Chapter 9 of 14

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Haryana (BSEH)Class 11 Mathematics★ Unit III · part of the 12-mark Coordinate Geometry block shared with Conic Sections — CBSE states no chapter-wise split

Straight Lines

How does GPS turn "the shortest distance from a point to a road" into a single formula? Straight Lines builds the full algebraic toolkit for a line — slope, five distinct equation forms, the angle between two lines, and the perpendicular-distance formula. One of the six forms coaching material lists, the Normal Form, is not part of the current book at all. CBSE Class 11 Mathematics Chapter 9.

⏱ 8 min readMedium difficulty✓ Free · NCERT-aligned

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Mind Map

The chapter at a glance

Every concept and exam-favourite branch — revisable in 5 minutes.

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Study Guide

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Key points, glossary and the questions most likely to show up.

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By the end of this chapter you'll be able to…

1Compute the slope of a line from two points or from its inclination, and determine whether two lines are parallel or perpendicular using their slopes
2Find the acute angle between two lines using the tan(theta) formula
3Write the equation of a line in point-slope, two-point, slope-intercept, or intercept form, choosing the right form for the information given
4Convert a line's equation between general form (Ax+By+C=0) and the other taught forms
5Compute the perpendicular distance from a point to a line, and the distance between two parallel lines
6Use collinearity tests (equal slopes, or zero triangle area) to verify that three points lie on a single line

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Why this chapter matters

This chapter builds the full algebraic toolkit for a line: slope, five distinct equation forms, the angle between two lines, and the perpendicular-distance formula. It is a direct prerequisite for Conic Sections and 3D Geometry. One of the six forms coaching material commonly lists, Normal Form, is not part of the current book at all.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 10 coordinate geometry: distance formula, section formula, midpoint formula, and area of a triangle — this chapter's own opening section recalls these without re-deriving them

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Trigonometric ratios, specifically tan, since slope is defined as the tangent of a line's inclination angle

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Comfort solving simultaneous linear equations, used throughout to find intersection points and unknowns in line equations

Straight Lines

1. Check this before you revise anything

Normal Form is not part of the current book. Coaching material commonly lists six equation forms for a line, including Normal Form ( $x cos ω + y sin ω = p$ ). CBSE's own formative-only block confirms this explicitly, listing "Normal form" under Straight Lines as formative, not summative — and the current 2026-27 book has no such section: no formula, no worked example, nowhere. Only five forms are actually taught: horizontal/vertical, point-slope, two-point, slope-intercept, and intercept form.

"General equation of a line" gets one defining sentence, not a full section — but it's still genuinely testable. The formative-only block also lists "General equation of a line" alongside Normal Form. True to that, the book never derives it as its own topic — it's introduced in a single sentence ("Any equation of the form $A x + B y + C = 0$ ... is called the general equation of a line") right before the distance formula, which needs that notation.

But Exercise 9.3's own Q1 and Q2 ask you to reduce a general-form equation back into slope-intercept or intercept form — so converting between general form and the five taught forms is real, examinable content from this book's own exercises, even though "general form" isn't taught as a destination in its own right.

2. What this chapter covers

Textbook section	Topic
9.1	Recall: distance formula, section formula, midpoint, area of a triangle (from earlier classes)
9.2	Slope of a line; parallel/perpendicular conditions; angle between two lines
9.3	Equation of a line: horizontal/vertical, point-slope, two-point, slope-intercept, intercept form
9.4	Distance of a point from a line; distance between two parallel lines

3. Recall from earlier classes

Four formulas carry forward from Class 10 and get reused throughout this chapter without re-derivation: the distance formula $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ ; the section formula for a point dividing $P (x_{1}, y_{1}) Q (x_{2}, y_{2})$ internally in ratio $m : n$ , at $(\frac{m x _{2} + n x _{1}}{m + n}, \frac{m y _{2} + n y _{1}}{m + n})$ ; the midpoint formula (the $m = n$ case of the above); and the area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ :

$Area = \frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$

If this area comes out to zero, the three points are collinear — they lie on a single straight line. This is the book's own bridge into the chapter: a straight line is what's left when a "triangle" collapses flat.

