Straight line | NEB Class 11 Mathematics Notes

Chapter Overview

This chapter is the backbone of coordinate geometry — the beautiful fusion of algebra and Euclidean geometry pioneered by French mathematician René Descartes. A straight line is the simplest curve, yet it appears in GPS navigation, structural engineering, data science (linear regression), and almost every branch of mathematics. In NEB board exams, straight-line questions appear across every section — from 1-mark MCQs to 5-mark long proofs. Master this chapter and coordinate geometry becomes easy.

What this chapter covers

Distance, Section, and Area FormulasSlope and Inclination of a LinePoint-Slope and Two-Point FormsSlope-Intercept Form y = mx + cDouble Intercept Form x/a + y/b = 1Normal Form x cosα + y sinα = pGeneral Form ax + by + c = 0Angle Between Two LinesConditions for Parallel and Perpendicular LinesLength of Perpendicular from a Point to a LineDistance Between Parallel LinesAngle Bisectors of Two LinesConcurrency of Three Lines

Learning Objectives

1

Apply the distance, section, midpoint, centroid, and area formulas correctly in all problem types

2

Write the equation of a straight line in any form (point-slope, two-point, slope-intercept, intercept, normal) and convert between them

3

Calculate the angle between two lines and state conditions for parallel (m₁ = m₂) and perpendicular (m₁m₂ = −1) lines

4

Find the perpendicular distance from any point to any line using the standard formula

5

Determine the distance between two parallel lines

6

Find the equations of angle bisectors of two lines and identify which one contains the origin or is acute

Every Line in the Plane is ax + by + c = 0

Key Insight

No matter what form you start with — slope-intercept, two-point, intercept, normal — every straight line can be written as ax + by + c = 0 (General Form). This single equation captures everything about the line.

Slope = −a/b. X-intercept = −c/a. Y-intercept = −c/b. Perpendicular distance from (x₁, y₁) = |ax₁ + by₁ + c| / √(a²+b²). All five forms are just different ways of writing this same equation.

In NEB exams: when asked to find a line parallel to ax+by+c=0, write ax+by+k=0. When perpendicular: bx−ay+k=0. Just change the constant or swap and negate the coefficients.

Key Definitions

Abscissa

The x-coordinate of a point. The horizontal distance from the y-axis.

Ordinate

The y-coordinate of a point. The vertical distance from the x-axis.

Inclination (θ)

The angle made by a line with the positive direction of the x-axis, measured counterclockwise. Range: 0° ≤ θ < 180°.

$$0^\circ \le \theta < 180^\circ$$

Slope (m)

The tangent of the angle of inclination. Measures how steep a line is. Positive slope = rising left to right. Negative = falling. Zero = horizontal. Undefined = vertical.

$$m = \tan \theta = \frac{y_2 - y_1}{x_2 - x_1}$$

X-Intercept (a)

The x-coordinate of the point where the line crosses the x-axis (where y = 0).

Y-Intercept (c or b)

The y-coordinate of the point where the line crosses the y-axis (where x = 0).

Centroid (G)

The point where the three medians of a triangle intersect. It divides each median in the ratio 2:1 from vertex.

$$G = \left(\frac{x_1+x_2+x_3}{3},\; \frac{y_1+y_2+y_3}{3}\right)$$

Orthocentre (H)

The point of intersection of the three altitudes of a triangle. An altitude is the perpendicular from a vertex to the opposite side.

Circumcentre

The intersection point of the perpendicular bisectors of the sides. It is equidistant from all three vertices.

Concurrent Lines

Three or more lines are concurrent if they all pass through the same single point. Test: solve any two lines, then check if the third passes through the solution point.

Parallel Lines vs Perpendicular Lines

The two most tested line relationships in NEB exams. Memorise both conditions cold.

Parallel Lines

$$m_1 = m_2$$

Equal slopes — both lines rise/fall at the same rate

Never intersect (or coincide if also same c)

Line parallel to ax+by+c=0 → write ax+by+k=0 (same coefficients, different constant)

Distance between them = $\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$ (only when a, b coefficients are identical)

Keywords: 'parallel to', 'same direction as', 'equidistant from'

VS

Perpendicular Lines

$$m_1 \cdot m_2 = -1$$

Slopes are negative reciprocals of each other: if m₁ = k, then m₂ = −1/k

Intersect at exactly 90°

Line perpendicular to ax+by+c=0 → write bx−ay+k=0 (swap a,b; negate one)

Special case: horizontal (m=0) is perpendicular to vertical (undefined slope)

Keywords: 'perpendicular to', 'at right angles', 'normal to'

Quick check: Are slopes equal? → Parallel. Do slopes multiply to −1? → Perpendicular. Neither? → Intersecting at some angle.

Complete Formula Sheet with Derivations

Distance Between Two Points

d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

Distance is always positive. Comes directly from Pythagoras' theorem.

1Draw a right triangle with $A(x_1, y_1)$ and $B(x_2, y_2)$ as two corners. The third corner is $C(x_2, y_1)$.

2Horizontal leg $AC = |x_2 - x_1|$. Vertical leg $BC = |y_2 - y_1|$.

