📌 Snapshot
- Slope, angle between lines, all the equation forms and distance formulas, explained step by step for CUET.
- Why slope is tan of the inclination, and how parallel and perpendicular tests follow from it.
- Five standard forms of a line plus the general form, with a table that tells you which one to use.
- Perpendicular distance from a point and between parallel lines, with the sign and scaling traps.
📖 Detailed Notes
Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XI, Chapter 9.
What does "10% gradient" on a road sign mean?
It means the road rises 10 metres for every 100 metres you travel horizontally. That ratio, rise over run, is exactly what mathematicians call the slope of a line. A line has the same steepness everywhere, so one number captures its direction. In this chapter you will use slope to decide whether lines are parallel or perpendicular, to find the angle between lines, and to write the equation of a line from whatever information you are given.
Tools you already have
Straight lines are coordinate geometry, so the formulas from earlier classes come back as building blocks.
| Result | Formula for points $P(x_1,y_1)$, $Q(x_2,y_2)$ |
|---|---|
| Distance | $PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ |
| Midpoint | $\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$ |
| Point dividing $PQ$ internally in ratio $m:n$ | $\left(\dfrac{mx_2+nx_1}{m+n},\ \dfrac{my_2+ny_1}{m+n}\right)$ |
| Area of triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ | $\dfrac12\left\lvert x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)\right\rvert$ |
| Three points are collinear | Area of the triangle they form is 0 |
Slope: how steep is the line?
The inclination of a line is the angle $\theta$ it makes with the positive x-axis, measured anticlockwise, with $0^\circ \le \theta \le 180^\circ$. Its slope is
$$m = \tan\theta \qquad (\theta \ne 90^\circ)$$
| Inclination $\theta$ | Slope $m$ | Looks like |
|---|---|---|
| $0^\circ$ | $0$ | Horizontal line |
| $30^\circ$ | $\dfrac{1}{\sqrt3}$ | Gentle rise |
| $45^\circ$ | $1$ | Diagonal |
| $60^\circ$ | $\sqrt3$ | Steep rise |
| $90^\circ$ | not defined | Vertical line |
| $120^\circ$ | $-\sqrt3$ | Steep fall |
| $135^\circ$ | $-1$ | Diagonal falling |
| $150^\circ$ | $-\dfrac{1}{\sqrt3}$ | Gentle fall |
Positive slope means the line rises to the right (acute inclination). Negative slope means it falls to the right (obtuse inclination). A horizontal line has slope 0, while a vertical line has no slope at all, because $\tan 90^\circ$ does not exist.
Slope from two points. Moving from $P(x_1,y_1)$ to $Q(x_2,y_2)$, the line rises by $y_2-y_1$ over a run of $x_2-x_1$. So
$$m = \dfrac{y_2 - y_1}{x_2 - x_1}, \qquad x_1 \ne x_2$$
You may subtract in either order, as long as the numerator and denominator use the same order.
Worked example: slope and direction. Find the slope of the line through $(3,-2)$ and $(-1,4)$, and say whether its inclination is acute or obtuse.
| Step | What we do | Result |
|---|---|---|
| 1 | Subtract $y$-values, then $x$-values in the same order | $\dfrac{4-(-2)}{-1-3}$ |
| 2 | Simplify | $\dfrac{6}{-4} = -\dfrac32$ |
| 3 | Sign of the slope decides the angle | Negative slope, so obtuse inclination |
Common wrong step: taking the numerator as $4-(-2) = 6$ but the denominator as $3-(-1) = 4$. Mixing the two orders flips the sign and gives $+\tfrac32$.
Parallel, perpendicular and the angle between lines
Parallel lines have equal inclination, hence equal slopes: $m_1 = m_2$.