4. Slope of a line

A line makes two supplementary angles with the x-axis; the one measured anticlockwise from the positive x-axis, satisfying $0° \leq θ \leq 180°$ , is its inclination. The slope is $m = tan θ$ ( $θ \neq = 90°$ ) — undefined exactly when the line is vertical. A horizontal line has slope $0$ .

Given two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ on a non-vertical line, splitting into the acute- and obtuse-inclination cases and working through the right triangle each time gives, in both cases, the same result:

$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}, x_{1} \neq = x_{2}$

Two non-vertical lines are parallel if and only if $m_{1} = m_{2}$ ; they are perpendicular if and only if $m_{1} m_{2} = - 1$ — both proved directly from how inclination and the tangent function behave (parallel lines share an inclination; perpendicular lines' inclinations differ by exactly $90°$ , and $tan (α + 90°) = - cot α$ ).

Worked, mirroring the textbook's own Example 3. The line through $(- 2, 6)$ and $(4, 8)$ is perpendicular to the line through $(8, 12)$ and $(x, 24)$ . Find $x$ . First slope: $m_{1} = \frac{8 - 6}{4 - ( - 2 )} = \frac{2}{6} = \frac{1}{3}$ . Second slope: $m_{2} = \frac{24 - 12}{x - 8} = \frac{12}{x - 8}$ . Perpendicularity requires $m_{1} m_{2} = - 1$ : $\frac{1}{3} \cdot \frac{12}{x - 8} = - 1 \Rightarrow \frac{4}{x - 8} = - 1 \Rightarrow x = 4$ .

5. Angle between two lines

For two non-vertical lines with slopes $m_{1}, m_{2}$ meeting at a point, the two adjacent angles between them, $θ$ and $φ = 180° - θ$ , satisfy $tan θ = \frac{m _{2} - m _{1}}{1 + m _{1} m _{2}}$ and $tan φ = - tan θ$ — one of these is positive (giving the acute angle), the other negative (the obtuse angle). Taking the acute angle explicitly:

$tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}, 1 + m_{1} m_{2} \neq = 0$

Worked, mirroring the textbook's own Example 2 — the formula run in reverse. If the angle between two lines is $\frac{π}{4}$ and one line's slope is $\frac{1}{2}$ , find the other slope $m$ . Substituting into the formula: $tan \frac{π}{4} = 1 = \frac{1/2 - m}{1 + m /2}$ , which splits into two linear equations depending on the sign, giving $m = 3$ or $m = - \frac{1}{3}$ — two valid answers, since two different lines can each make a $45°$ angle with the same given line, one on either side of it.

6. Various forms of the equation of a line

Horizontal/vertical lines. A horizontal line at distance $a$ from the x-axis is $y = a$ or $y = - a$ ; a vertical line at distance $b$ from the y-axis is $x = b$ or $x = - b$ .

Point-slope form, through a fixed point $(x_{0}, y_{0})$ with slope $m$ : derived directly from the two-point slope formula applied to $(x_{0}, y_{0})$ and a general point $(x, y)$ on the line.

$y - y_{0} = m (x - x_{0})$

Two-point form, through two fixed points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ : since a general point $(x, y)$ on the line must give the same slope whichever pair it's computed from,

$y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1})$

Slope-intercept form, given slope $m$ and y-intercept $c$ (the line meets the y-axis at $(0, c)$ , so this is just point-slope form applied there): $y = m x + c$ . The x-intercept version, given slope $m$ and x-intercept $d$ : $y = m (x - d)$ .