3By Pythagoras: $AB^2 = AC^2 + BC^2 = (x_2-x_1)^2 + (y_2-y_1)^2$. Take the positive square root: $$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$

Internal Section Formula

P = \left(\frac{m_1 x_2 + m_2 x_1}{m_1+m_2},\; \frac{m_1 y_2 + m_2 y_1}{m_1+m_2}\right)

P divides AB internally in the ratio m₁ : m₂. For external division, use a minus sign.

1Point $P(x, y)$ divides line joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in ratio $m_1 : m_2$.

2By similar triangles: $\frac{AP}{PB} = \frac{m_1}{m_2}$. The x-coordinate satisfies $\frac{x - x_1}{x_2 - x} = \frac{m_1}{m_2}$.

3Cross multiply and solve: $m_2(x - x_1) = m_1(x_2 - x) \Rightarrow x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$. Same logic gives y-coordinate.

Area of a Triangle

\Delta = \frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|

Always take the absolute value. If the result is 0, the three points are collinear.

1Consider triangle with vertices $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$. Drop perpendiculars to the x-axis.

2The area equals (trapezium under AB) + (trapezium under BC) − (trapezium under AC).

3After expanding and simplifying the trapezium areas: $\Delta = \frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$. The $|...|$ ensures area is always positive.

Point-Slope Form

$$y - y_1 = m(x - x_1)$$

Use when given slope m and one point (x₁, y₁). Most directly derived form.

1Let a line pass through fixed point $B(x_1, y_1)$ with slope $m$. Let $P(x, y)$ be any other point on the line.

2By the definition of slope: $m = \frac{y - y_1}{x - x_1}$.

3Cross-multiply both sides by $(x - x_1)$: $$y - y_1 = m(x - x_1)$$ This holds for every point P on the line — including when P = B, giving $0 = 0$.

Two-Point Form

\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}

Use when two points on the line are given. Equivalent to substituting the slope formula into point-slope form.

1A line through $A(x_1,y_1)$ and $B(x_2,y_2)$ has slope $m = \frac{y_2-y_1}{x_2-x_1}$.

2Substitute into the point-slope form $y - y_1 = m(x-x_1)$:

3$$y - y_1 = \frac{y_2-y_1}{x_2-x_1}(x - x_1) \quad\Rightarrow\quad \frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1}$$

Slope-Intercept Form

$$y = mx + c$$

m = slope, c = y-intercept. Fastest form to read off both values.

1A line with slope $m$ passes through the y-intercept point $(0, c)$.

2Using point-slope form with $(x_1, y_1) = (0, c)$: $y - c = m(x - 0)$.

3$$y = mx + c$$

Double Intercept Form

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

a = x-intercept, b = y-intercept. Use when both intercepts are given. Don't swap a and b!

1A line crosses the x-axis at $A(a, 0)$ and the y-axis at $B(0, b)$.

2Using two-point form with $(x_1,y_1)=(a,0)$ and $(x_2,y_2)=(0,b)$:

3$$\frac{y-0}{b-0} = \frac{x-a}{0-a} \Rightarrow \frac{y}{b} = \frac{a-x}{a} = 1 - \frac{x}{a} \Rightarrow \frac{x}{a} + \frac{y}{b} = 1$$

Normal Form

x \cos \alpha + y \sin \alpha = p

α = angle the normal (perpendicular) from origin makes with x-axis. p = perpendicular distance from origin to line (always positive).

1Let the foot of the perpendicular from the origin $O$ to the line be $N$. The distance $ON = p$ and the angle $\angle XON = \alpha$.

2So $N = (p\cos\alpha,\; p\sin\alpha)$. The line is perpendicular to $ON$, so its slope is $-\cos\alpha/\sin\alpha$.

3Using point-slope form at $N$: $y - p\sin\alpha = \frac{-\cos\alpha}{\sin\alpha}(x - p\cos\alpha)$. Simplify: $$x\cos\alpha + y\sin\alpha = p\cos^2\alpha + p\sin^2\alpha = p$$

Angle Between Two Lines

\tan \theta = \pm\frac{m_1 - m_2}{1 + m_1 m_2}

Use + for acute angle, − for obtuse. If m₁m₂ = −1, angle is 90°. If m₁ = m₂, angle is 0° (parallel).

1Let lines $L_1$ and $L_2$ have inclinations $\alpha_1$ and $\alpha_2$ (so $m_1 = \tan\alpha_1$, $m_2 = \tan\alpha_2$). The angle between them is $\theta = \alpha_1 - \alpha_2$.

2Apply $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}$:

3$$\tan\theta = \frac{m_1 - m_2}{1 + m_1 m_2}$$Since there are two angles between two lines (supplementary pair), we use $\pm$.

Perpendicular Distance from Point to Line

d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}

Most frequently tested formula in NEB. Always take absolute value in numerator.

1Convert $ax+by+c=0$ to normal form by dividing by $\sqrt{a^2+b^2}$: $\frac{a}{\sqrt{a^2+b^2}}x + \frac{b}{\sqrt{a^2+b^2}}y = \frac{-c}{\sqrt{a^2+b^2}}$.

2Comparing with $x\cos\alpha + y\sin\alpha = p$: we get $\cos\alpha = \frac{a}{\sqrt{a^2+b^2}}$, $\sin\alpha = \frac{b}{\sqrt{a^2+b^2}}$, $p = \frac{-c}{\sqrt{a^2+b^2}}$.