Perpendicular lines. Suppose one line has inclination $\alpha$ and the other $\alpha + 90^\circ$. Then $m_2 = \tan(\alpha+90^\circ) = -\cot\alpha = -\dfrac{1}{m_1}$, so
$$m_1 m_2 = -1$$
The slopes are negative reciprocals. This test is for non-vertical lines. A horizontal and a vertical line are perpendicular even though slope-product reasoning cannot be applied to the vertical one.
Worked example: a perpendicularity condition. The line through $(-2,6)$ and $(4,8)$ is perpendicular to the line through $(8,12)$ and $(x,24)$. Find $x$.
| Step | What we do | Result |
|---|---|---|
| 1 | Slope of the first line | $m_1 = \dfrac{8-6}{4-(-2)} = \dfrac{2}{6} = \dfrac13$ |
| 2 | Slope of the second line | $m_2 = \dfrac{24-12}{x-8} = \dfrac{12}{x-8}$ |
| 3 | Perpendicular means $m_1m_2=-1$ | $\dfrac13\cdot\dfrac{12}{x-8} = -1$ |
| 4 | Simplify | $\dfrac{4}{x-8} = -1 \Rightarrow x-8 = -4$ |
| 5 | Solve | $x = 4$ |
Common wrong step: setting $m_1m_2 = 1$. That would be the condition for the slopes to be reciprocals, which does not give perpendicular lines.
The angle between two lines. Let a line with slope $m_1$ have inclination $\alpha$ and a second line with slope $m_2$ have inclination $\beta$. The angle from the first to the second is $\theta = \beta - \alpha$, and the tangent of a difference gives
$$\tan\theta = \left\lvert \dfrac{m_2 - m_1}{1 + m_1 m_2} \right\rvert, \qquad 1 + m_1m_2 \ne 0$$
The modulus gives the acute angle; the obtuse angle between the same two lines is $180^\circ - \theta$. If $1 + m_1m_2 = 0$ the lines are perpendicular and the angle is $90^\circ$.
Worked example: a line at a given angle. A line makes an angle $\dfrac{\pi}{4}$ with a line of slope $\dfrac12$. Find the possible slopes of the first line.
| Step | What we do | Result |
|---|---|---|
| 1 | Use the angle formula with $\tan\dfrac\pi4 = 1$ | $\left\lvert \dfrac{m - \frac12}{1 + \frac m2} \right\rvert = 1$ |
| 2 | Remove the modulus using both signs | $\dfrac{m-\frac12}{1+\frac m2} = 1$ or $-1$ |
| 3 | Case $+1$ | $m - \tfrac12 = 1 + \tfrac m2 \Rightarrow m = 3$ |
| 4 | Case $-1$ | $m - \tfrac12 = -1 - \tfrac m2 \Rightarrow m = -\tfrac13$ |
| 5 | State | $m = 3$ or $m = -\dfrac13$ |
Common wrong step: dropping the modulus and keeping only one sign. A given line makes the same angle with two different lines through a point, so there are two answers.
Collinear points. $A$, $B$, $C$ are collinear when the slope of $AB$ equals the slope of $BC$, because both segments then lie on one line through $B$.
Pause and predict: find $k$ so that $(1,2)$, $(3,k)$ and $(5,6)$ lie on one line.
Check your answer: $k = 4$. Equal slopes need $\dfrac{k-2}{2} = \dfrac{6-k}{2}$, so $k-2 = 6-k$ and $k=4$. This is also the midpoint's $y$-value, since $(3,4)$ is the midpoint of $(1,2)$ and $(5,6)$.
