $\theta=0^\circ$.

Proof. Let the inclinations be $\alpha_1$ and $\alpha_2$, so $m_1=\tan\alpha_1$, $m_2=\tan\alpha_2$. Turning the first line through the angle $\alpha_2-\alpha_1$ brings it parallel to the second, so the two angles between the lines are $|\alpha_2-\alpha_1|$ and $180^\circ-|\alpha_2-\alpha_1|$. By the difference formula of Section 3.3 (valid because $\alpha_1,\alpha_2\ne90^\circ$ and, as $1+m_1m_2\ne0$, $\alpha_2-\alpha_1\ne\pm90^\circ$), $$\tan(\alpha_2-\alpha_1)=\dfrac{\tan\alpha_2-\tan\alpha_1}{1+\tan\alpha_1\tan\alpha_2}=\dfrac{m_2-m_1}{1+m_1m_2}.$$ Since $\tan(-x)=-\tan x$ and $\tan(180^\circ-x)=-\tan x$, the two angles between the lines have tangents $\pm\dfrac{m_2-m_1}{1+m_1m_2}$. The acute one has the positive tangent, which is the absolute value. $\blacksquare$

๐Ÿงญ Problem-solving strategy โ€” Using the angle formula
  1. Find both slopes. If one line is vertical, the formula does not apply: work with inclinations instead (the angle is $|90^\circ-\alpha|$ where $\alpha$ is the other line's inclination).
  2. If $1+m_1m_2=0$, stop: the lines are perpendicular.
  3. Otherwise compute $\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|$ and take $\tan^{-1}$ for the acute angle. Give the obtuse angle as $180^\circ-\theta$ if it is asked for.
  4. When the angle is given and a slope is unknown, remove the modulus by writing $\dfrac{m_2-m_1}{1+m_1m_2}=\pm\tan\theta$ and solve both equations โ€” there are usually two answers.
โœ๏ธ Example 5 โ€” Angles between lines
Find the acute and obtuse angles between the lines with slopes (a) $2$ and $-3$ (b) $\sqrt3$ and $\tfrac1{\sqrt3}$. In (b), check your answer using inclinations.
(a) $1+m_1m_2=1-6=-5\ne0$. $\tan\theta=\left|\dfrac{-3-2}{-5}\right|=1$, so $\theta=45^\circ$; the obtuse angle is $135^\circ$.
(b) $1+m_1m_2=1+1=2$. $\tan\theta=\left|\dfrac{\tfrac1{\sqrt3}-\sqrt3}{2}\right|=\left|\dfrac{1-3}{2\sqrt3}\right|=\dfrac1{\sqrt3}$, so $\theta=30^\circ$ and the obtuse angle is $150^\circ$.
(b) Check The inclinations are $\tan^{-1}\sqrt3=60^\circ$ and $\tan^{-1}\tfrac1{\sqrt3}=30^\circ$; they differ by $30^\circ$ โœ“.
โœ๏ธ Example 6 โ€” A slope from an angle
A line $l$ has slope $3$. Find the slopes of the lines that make an angle of $45^\circ$ with $l$. What do you notice about the two answers?
Set up Let the unknown slope be $m$. $\left|\dfrac{m-3}{1+3m}\right|=\tan45^\circ=1$, so $\dfrac{m-3}{1+3m}=1$ or $\dfrac{m-3}{1+3m}=-1$.
Case +1 $m-3=1+3m$, so $-2m=4$ and $m=-2$.
Case โˆ’1 $m-3=-1-3m$, so $4m=2$ and $m=\tfrac12$.
Check $m=-2$: $\left|\dfrac{-5}{-5}\right|=1$ โœ“; $m=\tfrac12$: $\left|\dfrac{-2.5}{2.5}\right|=1$ โœ“. And $(-2)\times\tfrac12=-1$: the two answers are perpendicular to each other. That is no accident โ€” the two lines at $45^\circ$ on either side of $l$ are $90^\circ$ apart.
Interactive 2 โ€” The Angle between Two Lines

Line 1 (red) and line 2 (blue) cross at the point $(1,1)$. Set their inclinations, or press Play to turn line 2. The purple arc marks the acute angle between them and the grey arc the obtuse one. The readouts compute $\left|\tfrac{m_2-m_1}{1+m_1m_2}\right|$ from the slopes and compare its $\tan^{-1}$ with the angle measured on the picture. Watch what happens when $1+m_1m_2$ reaches $0$.