3The perpendicular distance from $(x_1,y_1)$ to the line is $d = |x_1\cos\alpha + y_1\sin\alpha - p|$. Substituting: $$d = \left|\frac{ax_1 + by_1 + c}{\sqrt{a^2+b^2}}\right|$$

Distance Between Parallel Lines

d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}

Only valid when BOTH lines have identical a and b coefficients. Make sure to match coefficients first!

1Two parallel lines: $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ (same a, b by definition of parallel).

2Take any point on the first line, say $(-c_1/a, 0)$ (x-intercept). Find its distance to the second line.

3$$d = \frac{|a(-c_1/a) + b(0) + c_2|}{\sqrt{a^2+b^2}} = \frac{|-c_1 + c_2|}{\sqrt{a^2+b^2}} = \frac{|c_1 - c_2|}{\sqrt{a^2+b^2}}$$

Angle Bisectors of Two Lines

\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2+b_1^2}} = \pm\frac{a_2x + b_2y + c_2}{\sqrt{a_2^2+b_2^2}}

+ gives one bisector, − gives the other. To find the bisector containing the origin: make constants positive, then the + case contains the origin.

1A point $P(h,k)$ on the angle bisector is equidistant from both lines. Its signed distance from line 1 equals its signed distance from line 2 (or their negatives).

2Setting $\frac{a_1h+b_1k+c_1}{\sqrt{a_1^2+b_1^2}} = \frac{a_2h+b_2k+c_2}{\sqrt{a_2^2+b_2^2}}$ (the + case) gives one bisector.

3The − case gives the other bisector perpendicular to the first. Together, they bisect all four angles formed by the two lines.

Which Form of Line Equation Should I Use?

Given what the problem tells you, follow this decision tree to choose the right form immediately.

Are TWO POINTS on the line given?

Yes

Use Two-Point Form: (y−y₁)/(y₂−y₁) = (x−x₁)/(x₂−x₁)

No

Continue ↓

Are the X-intercept (a) and Y-intercept (b) given?

Yes

Use Double Intercept Form: x/a + y/b = 1

No

Continue ↓

Is the SLOPE (m) given along with ONE POINT?

Yes

Use Point-Slope Form: y − y₁ = m(x − x₁)

No

Continue ↓

Is slope (m) and y-intercept (c) given?

Yes

Use Slope-Intercept Form: y = mx + c

No

Use Normal Form (if perpendicular distance from origin p and angle α are given): x cosα + y sinα = p

The Four Quadrants and Sign Rules

The x-axis and y-axis divide the plane into four quadrants. The sign of coordinates changes in each quadrant, and this affects slope calculations and angle interpretations.

$$\text{Q-I: }(+,+) \quad \text{Q-II: }(-,+) \quad \text{Q-III: }(-,-) \quad \text{Q-IV: }(+,-)$$

Why it works (Derivation):

1Quadrant I (top-right): Both x and y positive. Lines here with positive slope go upward to the right. Inclination θ is between 0° and 90°.

2Quadrant II (top-left): x negative, y positive. Lines going from Q-I to Q-II typically have positive slope still. Inclination between 90° and 180° gives negative slope (tan of obtuse angle is negative).

3Key rule for slope: A line with positive slope (/) rises left to right. A line with negative slope (\) falls left to right. Horizontal line: slope = 0. Vertical line: slope undefined.

Example: Slope of a line with inclination 135°: m = tan(135°) = tan(180°−45°) = −tan(45°) = −1.

When to use: Slope sign tells you direction. Quadrant tells you where a point lies. Together they help set up equations correctly.

Memory Tricks — Never Forget These

Trick: Four quick memory hooks that cover the most common formula confusions in this chapter.

Steps to Remember:

Slope sign — The Hiker: Imagine walking left to right. Going uphill = positive slope. Going downhill = negative slope. Standing still (horizontal) = zero slope. Walking straight up a cliff (vertical) = undefined slope.
Perpendicular — Negative Reciprocal: If slope is $m$, the perpendicular slope is $-\frac{1}{m}$. Remember: multiply them: $m \times (-\frac{1}{m}) = -1$. The product is always −1.
Centroid — The Average: Centroid is just the average of all x-coordinates and the average of all y-coordinates. No other formula to memorise: $G = (\bar{x}, \bar{y})$.
Intercept Form — a is x, b is y: In $\frac{x}{a} + \frac{y}{b} = 1$, set y=0 to get x = a (x-intercept), set x=0 to get y = b (y-intercept). The variable tells you which intercept.
Parallel line trick: Same ax+by, just change c. Perpendicular line trick: swap a and b, negate one, change c. Example: parallel to 3x+4y=5 → 3x+4y=k. Perpendicular to 3x+4y=5 → 4x−3y=k.

Mnemonic: Hiker for slope | Negative reciprocal for perp | Average for centroid | Variable matches intercept in double-intercept form

Common Mistakes

These mistakes cause the most mark losses in NEB exams for this chapter.