Writing the equation of a line
Every form below is a way of using the data you have. Pick the form that matches it.
| You are given | Use this form | Equation |
|---|---|---|
| A horizontal line at distance $a$ from the x-axis | Horizontal line | $y = a$ (or $y=-a$ below the axis) |
| A vertical line at distance $b$ from the y-axis | Vertical line | $x = b$ (or $x=-b$ left of the axis) |
| A point $(x_0,y_0)$ and the slope $m$ | Point–slope | $y - y_0 = m\,(x - x_0)$ |
| Two points $(x_1,y_1)$, $(x_2,y_2)$ | Two–point | $y - y_1 = \dfrac{y_2-y_1}{x_2-x_1}\,(x - x_1)$ |
| Slope $m$ and y-intercept $c$ | Slope–intercept | $y = mx + c$ |
| Slope $m$ and x-intercept $d$ | Slope–intercept | $y = m\,(x - d)$ |
| x-intercept $a$ and y-intercept $b$ | Intercept | $\dfrac{x}{a} + \dfrac{y}{b} = 1$ |
| Anything, in the end | General | $Ax + By + C = 0$ |
The intercepts $a$ and $b$ are signed: the line crosses the x-axis at $(a,0)$ and the y-axis at $(0,b)$. Intercepts of $-3$ and $2$ give $\dfrac{x}{-3} + \dfrac{y}{2} = 1$, which simplifies to $2x - 3y + 6 = 0$.
Worked example: through two points. Find the equation of the line through $(1,-1)$ and $(3,5)$, then its intercepts.
| Step | What we do | Result |
|---|---|---|
| 1 | Slope | $m = \dfrac{5-(-1)}{3-1} = 3$ |
| 2 | Use the point–slope form with $(1,-1)$ | $y - (-1) = 3(x-1)$ |
| 3 | Simplify | $y + 1 = 3x - 3 \Rightarrow 3x - y - 4 = 0$ |
| 4 | x-intercept (put $y=0$) | $x = \dfrac43$ |
| 5 | y-intercept (put $x=0$) | $y = -4$ |
Common wrong step: writing $y - 1 = 3(x-1)$ because of the minus signs. Subtracting the point's $y$-coordinate $-1$ gives $y - (-1) = y + 1$.
Reading the general form. From $Ax + By + C = 0$ you can read everything:
$$m = -\dfrac{A}{B}, \qquad \text{x-intercept} = -\dfrac{C}{A}, \qquad \text{y-intercept} = -\dfrac{C}{B}$$
The slope comes from rewriting as $y = -\dfrac{A}{B}x - \dfrac{C}{B}$, and the intercepts come from putting $y=0$ and $x=0$.
Worked example: converting a general equation. For $3x - 4y + 12 = 0$, find the slope, the intercepts and the intercept form.
| Step | What we do | Result |
|---|---|---|
| 1 | Here $A=3$, $B=-4$, $C=12$ | Read off the coefficients with their signs |
| 2 | Slope $-\dfrac AB$ | $-\dfrac{3}{-4} = \dfrac34$ |
| 3 | x-intercept $-\dfrac CA$ | $-\dfrac{12}{3} = -4$ |
| 4 | y-intercept $-\dfrac CB$ | $-\dfrac{12}{-4} = 3$ |
| 5 | Intercept form | $\dfrac{x}{-4} + \dfrac{y}{3} = 1$ |
Common wrong step: calling the y-intercept $-C = -12$. The y-intercept is $-\dfrac CB$; $B$ is not 1 here.
Pause and predict: what is the slope of the line $x = 5$, and what is the equation of the line through $(2,3)$ perpendicular to it?
Check your answer: the line $x=5$ is vertical, so its slope is not defined. A line perpendicular to a vertical line is horizontal, so the required line is $y = 3$. The $m_1m_2=-1$ test cannot be used here.
Distance from a point to a line
The perpendicular distance from $P(x_1, y_1)$ to the line $Ax + By + C = 0$ is
$$d = \dfrac{\left\lvert Ax_1 + By_1 + C \right\rvert}{\sqrt{A^2 + B^2}}$$
Putting the point into the left side of the equation measures how far it is from satisfying the equation, and dividing by $\sqrt{A^2+B^2}$ converts that into a true length. The modulus is needed because distance is never negative. For example, the distance of $(3,-5)$ from $3x - 4y - 26 = 0$ is $\dfrac{\lvert 9 + 20 - 26 \rvert}{\sqrt{9+16}} = \dfrac35$.