Intercept form, given x-intercept $a$ and y-intercept $b$ (the line meets the axes at $(a, 0)$ and $(0, b)$ , so this is two-point form applied to those points):

$\frac{x}{a} + \frac{y}{b} = 1$

Worked, mirroring the textbook's own Example 7 — reading the right form off what's given. Find the equations of the lines with $tan θ = \frac{1}{2}$ and (i) y-intercept $- \frac{3}{2}$ , (ii) x-intercept $4$ . (i) Slope-intercept form: $y = \frac{1}{2} x - \frac{3}{2}$ , i.e. $x - 2 y - 3 = 0$ . (ii) x-intercept form: $y = \frac{1}{2} (x - 4)$ , i.e. $x - 2 y - 4 = 0$ .

7. Distance of a point from a line

For a line $A x + B y + C = 0$ and a point $(x_{1}, y_{1})$ , the perpendicular distance is derived by computing the area of the triangle formed by the point and the line's two axis-intercepts two different ways — once with the base-times-height formula using the unknown perpendicular distance, once with the coordinate area formula — then equating them:

$d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$

Distance between two parallel lines $y = m x + c_{1}$ and $y = m x + c_{2}$ (same slope, since parallel) is found by taking the perpendicular distance from a convenient point on one line — where it crosses the x-axis — to the other:

$d = \frac{∣ c _{1} - c _{2} ∣}{1 + m ^{2}}, or in general form: d = \frac{∣ C _{1} - C _{2} ∣}{A ^{2} + B ^{2}}$

Worked, mirroring the textbook's own Example 9. Find the distance of $(3, - 5)$ from $3 x - 4 y - 26 = 0$ . Here $A = 3, B = - 4, C = - 26$ : $d = \frac{∣3 ( 3 ) + ( - 4 ) ( - 5 ) - 26∣}{3 ^{2} + ( - 4 ) ^{2}} = \frac{∣9 + 20 - 26∣}{5} = \frac{3}{5}$ .

Summary

Slope $m = tan θ = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ ; horizontal lines have $m = 0$ , vertical lines have undefined slope.
Parallel lines have $m_{1} = m_{2}$ ; perpendicular lines have $m_{1} m_{2} = - 1$ ; three points are collinear exactly when the triangle they form has zero area.
Acute angle between two lines: $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$ .
Five forms are taught: horizontal/vertical ( $y = \pm a$ , $x = \pm b$ ), point-slope, two-point, slope-intercept ( $y = m x + c$ ), and intercept form ( $x / a + y / b = 1$ ) — Normal Form is formative-only and absent from the book.
General form $A x + B y + C = 0$ gets one defining sentence, not a full section, but converting into/out of it is directly tested in Exercise 9.3.
Distance from $(x_{1}, y_{1})$ to $A x + B y + C = 0$ : $d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$ ; distance between parallel lines $A x + B y + C_{1} = 0$ and $A x + B y + C_{2} = 0$ : $d = \frac{∣ C _{1} - C _{2} ∣}{A ^{2} + B ^{2}}$ .

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Slope from two points

m = (y2-y1)/(x2-x1), x1 not equal to x2

m = tan(theta), where theta is the line's inclination; undefined for a vertical line

Parallel and perpendicular conditions

Parallel: m1=m2. Perpendicular: m1.m2=-1

Both proved directly from how inclination behaves under the tangent function

Angle between two lines

tan(theta) = |(m1-m2)/(1+m1.m2)|, 1+m1.m2 not equal to 0

theta is the acute angle; the obtuse angle is 180-theta

Point-slope form

y - y0 = m(x - x0)

Use when one point (x0,y0) and the slope m are known

Two-point form

y - y1 = [(y2-y1)/(x2-x1)](x - x1)

Use when two points on the line are known; reduces to point-slope form once the slope is computed

Slope-intercept form

y = mx + c (y-intercept c); y = m(x-d) (x-intercept d)