Section Formula Signs: Students use a plus sign for external division. Remember: Internal = (+), External = (−) in the denominator. External division formula: $\left(\frac{m_1 x_2 - m_2 x_1}{m_1 - m_2},\; \frac{m_1 y_2 - m_2 y_1}{m_1 - m_2}\right)$.
Intercept form confusion: In $\frac{x}{a} + \frac{y}{b} = 1$, $a$ is the x-intercept and $b$ is the y-intercept. Never swap them.
Parallel distance without matching coefficients: Before using $d = \frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$, ensure the coefficients of x and y are identical in both equations. Rewrite if needed.
Missing absolute value: In both the perpendicular distance formula and the area formula, the absolute value $|...|$ is mandatory. Without it you risk a negative answer.
Vertical line slope: The slope of a vertical line is undefined (not 0 — that's horizontal). Don't write m = ∞.
Angle bisector — wrong sign for origin: To find the bisector containing the origin, first make the constant terms of both lines positive, THEN the $(+)$ case is the bisector containing the origin.

NEB Exam Traps — These Appear Every Year

NEB Exam Warning

Board examiners specifically test these edge cases. Knowing them can save 2–5 marks.

Asking for slope of line perpendicular to a vertical line — perpendicular to a vertical line is horizontal, slope = 0 (not undefined).
Parallel lines $2x+3y=5$ and $4x+6y=10$ — students say 'different equations so different lines'. They are the same line (second = 2 × first). Check for coincidence first.
Area formula gives zero — students panic. Zero area means the three points are collinear, not on a triangle. This is the correct test for collinearity.
Line through intersection of two lines: the general form is $L_1 + \lambda L_2 = 0$. NEB often asks for this — students forget the $\lambda$ parameter trick.
Centroid of a triangle is always inside the triangle. Circumcentre, orthocentre, incentre may be outside for obtuse/right triangles.
When both equations have the same coefficients but different constants (parallel), perpendicular distance formula works. But if slopes differ, find intersection point and use basic distance formula.

Question Recognition Guide

Keyword / Phrase in Question	Use This
Line passes through (x₁,y₁) and (x₂,y₂)	Two-Point Form
Slope m given, passes through (x₁,y₁)	Point-Slope Form: y−y₁ = m(x−x₁)
Slope m and y-intercept c given	Slope-Intercept Form: y = mx+c
x-intercept a and y-intercept b given	Double Intercept Form: x/a + y/b = 1
Perpendicular from origin = p, angle α given	Normal Form: x cosα + y sinα = p
'Parallel to line L' condition	Same ax+by, different constant: ax+by+k=0
'Perpendicular to line L' condition	Swap a,b; negate one: bx−ay+k=0
Distance from point to line	d = \|ax₁+by₁+c\| / √(a²+b²)
Distance between two parallel lines	d = \|c₁−c₂\| / √(a²+b²) (must match coefficients first)
Angle bisectors of two lines	(a₁x+b₁y+c₁)/√(a₁²+b₁²) = ±(a₂x+b₂y+c₂)/√(a₂²+b₂²)
Bisector containing the origin	Make constants positive, then use + sign case
Three lines concurrent?	Solve first two lines for intersection, substitute into third
Three points collinear?	Area formula = 0, OR slopes of AB and BC are equal
Line through intersection of L₁ and L₂	Use L₁ + λL₂ = 0, find λ from extra condition

Solved Examples (Easy → Hard)

1

Find the slope of the line passing through $(3, -2)$ and $(-1, 4)$.

Easy

Solution

Identify: $(x_1, y_1) = (3, -2)$ and $(x_2, y_2) = (-1, 4)$.

Apply slope formula:
$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-2)}{-1 - 3} = \frac{6}{-4} = -\frac{3}{2}$$

The negative slope means the line falls as we move left to right.

∴ Slope = −3/2

2

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

Easy

Solution

Step 1 — Parallel line form: A line parallel to $3x - 4y + 5 = 0$ has the same coefficients of x and y: $3x - 4y + k = 0$.

Step 2 — Find k using the given point: Substitute $(1, 2)$:
$3(1) - 4(2) + k = 0$
$3 - 8 + k = 0$
$k = 5$

Step 3 — Final equation: $3x - 4y + 5 = 0$.

(Note: this is the same as the original line — the point $(1,2)$ actually lies on $3x-4y+5=0$. If a different point had been used, we'd get a truly different parallel line.)

∴ 3x − 4y + 5 = 0

3

Find the area of a triangle with vertices $A(1,2)$, $B(3,4)$, and $C(0,-1)$.

Medium

Solution

Apply the area formula:
$$\Delta = \frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$$

Substituting $(x_1,y_1)=(1,2)$, $(x_2,y_2)=(3,4)$, $(x_3,y_3)=(0,-1)$:

$= \frac{1}{2}|1(4-(-1)) + 3((-1)-2) + 0(2-4)|$

$= \frac{1}{2}|1(5) + 3(-3) + 0|$

$= \frac{1}{2}|5 - 9| = \frac{1}{2}(4) = 2$ sq. units

∴ Area = 2 square units

4

Find the perpendicular distance from the point $(3, 4)$ to the line $5x - 12y + 2 = 0$.

Medium

Solution

Identify values: $a=5$, $b=-12$, $c=2$, $(x_1,y_1)=(3,4)$.