Distance between parallel lines. Take two lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$. Pick any point $(x_1,y_1)$ on the first, so $Ax_1+By_1 = -C_1$. Its distance to the second line is $\dfrac{\lvert -C_1 + C_2\rvert}{\sqrt{A^2+B^2}}$, hence
$$d = \dfrac{\lvert C_1 - C_2 \rvert}{\sqrt{A^2 + B^2}}$$
In slope form, for $y = mx + c_1$ and $y = mx + c_2$, the distance is $\dfrac{\lvert c_1 - c_2\rvert}{\sqrt{1+m^2}}$.
Worked example: unequal coefficients. Find the distance between $3x - 4y + 7 = 0$ and $6x - 8y + 5 = 0$.
| Step | What we do | Result |
|---|---|---|
| 1 | Check they are parallel | Both have slope $\dfrac34$ |
| 2 | Make the $x$ and $y$ coefficients match: halve the second equation | $3x - 4y + \dfrac52 = 0$ |
| 3 | Apply the formula with $C_1 = 7$, $C_2 = \dfrac52$ | $d = \dfrac{\lvert 7 - \frac52\rvert}{\sqrt{9+16}}$ |
| 4 | Simplify | $\dfrac{9/2}{5} = \dfrac{9}{10}$ |
Common wrong step: using $\lvert 7-5\rvert / \sqrt{36+64} = \tfrac15$ straight from the given equations. The formula needs identical coefficients of $x$ and $y$ in both equations.
Beyond the textbook. For the exam: slope is $\tan\theta$, and a road gradient is just slope written as a percentage. In reality: a 10% gradient means $m = 0.1$, so the road makes an angle of $\tan^{-1}0.1 \approx 5.7^\circ$ with the horizontal, not $10^\circ$. Roof pitch and wheelchair ramps are specified the same way, as rise over run.
Beyond the textbook. For the exam: the distance formula always uses the modulus. In reality: the expression $\dfrac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}$ without the modulus is a signed distance. Two points lie on the same side of the line when it has the same sign for both and on opposite sides when the signs differ. Computer graphics and linear programming use this test to decide on which side of a boundary a point falls.
Beyond the textbook. For the exam: $y = mx + c$ is a form to convert to and from. In reality: in statistics, a line of best fit through a scatter plot is written in this form, and the slope $m$ is the average change in $y$ for each unit increase in $x$. A taxi fare of ₹50 plus ₹12 per kilometre is the line $y = 12x + 50$: the slope is the rate and the intercept is the fixed charge.
Exam traps
| The trap | Why it's tempting | The truth |
|---|---|---|
| Distance can be negative | The numerator can be negative | $d = \dfrac{\lvert Ax_1+By_1+C\rvert}{\sqrt{A^2+B^2}}$ is always non-negative |
| Slope of $Ax+By+C=0$ is $\dfrac AB$ or $-\dfrac BA$ | The coefficients are right there | It is $-\dfrac AB$ |
| Horizontal line has undefined slope; vertical has 0 | Mixing up which is which | Horizontal: slope 0. Vertical: slope not defined |
| Angle formula without the modulus | It looks complete | Without $\lvert\ \rvert$ you may get the obtuse angle; with it you get the acute angle |
| $m_1m_2=-1$ proves $x=4$ and $y=7$ are perpendicular | The lines look perpendicular | One slope is not defined, so the product test does not apply; they are perpendicular by inspection |
| Parallel distance from unequal coefficients | Formulas are easy to apply blindly | Rescale so $A$ and $B$ match first |