Directly readable: m is the slope, c is where the line crosses the y-axis

Intercept form

x/a + y/b = 1

a = x-intercept, b = y-intercept; undefined for lines through the origin, which have no intercepts to speak of

Distance from a point to a line

d = |Ax1 + By1 + C| / sqrt(A^2+B^2)

For line Ax+By+C=0 and point (x1,y1); the absolute value ensures a non-negative distance

Distance between two parallel lines

d = |C1-C2| / sqrt(A^2+B^2)

Both lines must be written with identical A and B coefficients before subtracting C1-C2

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Expecting Normal Form (x.cos(omega)+y.sin(omega)=p) as one of the forms this chapter teaches

✓ Normal Form is formative-only and doesn't appear anywhere in the current book. Only five forms are taught: horizontal/vertical, point-slope, two-point, slope-intercept, and intercept form.

WATCH OUT

✗ Confusing the parallel condition (m1=m2) with the perpendicular condition (m1.m2=-1)

✓ Parallel lines have identical slopes. Perpendicular lines have slopes that are negative reciprocals of each other, so their product is always -1.

WATCH OUT

✗ Forgetting the absolute value in the distance-from-a-point-to-a-line formula

✓ Ax1+By1+C can come out negative depending on which side of the line the point sits on. Distance is always non-negative, so the numerator must be wrapped in absolute value bars.

WATCH OUT

✗ Using mismatched A and B coefficients when applying the distance-between-parallel-lines formula

✓ Both lines must be written with the exact same A and B (multiply one equation through by a constant if needed) before the formula d=|C1-C2|/sqrt(A^2+B^2) applies.

WATCH OUT

✗ Reporting only one slope when 'find the other line given the angle between them' problems are asked

✓ The absolute value in the angle formula splits into two cases, usually giving two valid slopes for the second line — one on each side of the first line. State both unless the question restricts to one.

WATCH OUT

✗ Plugging a vertical line's undefined slope directly into a slope-based formula

✓ Vertical lines (x=k) have no defined slope. Handle them separately as x=k rather than trying to force a numeric m into the point-slope or angle formulas.

WATCH OUT

✗ Treating 'general equation of a line' as a topic with its own derivation to memorise

✓ The book introduces Ax+By+C=0 in a single sentence, purely as notation for the distance formula. What is genuinely tested is reducing a general-form equation back into slope-intercept or intercept form, per Exercise 9.3's own questions.

WATCH OUT

✗ Computing the area-of-a-triangle collinearity test with the wrong sign convention and concluding three genuinely collinear points aren't

✓ The formula already has an absolute value around the whole expression — a zero result (with or without the bars) always means collinear; a small nonzero rounding-type value from correct exact arithmetic should be re-checked, not dismissed.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 9.1

Exercise 9.1

Slope from two points, inclination, collinearity, area of a triangle, and the recall formulas from earlier classes

Questions

Ex 9.2

Exercise 9.2

Equation of a line in its various forms — point-slope, two-point, slope-intercept, intercept form, and applied word problems

Questions

Ex 9.3

Exercise 9.3

Reducing to/from general form, distance from a point to a line, angle between lines, and distance between parallel lines

Questions

Miscellaneous Ex

Miscellaneous Exercise

Mixed problems on concurrency, reflection, perpendicular feet, family of lines, and applied distance problems

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Straight Lines?

10 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

10 questions~7 min worth ~12 marks in Haryana (BSEH) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Slope m=tan(theta)=(y2-y1)/(x2-x1); zero for horizontal lines, undefined for vertical lines
•Parallel: m1=m2. Perpendicular: m1.m2=-1. Collinear: equal slopes across pairs, or triangle area = 0
•Acute angle between two lines: tan(theta)=|(m1-m2)/(1+m1.m2)|
•Five forms taught: horizontal/vertical (y=+-a, x=+-b), point-slope, two-point, slope-intercept (y=mx+c), intercept (x/a+y/b=1)
•Normal Form is formative-only and not in the current book; General form Ax+By+C=0 gets one sentence, mainly as distance-formula notation
•Distance from (x1,y1) to Ax+By+C=0: d=|Ax1+By1+C|/sqrt(A^2+B^2)
•Distance between parallel lines Ax+By+C1=0 and Ax+By+C2=0: d=|C1-C2|/sqrt(A^2+B^2), same A and B required
•Family of lines through the intersection of L1=0 and L2=0: L1+lambda.L2=0 for varying lambda, useful for concurrency and 'passes through intersection and satisfies X' problems