30ยฐ105ยฐ
$m_1$โ€”
$m_2$โ€”
$1+m_1m_2$โ€”
formula angleโ€”
measured angleโ€”
โš ๏ธ Common mistakes with slopes

1. Mixing the order of the points. $\dfrac{y_2-y_1}{x_2-x_1}$ is right and so is $\dfrac{y_1-y_2}{x_1-x_2}$, but $\dfrac{y_2-y_1}{x_1-x_2}$ has the wrong sign. Take both differences in the same order.

2. "Run over rise". Slope is change in $y$ divided by change in $x$, never the other way round.

3. A vertical line has slope $0$. No: a horizontal line has slope $0$; a vertical line has no slope at all.

4. Perpendicular slope $=-m$ or $\tfrac1m$. You need both steps: the negative reciprocal, $-\tfrac1m$. Check that the product is $-1$.

5. Dropping the modulus โ€” or keeping it when solving. The formula with $|\ |$ gives the acute angle. When the angle is known and a slope is unknown, the modulus hides two equations; solving only one loses an answer.

6. Using the angle formula when $1+m_1m_2=0$. The fraction is undefined โ€” and that is the signal that the lines are perpendicular, not an error.

๐Ÿš€ Beyond the board syllabus (JEE Main ยท JEE Advanced)

In JEE Main and JEE Advanced 2026 (both list the centroid, orthocentre and circumcentre of a triangle); not in the CBSE year-end exam. Both often ask for the special points of a triangle, and slopes find them quickly.

Centroid, orthocentre and circumcentre. For a triangle $ABC$: the centroid $G$ (where the medians meet) is $\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)$; the orthocentre $H$ is where the altitudes meet, so $AH\perp BC$ and $BH\perp CA$; the circumcentre $O$ is equidistant from $A$, $B$, $C$. For every triangle, $G$ lies on the segment $HO$ with $HG:GO=2:1$ (the Euler line).

Worked example. Find $G$, $H$ and $O$ for $A(0,0)$, $B(6,0)$, $C(2,4)$, and check the $2:1$ property.
Solution. $G=\left(\tfrac83,\tfrac43\right)$. Let $H=(h,k)$. $m_{BC}=\tfrac{4}{2-6}=-1$, so $m_{AH}=1$: $\tfrac kh=1$, $k=h$. $m_{CA}=2$, so $m_{BH}=-\tfrac12$: $\tfrac{k}{h-6}=-\tfrac12$, $2k=6-h$. With $k=h$: $h=k=2$, so $H=(2,2)$. For $O=(a,b)$: $OA=OB$ gives $a=3$; $OA=OC$ gives $9+b^2=1+(b-4)^2$, so $8b=8$, $b=1$: $O=(3,1)$. Check: the point dividing $HO$ in the ratio $2:1$ is $\left(\tfrac{2\cdot3+1\cdot2}{3},\tfrac{2\cdot1+1\cdot2}{3}\right)=\left(\tfrac83,\tfrac43\right)=G$ โœ“.