Apply perpendicular distance formula:
$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}$$

$= \frac{|5(3) + (-12)(4) + 2|}{\sqrt{5^2+(-12)^2}}$

$= \frac{|15 - 48 + 2|}{\sqrt{25+144}}$

$= \frac{|-31|}{\sqrt{169}} = \frac{31}{13}$ units

∴ d = 31/13 units

5

Find the equation of the bisector of the acute angle between $x + y - 1 = 0$ and $x - 7y - 4 = 0$.

Hard

Solution

Step 1 — Set up the bisector formula:
$$\frac{x+y-1}{\sqrt{1^2+1^2}} = \pm\frac{x-7y-4}{\sqrt{1^2+(-7)^2}}$$

$\frac{x+y-1}{\sqrt{2}} = \pm\frac{x-7y-4}{5\sqrt{2}}$

Step 2 — Cross multiply: $5(x+y-1) = \pm(x-7y-4)$

(+) case: $5x+5y-5 = x-7y-4 \Rightarrow 4x+12y-1=0$ (slope $m_1 = -1/3$)

(−) case: $5x+5y-5 = -x+7y+4 \Rightarrow 6x-2y-9=0$ (slope $m_2 = 3$)

Step 3 — Determine which is acute: Original line $x+y-1=0$ has slope $m=-1$. Find angle with $6x-2y-9=0$ (slope $m_2=3$):
$\tan\theta = \left|\frac{3-(-1)}{1+3(-1)}\right| = \left|\frac{4}{-2}\right| = 2 > 1$, so $\theta > 45°$ → obtuse bisector.

Therefore, the acute bisector is $4x + 12y - 1 = 0$.

∴ Acute angle bisector: 4x + 12y − 1 = 0

MCQ Practice

Mixed Level — MCQ

1

What is the slope of a line perpendicular to the x-axis?

Correct!

Incorrect. Correct: Undefined

Solution

A line perpendicular to the x-axis is a vertical line.
Inclination $\theta = 90°$, so $m = \tan 90°$ = undefined.
(This is why the equation of such lines is $x = a$, not $y = mx+c$.)

∴ Undefined

2

If the area of a triangle with vertices $(k, 0)$, $(0, 0)$, and $(0, 2)$ is 4 sq. units, find $k$.

Correct!

Incorrect. Correct: 4

Solution

Using area formula with $(x_1,y_1)=(k,0)$, $(x_2,y_2)=(0,0)$, $(x_3,y_3)=(0,2)$:

$\Delta = \frac{1}{2}|k(0-2) + 0(2-0) + 0(0-0)|$
$= \frac{1}{2}|-2k| = |k|$

Setting $|k| = 4 \Rightarrow k = \pm 4$.
Typically $k = 4$ is the expected answer.

∴ 4

3

The centroid of a triangle with vertices $(1,2)$, $(3,4)$, and $(5,6)$ is:

Correct!

Incorrect. Correct: $(3, 4)$

Solution

Centroid formula — average of coordinates:
$$G = \left(\frac{1+3+5}{3},\; \frac{2+4+6}{3}\right) = \left(\frac{9}{3},\; \frac{12}{3}\right) = (3, 4)$$

∴ $(3, 4)$

4

A line has x-intercept $a=3$ and y-intercept $b=-2$. Its equation is:

Correct!

Incorrect. Correct: $2x - 3y = 6$

Solution

Using double intercept form $\frac{x}{a} + \frac{y}{b} = 1$ with $a=3$, $b=-2$:

$\frac{x}{3} + \frac{y}{-2} = 1$

Multiply through by $-6$:
$-2x + 3y = -6 \Rightarrow \boxed{2x - 3y = 6}$

∴ $2x - 3y = 6$

5

Distance between parallel lines $3x + 4y = 10$ and $3x + 4y = 0$:

Correct!

Incorrect. Correct: 2

Solution

Both lines have the same coefficients of x and y ($a=3$, $b=4$).
Constants are $c_1=−10$ and $c_2=0$ (rewriting as $3x+4y−10=0$ and $3x+4y+0=0$).

$$d = \frac{|c_1-c_2|}{\sqrt{a^2+b^2}} = \frac{|-10-0|}{\sqrt{9+16}} = \frac{10}{5} = 2$$

∴ 2

6

If $m_1 \cdot m_2 = -1$, the lines are:

Correct!

Incorrect. Correct: Perpendicular

Solution

The condition $m_1 \cdot m_2 = -1$ is the definition of perpendicular lines.

If $m_1 = k$, then $m_2 = -1/k$ (negative reciprocal).

This ensures the angle between them satisfies $\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$ → denominator = 0 → $\theta = 90°$.

∴ Perpendicular

7

The equation $y = 5$ represents a line:

Correct!

Incorrect. Correct: Parallel to x-axis

Solution

$y = 5$ means the y-coordinate is always 5, regardless of x.

This is a horizontal line passing through the point $(0, 5)$, parallel to the x-axis.

Its slope is 0 (horizontal), its inclination is 0°.

∴ Parallel to x-axis

8

The ratio in which the x-axis divides the segment joining $(2,-3)$ and $(5,6)$ is:

Correct!

Incorrect. Correct: 1:2

Solution

On the x-axis, $y = 0$. Let the ratio be $m_1 : m_2 = k : 1$.