Haryana (BSEH) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Part of Unit III's 12-mark Coordinate Geometry block (shared with Conic Sections; no chapter-wise split, per CBSE)

Question type	Marks each	Typical count	What it tests
Slope, Parallel/Perpendicular Conditions, and Collinearity	2-3	1	Computing slope from two points or inclination, checking parallel/perpendicular, verifying collinearity
Equation of a Line in Point-Slope, Two-Point, Slope-Intercept, and Intercept Form	3-4	1-2	Point-slope, two-point, slope-intercept, and intercept form; reducing general form into a named form
Angle Between Two Lines	3-4	0-1	Finding the acute angle between two lines, or the unknown slope given the angle and one line
Distance From a Point to a Line and Between Parallel Lines	4-6	1	Perpendicular distance formula, distance between parallel lines, applied word problems

Prep strategy

State the slope explicitly (or 'undefined' for a vertical line) before choosing which equation form to use
For every distance question, write the line in general form Ax+By+C=0 first, even if it wasn't given that way
Keep the five taught forms and their trigger conditions (what's given: one point+slope? two points? both intercepts?) in a quick mental checklist

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

GPS and shortest-distance-to-a-road calculations

Roads and paths modelled as lines in a coordinate system use exactly the distance-from-a-point-to-a-line formula to compute the shortest route from a location to the nearest road.

Linear regression in data science

The best-fit line through scattered data, written in slope-intercept form y=mx+c, is found by minimising the sum of squared perpendicular-style distances from each data point to the line.

Civil engineering and road gradients

The slope formula directly measures a road's gradient (rise over run); a slope of 0.08 means an 8 cm rise per metre of horizontal run, a figure that matters for drainage and vehicle safety design.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Identify what's given (points? slope? intercepts?) before picking a form — this alone earns method marks even before computing anything

For every distance question, convert to general form Ax+By+C=0 first, since both distance formulas are stated in terms of A, B, C

For parallel/perpendicular checks, compute both slopes first: m1=m2 means parallel, m1.m2=-1 means perpendicular — don't guess from a rough sketch alone

For 'two lines meeting at an angle' problems, always check both sign cases of the absolute value before finalising a single answer

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Family of lines: every line through the intersection of L1=0 and L2=0 can be written as L1+lambda.L2=0 for varying lambda — used heavily in JEE to find a line satisfying one extra condition without solving for the intersection point first

STRETCH

Reflection of a line in another line, and the angle bisectors between two given lines (using the equal-distance-from-both-lines condition), extend the single-point reflection technique to whole lines

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainFamily of lines through an intersection pointParameter-matching technique

Find the equation of the line through the intersection of $2 x + 3 y - 5 = 0$ and $3 x - 2 y - 1 = 0$ , and parallel to the line $4 x + y - 2 = 0$ .

Stuck? Show the approach

Write the family of lines through the intersection as $L_{1} + λ L_{2} = 0$ , express its slope in terms of $λ$ , and match it to the given line's slope.

Show the full solution

Family: $(2 x + 3 y - 5) + λ (3 x - 2 y - 1) = 0$ , i.e. $(2 + 3 λ) x + (3 - 2 λ) y - (5 + λ) = 0$ , with slope $- \frac{2 + 3 λ}{3 - 2 λ}$ . Setting this equal to $- 4$ (the slope of $4 x + y - 2 = 0$ ): $2 + 3 λ = 4 (3 - 2 λ) \Rightarrow 11 λ = 10 \Rightarrow λ = 10/11$ . Substituting back and simplifying gives $4 x + y - 5 = 0$ .