Exercises

I
Recall, slopes and inclinations.
  • 1 Find the distance between $(-3,4)$ and $(5,-2)$.
  • 2 Find the mid-point of the segment joining $(-3,4)$ and $(5,-2)$.
  • 3 Find the point dividing the join of $(2,-1)$ and $(8,5)$ internally in the ratio $2:1$.
  • 4 Find the slope of the line through $(2,3)$ and $(6,11)$.
  • 5 Find the slope of the line through $(-4,1)$ and $(2,-2)$.
  • 6 Find the slope, if it exists, of the line through $(3,-5)$ and $(3,7)$.
  • 7 Find the slope of the line through $(-1,4)$ and $(5,4)$. What kind of line is it?
  • 8 Find the slopes of lines with inclinations $60^\circ$, $150^\circ$ and $0^\circ$.
  • 9 Find the inclinations of lines with slopes $-\sqrt3$ and $1$.
1 $\sqrt{8^2+(-6)^2}=\sqrt{100}=10$ units.
2 $\left(\tfrac{-3+5}{2},\tfrac{4-2}{2}\right)=(1,1)$.
3 $\left(\tfrac{2\cdot8+1\cdot2}{3},\tfrac{2\cdot5+1\cdot(-1)}{3}\right)=(6,3)$.
4 $\tfrac{11-3}{6-2}=\tfrac84=2$.
5 $\tfrac{-2-1}{2-(-4)}=\tfrac{-3}{6}=-\tfrac12$.
6 The run is $3-3=0$: the line is vertical and has no slope.
7 $\tfrac{4-4}{5+1}=0$: a horizontal line (parallel to the $x$-axis).
8 $\tan60^\circ=\sqrt3$; $\tan150^\circ=-\tan30^\circ=-\tfrac1{\sqrt3}$; $\tan0^\circ=0$.
9 $\tan\theta=-\sqrt3$ with $0^\circ\le\theta\lt180^\circ$ gives $\theta=120^\circ$; $\tan\theta=1$ gives $\theta=45^\circ$.
II
Unknowns and applications.
  • 1 Find $k$ if the line through $(k,2)$ and $(5,-4)$ has slope $-\tfrac32$.
  • 2 Find $a$ if the line through $(-2,a)$ and $(4,2a+3)$ has inclination $45^\circ$.
  • 3 Show that the line through $(a,0)$ and $(0,b)$, where $a\ne0$, has slope $-\tfrac ba$.
  • 4 A wheelchair ramp must rise $0.9\ \text{m}$ and may be no steeper than slope $\tfrac1{12}$. What is the shortest horizontal run it can have?
  • 5 A railway gradient is "1 in 150" (a rise of $1\ \text{m}$ per $150\ \text{m}$ of horizontal distance). How much does the track rise over $3\ \text{km}$ of horizontal distance, and what angle does it make with the horizontal (to two decimal places of a degree)?
  • 6 The cross-section of a roof has eaves at $(0,0)$ and $(8,0)$ and its ridge at $(4,3)$ (in metres). Find the slopes of the two rafters and the pitch (the angle each makes with the horizontal).
1 $\dfrac{-4-2}{5-k}=-\dfrac32$, so $5-k=4$ and $k=1$. Check: $\tfrac{-6}{4}=-\tfrac32$ โœ“.
2 $\dfrac{(2a+3)-a}{4-(-2)}=1$, so $a+3=6$ and $a=3$. Check: $(-2,3)$ to $(4,9)$ has slope $\tfrac66=1$ โœ“.
3 $\dfrac{b-0}{0-a}=-\dfrac ba$.
4 $\dfrac{0.9}{\text{run}}\le\dfrac1{12}$ means $\text{run}\ge12\times0.9=10.8\ \text{m}$.