Using section formula for y:
$0 = \frac{k(6) + 1(-3)}{k+1} \Rightarrow 6k - 3 = 0 \Rightarrow k = \frac{1}{2}$

Ratio = $\frac{1}{2} : 1 = 1 : 2$

∴ 1:2

9

The length of the perpendicular from $(0,0)$ to $x + y + 1 = 0$ is:

Correct!

Incorrect. Correct: $1/\sqrt{2}$

Solution

With $(x_1,y_1)=(0,0)$ and line $x+y+1=0$ (so $a=1, b=1, c=1$):

$$d = \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}} = \frac{|0+0+1|}{\sqrt{1+1}} = \frac{1}{\sqrt{2}}$$

∴ $1/\sqrt{2}$

10

Which form of the line equation is $x\cos\alpha + y\sin\alpha = p$?

Correct!

Incorrect. Correct: Normal form

Solution

$x\cos\alpha + y\sin\alpha = p$ is the Normal Form.

Here: $\alpha$ = angle that the perpendicular from origin to the line makes with x-axis. $p$ = length of that perpendicular (positive). The coefficients are $\cos\alpha$ and $\sin\alpha$, which always satisfy $\cos^2\alpha + \sin^2\alpha = 1$.

∴ Normal form

Short Questions (2 Marks Each)

1

Find the equation of a line with slope 2 passing through $(-1, 3)$.

[2 marks]

Solution

Using point-slope form $y - y_1 = m(x - x_1)$ with $m=2$ and $(x_1,y_1)=(-1,3)$:

$y - 3 = 2(x - (-1))$
$y - 3 = 2x + 2$
$$2x - y + 5 = 0$$

∴ 2x − y + 5 = 0

2

Find the x and y intercepts of $4x - 5y = 20$.

[2 marks]

Solution

Divide both sides by 20 to get intercept form:
$$\frac{x}{5} + \frac{y}{-4} = 1$$

Comparing with $\frac{x}{a} + \frac{y}{b} = 1$: x-intercept = 5, y-intercept = −4.

Check: set y=0 → x=5 ✓. Set x=0 → y=−4 ✓.

∴ x-intercept = 5, y-intercept = −4

3

Determine if lines $2x + 3y = 7$ and $4x + 6y = 5$ are parallel.

[2 marks]

Solution

Slope of $2x+3y=7$: $m_1 = -\frac{2}{3}$

Slope of $4x+6y=5$: $m_2 = -\frac{4}{6} = -\frac{2}{3}$

Since $m_1 = m_2 = -\frac{2}{3}$ and the lines are not the same (7/2 ≠ 5/4), the lines are parallel.

∴ Yes, the lines are parallel (both have slope −2/3)

4

Find the inclination of a line with slope $\sqrt{3}$.

[2 marks]

Solution

$m = \tan\theta = \sqrt{3}$

We know $\tan 60° = \sqrt{3}$.

$$\theta = 60°$$

The line makes a 60° angle with the positive x-axis.

∴ θ = 60°

5

If lines $3x - ay + 2 = 0$ and $2x + 3y - 1 = 0$ are perpendicular, find $a$.

[2 marks]

Solution

Slope of first line: $m_1 = \frac{3}{a}$
Slope of second line: $m_2 = -\frac{2}{3}$

Condition for perpendicularity: $m_1 m_2 = -1$

$\frac{3}{a} \times \left(-\frac{2}{3}\right) = -1$

$-\frac{2}{a} = -1 \Rightarrow a = 2$

∴ a = 2

6

Find the distance between $(2,3)$ and $(-1,-1)$.

[2 marks]

Solution

$$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$

$= \sqrt{(-1-2)^2 + (-1-3)^2}$
$= \sqrt{(-3)^2 + (-4)^2}$
$= \sqrt{9 + 16} = \sqrt{25} = 5$ units

∴ d = 5 units

7

Find the mid-point of the line joining $(4,8)$ and $(2,10)$.

[2 marks]

Solution

Mid-point divides the segment in ratio 1:1. Using section formula with $m_1=m_2=1$:

$$M = \left(\frac{4+2}{2},\; \frac{8+10}{2}\right) = (3,\; 9)$$

∴ M = (3, 9)

8

Find the equation of a line parallel to the y-axis passing through $(4,5)$.

[2 marks]

Solution

A line parallel to the y-axis is a vertical line. All vertical lines have equations of the form $x = \text{constant}$.

Since the line passes through $(4, 5)$, the x-coordinate is always 4:

$$x = 4$$

∴ x = 4

9

Are the points $(1,1)$, $(2,2)$, and $(3,3)$ collinear?

[2 marks]

Solution

Method — Equal slopes:
Slope of AB: $m_{AB} = \frac{2-1}{2-1} = 1$
Slope of BC: $m_{BC} = \frac{3-2}{3-2} = 1$

Since $m_{AB} = m_{BC} = 1$ and they share point B, the three points are collinear.

(All three points lie on the line $y = x$.)

∴ Yes, the points are collinear

10

Find the area of a triangle with vertices $(0,0)$, $(4,0)$, and $(0,3)$.