Answer: 4x + y - 5 = 0

The trap

Forgetting to simplify the messy fractional coefficients at the end (by clearing denominators and dividing by the common factor) leaves an unrecognisably scaled but technically-correct equation — always reduce to lowest integer coefficients.

JEE MainReflection of a point in a linePerpendicular-foot technique

Find the image (reflection) of the point $(1, 2)$ in the line $x + y = 1$ .

Stuck? Show the approach

Use the standard reflection formula: the image of $(x_{1}, y_{1})$ in $A x + B y + C = 0$ is obtained by moving twice the signed perpendicular distance along the line's normal direction.

Show the full solution

With $A = 1, B = 1, C = - 1$ : $A x_{1} + B y_{1} + C = 1 + 2 - 1 = 2$ , and $A^{2} + B^{2} = 2$ . Image $x = 1 - \frac{2 ( 1 ) ( 2 )}{2} = 1 - 2 = - 1$ . Image $y = 2 - \frac{2 ( 1 ) ( 2 )}{2} = 2 - 2 = 0$ .

Answer: (-1, 0)

The trap

Using the foot-of-perpendicular formula (which stops halfway) instead of the full reflection formula (which goes twice as far) gives the midpoint of the segment, not the actual mirror image.

JEE MainTwo lines meeting at a given angle, with two valid slopesAngle formula, both branches

Two lines through the point $(2, 3)$ intersect each other at an angle of $45°$ . If the slope of one line is 3, find the slopes of the other line, and hence its two possible equations.

Stuck? Show the approach

Apply the angle-between-two-lines formula with $m_{1} = 3$ and $θ = 45°$ , solving both sign cases of the absolute value separately.

Show the full solution

$\frac{m _{2} - 3}{1 + 3 m _{2}} = 1$ . Case $+ 1$ : $m_{2} - 3 = 1 + 3 m_{2} \Rightarrow m_{2} = - 2$ . Case $- 1$ : $m_{2} - 3 = - 1 - 3 m_{2} \Rightarrow m_{2} = 1/2$ . Through $(2, 3)$ : with $m_{2} = - 2$ , $y - 3 = - 2 (x - 2) \Rightarrow 2 x + y - 7 = 0$ . With $m_{2} = 1/2$ , $y - 3 = \frac{1}{2} (x - 2) \Rightarrow x - 2 y + 4 = 0$ .

Answer: m = -2 (line 2x+y-7=0) or m = 1/2 (line x-2y+4=0)

The trap

Reporting only one of the two slopes is the most common mark loss here — a 45 degree angle can open on either side of the given line, giving two genuinely different second lines.

JEE MainConcurrency of three linesIntersection-then-substitute

Find the value of $λ$ for which the lines $x + y - 1 = 0$ , $2 x - y + 4 = 0$ , and $3 x + λ y - 5 = 0$ are concurrent.

Stuck? Show the approach

Find the intersection point of the first two lines, then require that point to satisfy the third line's equation.

Show the full solution

Adding $x + y = 1$ and $2 x - y = - 4$ : $3 x = - 3 \Rightarrow x = - 1$ , and $y = 1 - x = 2$ . So the first two lines meet at $(- 1, 2)$ . Substituting into the third: $3 (- 1) + λ (2) - 5 = 0 \Rightarrow 2 λ = 8 \Rightarrow λ = 4$ .

Answer: lambda = 4

The trap

Solving the wrong pair of lines first (e.g. lines 1 and 3, which both contain the unknown lambda) turns a simple substitution into an unnecessary second unknown-in-unknown system.