| In $\dfrac xa+\dfrac yb=1$, $a$ and $b$ are always positive | Intercepts are lengths on the diagram | They are signed coordinates: $a=-3$ is allowed |
| Inclination of 200° | Falling lines look like larger angles | Inclination lies between $0^\circ$ and $180^\circ$ |
Formula sheet
| Result | Formula | When to use |
|---|---|---|
| Slope | $m = \tan\theta = \dfrac{y_2-y_1}{x_2-x_1}$ | Direction of a line |
| Parallel / perpendicular | $m_1=m_2$ / $m_1m_2=-1$ | Relating two lines |
| Angle between lines | $\tan\theta = \left\lvert\dfrac{m_2-m_1}{1+m_1m_2}\right\rvert$ | Acute angle between two lines |
| Collinearity | Slope $AB$ = slope $BC$, or area $=0$ | Points on one line |
| Horizontal / vertical | $y=a$ / $x=b$ | Lines parallel to the axes |
| Point–slope | $y-y_0=m(x-x_0)$ | Point and slope given |
| Two–point | $y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$ | Two points given |
| Slope–intercept | $y=mx+c$ | Slope and y-intercept given |
| Intercept form | $\dfrac xa+\dfrac yb=1$ | Both intercepts given |
| General form | $Ax+By+C=0$, $m=-\dfrac AB$ | Final answers, reading slope and intercepts |
| Intercepts of general form | $-\dfrac CA$ and $-\dfrac CB$ | Axis crossings |
| Point to line distance | $\dfrac{\lvert Ax_1+By_1+C\rvert}{\sqrt{A^2+B^2}}$ | Shortest distance to a line |
| Distance from the origin | $\dfrac{\lvert C\rvert}{\sqrt{A^2+B^2}}$ | Special case $x_1=y_1=0$ |
| Parallel lines | $\dfrac{\lvert C_1-C_2\rvert}{\sqrt{A^2+B^2}}$ | Equal $A$, $B$ in both |
60-second revision
| Topic | Lock this in |
|---|---|
| Slope | $\tan\theta$; horizontal 0, vertical not defined |
| Two points | Rise over run; same order top and bottom |
| Parallel / perpendicular | Equal slopes / product $-1$ |
| Angle | Modulus gives acute angle; $1+m_1m_2=0$ gives $90^\circ$ |
| Forms | Choose by data: point and slope, two points, intercepts |
| General form | Slope $-\dfrac AB$, intercepts $-\dfrac CA$ and $-\dfrac CB$ |
| Intercepts | Signed; $x$-intercept sets $y=0$, $y$-intercept sets $x=0$ |
| Point distance | Numerator with modulus, denominator $\sqrt{A^2+B^2}$ |
| Parallel distance | Same $A$, $B$ first, then $\lvert C_1-C_2\rvert$ over the root |
| Collinear | Equal slopes through a common point |
Where this chapter connects
- Previous: Sequences and Series used constant ratios; slope is the geometric version of a constant rate of change.
- Next: Conic Sections moves from lines to curves, and it uses distances and equations the same way.
- Same strand: Linear Inequalities shades the regions on each side of a line such as $Ax+By+C=0$.
- Class XII link: Three Dimensional Geometry extends lines to space, and Linear Programming uses lines as constraint boundaries.
🎯 Practice MCQs
First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed
Q1. The slope of the line through the points $(3,-2)$ and $(-1,4)$ is
▸ Show answer & explanation
Answer: B
$m = \dfrac{4-(-2)}{-1-3} = \dfrac{6}{-4} = -\dfrac32$. Option (A) comes from mixing the subtraction orders and option (D) from inverting the ratio.
Q2. The slope of a line whose inclination is $120^\circ$ is
▸ Show answer & explanation
Answer: B
$m = \tan120^\circ = \tan(180^\circ - 60^\circ) = -\tan60^\circ = -\sqrt3$. An obtuse inclination always gives a negative slope.