5 Rise $=\tfrac{3000}{150}=20\ \text{m}$. Angle $=\tan^{-1}\tfrac1{150}\approx0.38^\circ$.
6 Left rafter $(0,0)\to(4,3)$: slope $\tfrac34$; right rafter $(4,3)\to(8,0)$: slope $-\tfrac34$. Pitch $=\tan^{-1}\tfrac34\approx36.87^\circ$ on both sides (the right rafter has inclination $180^\circ-36.87^\circ=143.13^\circ$).
III
Parallel and perpendicular lines.
  • 1 Are the line through $(2,-3)$ and $(-4,0)$ and the line through $(1,5)$ and $(3,4)$ parallel? Are they the same line?
  • 2 Show that the line through $(0,0)$ and $(2,3)$ is perpendicular to the line through $(1,1)$ and $(4,-1)$.
  • 3 Write the slope of a line (a) parallel (b) perpendicular to a line of slope $-\tfrac45$.
  • 4 Find $x$ if the line through $(x,-1)$ and $(2,5)$ is perpendicular to a line of slope $\tfrac32$.
  • 5 Find $p$ if the line through $(p,1)$ and $(2p,5)$ is parallel to the line through $(1,1)$ and $(3,3)$.
  • 6 Show that $A(-2,-1)$, $B(4,0)$, $C(3,3)$, $D(-3,2)$ form a parallelogram. Is it a rectangle?
  • 7 Show that the triangle with vertices $P(-2,1)$, $Q(2,3)$, $R(4,-1)$ is right-angled and isosceles.
  • 8 $A(0,0)$, $B(5,0)$, $C(8,4)$, $D(3,4)$. Show that $ABCD$ is a rhombus and that its diagonals are perpendicular.
1 Slopes $\tfrac{0-(-3)}{-4-2}=-\tfrac12$ and $\tfrac{4-5}{3-1}=-\tfrac12$: parallel. Not the same line: the first line through $(2,-3)$ with slope $-\tfrac12$ is at height $-3+\tfrac12=-2.5$ when $x=1$, not $5$.
2 Slopes $\tfrac32$ and $\tfrac{-1-1}{4-1}=-\tfrac23$; product $-1$, so perpendicular.
3 (a) $-\tfrac45$ (b) $\tfrac54$.
4 The slope must be $-\tfrac23$: $\dfrac{5-(-1)}{2-x}=-\dfrac23$, so $2-x=-9$ and $x=11$. Check: $\tfrac{6}{-9}=-\tfrac23$ โœ“.
5 $\dfrac{5-1}{2p-p}=\dfrac4p=1$, so $p=4$ (and $p\ne0$). The points $(4,1)$, $(8,5)$ are not on the line through $(1,1)$ and $(3,3)$ (on that line the two coordinates are equal), so the lines are distinct.
6 $m_{AB}=\tfrac16$, $m_{BC}=\tfrac{3}{-1}=-3$, $m_{CD}=\tfrac{-1}{-6}=\tfrac16$, $m_{DA}=\tfrac{-3}{1}=-3$: opposite sides parallel, so a parallelogram. $\tfrac16\times(-3)=-\tfrac12\ne-1$, so no right angle: not a rectangle.
7 $m_{PQ}=\tfrac24=\tfrac12$, $m_{QR}=\tfrac{-4}{2}=-2$: product $-1$, right angle at $Q$. $PQ=\sqrt{16+4}=\sqrt{20}$ and $QR=\sqrt{4+16}=\sqrt{20}$: isosceles.
8 $AB=5$, $BC=\sqrt{9+16}=5$, $CD=5$, $DA=\sqrt{9+16}=5$: a rhombus. Diagonal slopes $m_{AC}=\tfrac48=\tfrac12$, $m_{BD}=\tfrac{4}{-2}=-2$: product $-1$, perpendicular.
IV
The angle between two lines.