[2 marks]

Solution

$$\Delta = \frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$$

$= \frac{1}{2}|0(0-3) + 4(3-0) + 0(0-0)|$

$= \frac{1}{2}|0 + 12 + 0| = 6$ sq. units

Note: This is a right triangle with legs 4 and 3, so area = ½ × 4 × 3 = 6. ✓

∴ Area = 6 square units

Long Questions (4–5 Marks Each)

1

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

[4 marks]

Solution

Step 1 — Find intersection:
$x + y = 4$ ... (i) $2x - y = 2$ ... (ii)
Adding: $3x = 6 \Rightarrow x = 2$, then $y = 2$.
Intersection point: $(2, 2)$.

Step 2 — Find required slope:
Slope of $3x+y=5$: $m_1 = -3$.
Slope of perpendicular: $m = \frac{1}{3}$.

Step 3 — Write equation:
$y - 2 = \frac{1}{3}(x - 2)$
$3y - 6 = x - 2$
$$x - 3y + 4 = 0$$

∴ x − 3y + 4 = 0

2

Show that the lines $x+y-1=0$, $2x+3y-5=0$, and $3x+5y-9=0$ are concurrent.

[4 marks]

Solution

Step 1 — Solve first two lines:
$x+y=1$ ... (i) $2x+3y=5$ ... (ii)

From (i): $x = 1-y$. Substitute into (ii):
$2(1-y)+3y = 5 \Rightarrow 2+y=5 \Rightarrow y=3$
Then $x = 1-3 = -2$.
Intersection: $(-2, 3)$.

Step 2 — Verify in third line:
Substitute $(-2, 3)$ into $3x+5y-9$:
$3(-2) + 5(3) - 9 = -6 + 15 - 9 = 0$ ✓

Since the third line passes through the intersection of the first two, all three lines are concurrent at $(-2, 3)$.

∴ Concurrent at (−2, 3) — proved

3

A line passes through $(2,3)$ such that its segment between the axes is bisected at this point. Find its equation.

[4 marks]

Solution

Step 1 — Setup:
Let the line cut the x-axis at $A(a, 0)$ and y-axis at $B(0, b)$.

Step 2 — Use midpoint condition:
Midpoint of AB = $\left(\frac{a+0}{2},\; \frac{0+b}{2}\right) = \left(\frac{a}{2},\; \frac{b}{2}\right)$.

Given this equals $(2, 3)$:
$\frac{a}{2} = 2 \Rightarrow a = 4$ and $\frac{b}{2} = 3 \Rightarrow b = 6$.

Step 3 — Write equation using double intercept form:
$\frac{x}{4} + \frac{y}{6} = 1$

Multiply by 12: $3x + 2y = 12$

∴ 3x + 2y = 12

4

Find the coordinates of the foot of the perpendicular from $(1,2)$ to the line $x-y+3=0$.

[5 marks]

Solution

Step 1 — Slope of given line:
$x - y + 3 = 0 \Rightarrow m_1 = 1$.

Step 2 — Slope of perpendicular:
$m_2 = -1$ (negative reciprocal).

Step 3 — Equation of perpendicular from $(1,2)$:
$y - 2 = -1(x - 1)$
$y - 2 = -x + 1$
$x + y = 3$ ... (i)

Step 4 — Solve with original line:
$x - y = -3$ ... (ii) [from $x-y+3=0$]

Adding (i) and (ii): $2x = 0 \Rightarrow x = 0$, $y = 3$.

Foot of perpendicular = $(0, 3)$.

∴ Foot of perpendicular = (0, 3)

5

Find the distance between the parallel lines $5x+12y-13=0$ and $5x+12y+26=0$.

[4 marks]

Solution

Step 1 — Verify parallel:
Both lines have coefficients $a=5$, $b=12$. Slopes are equal: $m = -5/12$ for both. ✓ Parallel.

Step 2 — Apply distance formula:
Rewrite as $5x+12y+(-13)=0$ (so $c_1=-13$) and $5x+12y+26=0$ (so $c_2=26$).

$$d = \frac{|c_1-c_2|}{\sqrt{a^2+b^2}} = \frac{|-13-26|}{\sqrt{25+144}} = \frac{|-39|}{\sqrt{169}} = \frac{39}{13} = 3 \text{ units}$$

∴ d = 3 units

6

Find the equation of the median through $A$ for a triangle with vertices $A(1,2)$, $B(3,4)$, $C(5,-2)$.

[4 marks]

Solution

Step 1 — Find midpoint M of BC:
$$M = \left(\frac{3+5}{2},\; \frac{4+(-2)}{2}\right) = (4, 1)$$

Step 2 — Find equation of line AM:
$A(1,2)$ and $M(4,1)$. Slope $= \frac{1-2}{4-1} = -\frac{1}{3}$

Using point-slope form at $A(1,2)$:
$y - 2 = -\frac{1}{3}(x-1)$
$3y - 6 = -x + 1$
$$x + 3y - 7 = 0$$

∴ Median through A: x + 3y − 7 = 0

7

Find the equation of a line making $45°$ with $2x - y + 3 = 0$ and passing through $(1, 0)$.

[5 marks]

Solution

Step 1 — Slope of given line: $m_1 = 2$.