Q3. The line through $(-2,6)$ and $(4,8)$ is perpendicular to the line through $(8,12)$ and $(x,24)$. Then $x$ equals
▸ Show answer & explanation
Answer: B
The first slope is $\dfrac{2}{6}=\dfrac13$, the second is $\dfrac{12}{x-8}$. Setting the product equal to $-1$ gives $x-8=-4$, so $x=4$. Check: the second slope is $\dfrac{12}{-4}=-3$ and $\dfrac13\times(-3)=-1$.
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Q4. The acute angle between two lines with slopes 2 and $\dfrac13$ is
▸ Show answer & explanation
Answer: B
$\tan\theta = \left\lvert \dfrac{\frac13 - 2}{1 + 2\cdot\frac13} \right\rvert = \left\lvert \dfrac{-\frac53}{\frac53} \right\rvert = 1$, so $\theta = 45^\circ$.
Q5. The equation of the line through $(-2,3)$ with slope $-4$ is
▸ Show answer & explanation
Answer: A
$y - 3 = -4(x+2)$ gives $y - 3 = -4x - 8$, so $4x + y + 5 = 0$. Check with $(-2,3)$: $-8 + 3 + 5 = 0$.
Q6. The equation of the line with x-intercept 4 and y-intercept $-3$ is
▸ Show answer & explanation
Answer: A
$\dfrac{x}{4} + \dfrac{y}{-3} = 1$. Multiplying by 12 gives $3x - 4y = 12$, that is $3x - 4y - 12 = 0$. Check: $(4,0)$ gives $12-12=0$ and $(0,-3)$ gives $12-12=0$.
Q7. The slope of the line $5x - 2y + 7 = 0$ is
▸ Show answer & explanation
Answer: B
The slope is $-\dfrac AB = -\dfrac{5}{-2} = \dfrac52$. Rewriting gives $y = \dfrac52x + \dfrac72$, which confirms it.
Q8. For the line $3x - 4y + 12 = 0$, consider the statements. I. The x-intercept is $-4$. II. The y-intercept is 3. III. The slope is $-\dfrac34$. Which statements are correct?
▸ Show answer & explanation
Answer: A
Putting $y=0$ gives $x=-4$, so I is true. Putting $x=0$ gives $y=3$, so II is true. The slope is $-\dfrac{3}{-4} = \dfrac34$, so III is false.
Q9. The distance of the origin from the line $5x - 12y + 26 = 0$ is
▸ Show answer & explanation
Answer: A
$d = \dfrac{\lvert 26\rvert}{\sqrt{25+144}} = \dfrac{26}{13} = 2$. Option (B) forgets to divide by the square root.
Q10. The distance between the parallel lines $3x - 4y + 7 = 0$ and $6x - 8y + 5 = 0$ is
▸ Show answer & explanation
Answer: C
Rewrite the second line as $3x - 4y + \dfrac52 = 0$ so that the coefficients match. Then $d = \dfrac{\lvert 7 - \frac52\rvert}{5} = \dfrac{9/2}{5} = \dfrac{9}{10}$. Option (B) uses the unequal coefficients directly.
Q11. Assertion (A): The lines $x = 4$ and $y = 7$ are perpendicular. Reason (R): The product of the slopes of the lines $x = 4$ and $y = 7$ is $-1$.
▸ Show answer & explanation
Answer: C
The line $x=4$ is vertical and $y=7$ is horizontal, so they are perpendicular and A is true. But the vertical line has no slope, so there is no product of slopes to equal $-1$, and R is false. The perpendicularity here is seen directly from the picture.
Q12. Match List I with List II. List I: (i) a horizontal line (ii) a vertical line (iii) a line with inclination $45^\circ$ (iv) two parallel lines. List II: (p) slope is 1 (q) slope is 0 (r) slope is not defined (s) slopes are equal
▸ Show answer & explanation
Answer: A
A horizontal line has slope $\tan0^\circ=0$, a vertical line has no slope since $\tan90^\circ$ does not exist, $\tan45^\circ=1$, and parallel lines have equal inclinations and so equal slopes.
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