  • 1 Find the acute angle between lines with slopes $1$ and $0$.
  • 2 Find the acute angle between lines with slopes $\sqrt3$ and $-\sqrt3$.
  • 3 Find the acute angle between lines with slopes $\tfrac12$ and $3$.
  • 4 Find the angle between lines with slopes $2$ and $-\tfrac12$.
  • 5 Find, to two decimal places of a degree, the acute angle between lines with slopes $2$ and $3$.
  • 6 Find the slopes of the lines that make an angle of $45^\circ$ with a line of slope $4$.
  • 7 The tangent of the acute angle between two lines is $\tfrac13$, and one of them has slope $2$. Find the possible slopes of the other.
  • 8 Find the three angles of the triangle with vertices $A(0,0)$, $B(4,0)$, $C(1,3)$ using slopes.
  • 9 Two straight roads leave a town at the origin, one towards the point $(6,2)$ and the other towards $(2,5)$ (in km). Find the angle between the roads.
Check: the angle between lines with slopes $m_1$ and $m_2$
$\left|\tfrac{m_2-m_1}{1+m_1m_2}\right|$โ€”
acute angleโ€”
from inclinationsโ€”
1 $\left|\tfrac{0-1}{1+0}\right|=1$: $45^\circ$.
2 $\left|\tfrac{-\sqrt3-\sqrt3}{1-3}\right|=\left|\tfrac{-2\sqrt3}{-2}\right|=\sqrt3$: $60^\circ$ (inclinations $60^\circ$ and $120^\circ$ differ by $60^\circ$ โœ“).
3 $\left|\tfrac{3-\frac12}{1+\frac32}\right|=\tfrac{5/2}{5/2}=1$: $45^\circ$.
4 $1+m_1m_2=1-1=0$: the lines are perpendicular, $90^\circ$.
5 $\left|\tfrac{3-2}{1+6}\right|=\tfrac17$: $\theta=\tan^{-1}\tfrac17\approx8.13^\circ$.
6 $\tfrac{m-4}{1+4m}=1$ gives $m-4=1+4m$, $m=-\tfrac53$; $\tfrac{m-4}{1+4m}=-1$ gives $m-4=-1-4m$, $m=\tfrac35$. (Again the two answers multiply to $-1$.)
7 $\tfrac{m-2}{1+2m}=\pm\tfrac13$. With $+$: $3m-6=1+2m$, $m=7$. With $-$: $3m-6=-1-2m$, $m=1$. Check: $\tfrac{5}{15}=\tfrac13$ and $\left|\tfrac{-1}{3}\right|=\tfrac13$ โœ“.
8 $m_{AB}=0$, $m_{AC}=3$, $m_{BC}=\tfrac{3}{1-4}=-1$. At $A$: $\tan A=|3-0|=3$, $A\approx71.57^\circ$. At $B$: $\tan B=|-1-0|=1$, $B=45^\circ$. At $C$: $\tan C=\left|\tfrac{-1-3}{1-3}\right|=2$, $C\approx63.43^\circ$. Sum $71.57^\circ+45^\circ+63.43^\circ=180^\circ$ โœ“. (Every angle of this triangle is acute, so the acute angle between two sides is the angle of the triangle.)
9 Slopes $\tfrac13$ and $\tfrac52$: $\left|\tfrac{\frac52-\frac13}{1+\frac56}\right|=\tfrac{13/6}{11/6}=\tfrac{13}{11}$, so $\theta=\tan^{-1}\tfrac{13}{11}\approx49.76^\circ$. Both roads head into the first quadrant, so this acute angle is the angle between them.
๐ŸŽฏ Exam corner Board ยท 1 markBoard ยท 2 marksBoard ยท 3 marksBoard ยท Case studyJEE MainJEE Advanced