JEE AdvancedMulti-step construction: intersection, perpendicularity, and an axis interceptSequential coordinate-geometry construction

Lines $L_{1} : 3 x + 4 y = 9$ and $L_{2} : 4 x - 3 y = 12$ intersect at point $P$ . A line through $P$ , perpendicular to the line $7 x - y + 3 = 0$ , meets the y-axis at point $Q$ . Find the coordinates of $Q$ .

Stuck? Show the approach

Find $P$ by solving $L_{1}$ and $L_{2}$ simultaneously, find the required slope as the negative reciprocal of $7 x - y + 3 = 0$ 's slope, build the new line through $P$ , then set $x = 0$ to find where it crosses the y-axis.

Show the full solution

Solving $3 x + 4 y = 9$ and $4 x - 3 y = 12$ : multiplying by 3 and 4 respectively and adding eliminates $y$ , giving $25 x = 75 \Rightarrow x = 3$ , and then $y = 0$ , so $P = (3, 0)$ . The line $7 x - y + 3 = 0$ has slope 7, so the perpendicular line has slope $- 1/7$ . Through $P$ : $y - 0 = - \frac{1}{7} (x - 3) \Rightarrow x + 7 y - 3 = 0$ . Setting $x = 0$ : $y = 3/7$ .

Answer: Q = (0, 3/7)

The trap

Using the slope of $L_{1}$ or $L_{2}$ instead of the slope of $7 x - y + 3 = 0$ for the perpendicularity condition is the most common mix-up in a problem with three different lines in play.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 BoardVery High

JEE MainVery High

JEE AdvancedHigh

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

No. Normal Form is formative-only per CBSE's own curriculum listing, and the current 2026-27 book has no section, formula, or exercise question on it at all. Only five forms are taught here.

No — the book introduces Ax+By+C=0 in a single defining sentence, mainly as the notation the distance formula needs. What is genuinely tested is converting a general-form equation back into slope-intercept or intercept form, which Exercise 9.3 asks directly.

One point plus slope: point-slope form. Two points: two-point form. Slope plus a y-intercept: slope-intercept form. Both intercepts: intercept form. If the question gives the equation already in general form and asks for slope or intercepts, reduce it into the matching named form first.

The formula involves an absolute value, which splits into two cases. Geometrically, a line making a given angle with a fixed line can lie on either side of it, producing two distinct valid slopes unless the question restricts to one.

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Students who read Straight Lines usually read these next.

Sets

Every later chapter in this book — functions, trigonometry, probability, even calculus — is written in a language, and that language is set notation. This chapter builds that language from a single idea (a well-defined collection of objects) up through subsets, Venn diagrams, and the operations that combine sets, ending with De Morgan's laws. One formula that shows up in almost every coaching handout for this chapter is not actually in the 2026-27 book at all. CBSE Class 11 Mathematics Chapter 1.

Relations and Functions

A relation is any association between two sets at all — as loose or as strange as you like. A function is the one, very specific kind of relation where every input has exactly one output, and that single restriction is what makes it possible to build the rest of mathematics on top of it. One classification almost every coaching handout drills for this chapter — one-one, onto, bijective — has actually moved to Class 12. CBSE Class 11 Mathematics Chapter 2.

Trigonometric Functions

The unit-circle definition extends sine and cosine, first met as ratios inside a right triangle, to every real number — including negative numbers and numbers far larger than 360 degrees. One topic many coaching sheets drill hard for this chapter, the general solution of trigonometric equations, is graded formatively only under the 2026-27 CBSE syllabus and never appears in this chapter's own exercises. CBSE Class 11 Mathematics Chapter 3.

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The equation x² + 1 = 0 has no real solution, so this chapter builds a new number, i = √(-1), and a new algebra around it. Despite the chapter's own name, actually solving ax² + bx + c = 0 for complex roots — and the polar form its own section heading promises — are both graded formatively only under the 2026-27 CBSE syllabus. CBSE Class 11 Mathematics Chapter 4.

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