How this is examined: Board papers ask for a slope or an inclination (1โ€“2 marks), use slopes to prove that a figure is a parallelogram, rectangle or right triangle, and find an unknown coordinate from a parallel or perpendicular condition (2โ€“3 marks); the angle formula appears in both directions. JEE expects the two-answer case of the angle formula and quick work with right-angle conditions.

E1. Board ยท 1 mark The slope of a line perpendicular to the line through $(1,2)$ and $(3,8)$ is (A) $3$   (B) $-3$   (C) $\tfrac13$   (D) $-\tfrac13$

Idea The given line has slope $\tfrac{8-2}{3-1}=3$; a perpendicular line has slope $-\tfrac13$, since $3\times\left(-\tfrac13\right)=-1$.
Trap (B) changes the sign but forgets to take the reciprocal; (A) is the slope of the line itself, i.e. of a parallel line.
Answer (D).

E2. Board ยท 2 marks Find the inclination of the line through $(1,0)$ and $(2,-\sqrt3)$, and the slope of a line perpendicular to it.

Slope $m=\dfrac{-\sqrt3-0}{2-1}=-\sqrt3$, so $\tan\theta=-\sqrt3$ with $0^\circ\le\theta\lt180^\circ$: $\theta=120^\circ$. (1 mark)
Answer Inclination $120^\circ$; perpendicular slope $-\dfrac{1}{-\sqrt3}=\dfrac1{\sqrt3}$ (inclination $30^\circ$, and $120^\circ-30^\circ=90^\circ$ โœ“). (1 mark)

E3. Board ยท 3 marks The triangle with vertices $A(a,4)$, $B(1,1)$ and $C(5,3)$ is right-angled at $B$. Find $a$, and then find $\tan C$.

Slopes $m_{BC}=\tfrac{3-1}{5-1}=\tfrac12$, so $BA\perp BC$ needs $m_{BA}=-2$. (1 mark)
Find a $\dfrac{4-1}{a-1}=-2$, so $a-1=-\tfrac32$ and $a=-\tfrac12$. (1 mark)
Answer $m_{CA}=\dfrac{4-3}{-\frac12-5}=-\dfrac2{11}$ and $m_{CB}=\tfrac12$: $\tan C=\left|\dfrac{-\frac2{11}-\frac12}{1-\frac1{11}}\right|=\dfrac{15/22}{10/11}=\dfrac34$. Check: in the right triangle, $\tan C=\tfrac{AB}{BC}=\tfrac{\sqrt{11.25}}{\sqrt{20}}=\sqrt{0.5625}=0.75$ โœ“. (1 mark)

E4. Board ยท Case study A school builds a two-part ramp. Taking the foot of the ramp as the origin and measuring in metres, the first part runs straight from $O(0,0)$ to a landing at $L(10.8,\ 0.9)$, and the second part runs back from $L$ to the door at $D(4.8,\ 1.4)$. (i) Find the slope of $OL$ and say whether it meets the "no steeper than $1$ in $12$" rule. (1 mark) (ii) Find the slope of $LD$. (1 mark) (iii) Find the acute angle between the two parts of the ramp, correct to two decimal places of a degree. (2 marks)

(i) $m_{OL}=\tfrac{0.9}{10.8}=\tfrac1{12}$: exactly the limit, so it meets the rule. (1 mark)
(ii) $m_{LD}=\dfrac{1.4-0.9}{4.8-10.8}=\dfrac{0.5}{-6}=-\dfrac1{12}$ โ€” equally steep, climbing to the left. (1 mark)
(iii) Formula $\tan\theta=\left|\dfrac{-\frac1{12}-\frac1{12}}{1-\frac1{144}}\right|=\dfrac{1/6}{143/144}=\dfrac{24}{143}$. (1 mark)
Answer $\theta=\tan^{-1}\tfrac{24}{143}\approx9.53^\circ$. Check: each part makes $\tan^{-1}\tfrac1{12}\approx4.76^\circ$ with the horizontal, on opposite sides, and $2\times4.764^\circ\approx9.53^\circ$ โœ“. (1 mark)

E5. JEE Main A line makes an angle of $45^\circ$ with the line of slope $2$. The sum of all possible values of its slope is (A) $-\tfrac83$   (B) $\tfrac83$   (C) $-3$   (D) $\tfrac13$

Method $\dfrac{m-2}{1+2m}=\pm1$. With $+1$: $m-2=1+2m$, $m=-3$. With $-1$: $m-2=-1-2m$, $m=\tfrac13$.
Trap (C) and (D) are the two individual slopes โ€” each comes from solving only one sign of the modulus.
โฑ Fast route The two lines at $45^\circ$ to a given line are perpendicular, so the slopes $m_1,m_2$ satisfy $m_1m_2=-1$ and are the roots of $(m-2)^2=(1+2m)^2$, i.e. $3m^2+8m-3=0$: sum $-\tfrac83$ at once.
Answer (A) $-3+\tfrac13=-\tfrac83$.

E6. JEE Advanced Numerical answer. The points $A(1,1)$, $B(5,3)$ and $C(k,7)$ form a right-angled triangle. Find the sum of all real values of $k$.

Right angle at A $m_{AB}=\tfrac12$, so $m_{AC}=\tfrac{6}{k-1}=-2$: $k-1=-3$, $k=-2$.
Right angle at B $m_{BC}=\tfrac{4}{k-5}=-2$: $k-5=-2$, $k=3$.
Right angle at C $\dfrac{6}{k-1}\cdot\dfrac{4}{k-5}=-1$ gives $(k-1)(k-5)=-24$, i.e. $k^2-6k+29=0$, whose discriminant $36-116\lt0$: no real $k$. (The vertical cases $k=1$ and $k=5$ give no right angle, since $AB$ is not horizontal.)
Answer $k\in\{-2,\,3\}$; the sum is $\mathbf{1}$.
๐Ÿ”ฌ Researchโ€“Teaching NexusMini project ยท about 60 minutes

Research question. How steep are the ramps, staircases and slopes around your school โ€” and do the ramps meet the "no steeper than $1$ in $12$" accessibility guideline?

Why it matters. Slope is a safety number. A ramp that is too steep is dangerous for a wheelchair user; staircases, roads, railway lines and drainage pipes all have slope limits written into building codes, and a drain with too little slope does not empty. Engineers check these numbers with exactly the rise-over-run calculation of this section.

What to do.

  1. Choose at least five sloping things: a ramp, a staircase (use one step: riser height and tread depth), a sloping road or path, a slide, a roof edge you can measure safely.
  2. For each, measure a rise and the matching horizontal run with a tape measure (for a long ramp, hold a straight plank level with a spirit level or a phone level app and measure the gap to the ramp). Compute the slope and $\theta=\tan^{-1}m$.
  3. Where possible, check the angle with a phone inclinometer app and compare with your calculation.
  4. Draw a scale diagram of the steepest and the gentlest item on squared paper, and mark the rise-and-run triangle.
  5. For every ramp, decide whether it meets the $1$ in $12$ guideline; if not, compute how long a compliant ramp of the same rise would need to be.

Tools: Tape measure, a spirit level or phone level app, a calculator or spreadsheet, squared paper

Hand in: A table (item, rise, run, slope, angle, app angle), the two scale drawings, and a short paragraph recommending any ramp improvements.

Going further. Find out what slope limits are used for highways in hilly areas and for main-line railways, and express each as a slope, an angle and a "1 in $n$" figure. Why do engineers prefer "1 in $n$" or a percentage to an angle?

Instructor note: typical school stairs have risers of about $15\ \text{cm}$ and treads of about $30\ \text{cm}$ โ€” slope $0.5$, angle $\approx26.6^\circ$, far steeper than any ramp. A $1$ in $12$ ramp has $\theta\approx4.76^\circ$; a ramp of rise $0.45\ \text{m}$ needs at least $5.4\ \text{m}$ of run. Phone-app angles usually agree with the tape-measure value to within about $0.5^\circ$; larger differences point to an uneven surface or a sloping "horizontal" reference.

๐Ÿงช Self-check quiz

0/5 correct

Q1. The slope of the line through $(2,-1)$ and $(-2,7)$ is

Q2. The inclination of a line with slope $-1$ is

Q3. Lines with slopes $\tfrac23$ and $k$ are perpendicular. Then $k$ is

Q4. The acute angle between the lines with slopes $\sqrt3$ and $\tfrac1{\sqrt3}$ is

Q5. Which line has no slope?

โ† Chapter 9 โ€” Straight Lines9.2 Various Forms of the Equation of a Line โ†’