Step 2 — Use angle condition:
$\tan 45° = \left|\frac{m - m_1}{1 + m\cdot m_1}\right| = \left|\frac{m-2}{1+2m}\right| = 1$

Two cases:
Case (+): $m-2 = 1+2m \Rightarrow -m=3 \Rightarrow m=-3$
Case (−): $m-2 = -(1+2m) \Rightarrow m-2=-1-2m \Rightarrow 3m=1 \Rightarrow m=\frac{1}{3}$

Step 3 — Write equations through $(1,0)$:
$m=-3$: $y-0=-3(x-1) \Rightarrow 3x+y-3=0$
$m=1/3$: $y-0=\frac{1}{3}(x-1) \Rightarrow x-3y-1=0$

∴ 3x + y − 3 = 0 OR x − 3y − 1 = 0

8

Find the ratio in which $x - y - 2 = 0$ divides the segment joining $(3,-1)$ and $(8,9)$.

[4 marks]

Solution

Step 1 — Let ratio be $k:1$.

Step 2 — Find coordinates using section formula:
$x = \frac{k(8) + 1(3)}{k+1} = \frac{8k+3}{k+1}$
$y = \frac{k(9) + 1(-1)}{k+1} = \frac{9k-1}{k+1}$

Step 3 — Substitute into line equation $x-y-2=0$:
$\frac{8k+3}{k+1} - \frac{9k-1}{k+1} - 2 = 0$

$\frac{8k+3-9k+1}{k+1} = 2$

$\frac{4-k}{k+1} = 2 \Rightarrow 4-k = 2k+2 \Rightarrow k = \frac{2}{3}$

Ratio = $\frac{2}{3} : 1 = 2 : 3$

∴ The line divides the segment in ratio 2:3 (internally)

9

Prove that the points $(1,1)$, $(-1,-1)$, and $(-\sqrt{3}, \sqrt{3})$ form an equilateral triangle.

[5 marks]

Solution

Let $A=(1,1)$, $B=(-1,-1)$, $C=(-\sqrt{3},\sqrt{3})$.

AB:
$AB = \sqrt{(-1-1)^2+(-1-1)^2} = \sqrt{4+4} = 2\sqrt{2}$

BC:
$BC = \sqrt{(-\sqrt{3}-(-1))^2+(\sqrt{3}-(-1))^2} = \sqrt{(1-\sqrt{3})^2+(1+\sqrt{3})^2}$
$= \sqrt{1-2\sqrt{3}+3+1+2\sqrt{3}+3} = \sqrt{8} = 2\sqrt{2}$

CA:
$CA = \sqrt{(1-(-\sqrt{3}))^2+(1-\sqrt{3})^2} = \sqrt{(1+\sqrt{3})^2+(1-\sqrt{3})^2} = \sqrt{8} = 2\sqrt{2}$

Since $AB = BC = CA = 2\sqrt{2}$, all sides are equal. Therefore the triangle is equilateral.

∴ AB = BC = CA = 2√2 — equilateral triangle proved

10

Find the equation of the bisector of the angle containing the origin for $3x-4y+6=0$ and $5x+12y-10=0$.

[5 marks]

Solution

Step 1 — Check sign of constants:
$3x-4y+6=0$: constant $c_1=+6>0$ ✓
$5x+12y-10=0$: constant $c_2=-10<0$.

Make both constants positive: rewrite second as $-5x-12y+10=0$.

Step 2 — Bisector containing origin is the (+) case:
$$\frac{3x-4y+6}{\sqrt{9+16}} = \frac{-5x-12y+10}{\sqrt{25+144}}$$

$\frac{3x-4y+6}{5} = \frac{-5x-12y+10}{13}$

Step 3 — Cross multiply:
$13(3x-4y+6) = 5(-5x-12y+10)$
$39x-52y+78 = -25x-60y+50$
$64x+8y+28 = 0$
$$16x + 2y + 7 = 0$$

∴ Bisector containing origin: 16x + 2y + 7 = 0

Past NEB Board Questions

1

Find the equation of a straight line passing through the intersection of $2x+y-4=0$ and $x-y+1=0$, and parallel to $3x+5y=7$.

NEB 2079 [4 marks]

Solution

Step 1 — Find intersection:
$2x+y=4$ and $x-y=-1$. Add: $3x=3 \Rightarrow x=1$, $y=2$. Point: $(1,2)$.

Step 2 — Parallel line form:
Parallel to $3x+5y=7$: write $3x+5y+k=0$.

Step 3 — Find k:
$3(1)+5(2)+k=0 \Rightarrow 3+10+k=0 \Rightarrow k=-13$.

$$3x + 5y - 13 = 0$$

∴ 3x + 5y − 13 = 0

2

Find the equation of the angle bisectors of the lines $3x-4y+7=0$ and $12x+5y-2=0$. Also find which bisector passes through the origin.

NEB 2078 [5 marks]

Solution

Step 1 — Set up bisector equation:
$$\frac{3x-4y+7}{\sqrt{9+16}} = \pm\frac{12x+5y-2}{\sqrt{144+25}}$$

$\frac{3x-4y+7}{5} = \pm\frac{12x+5y-2}{13}$

(+) case: $13(3x-4y+7) = 5(12x+5y-2)$
$39x-52y+91 = 60x+25y-10$
$21x+77y-101=0$

(−) case: $13(3x-4y+7) = -5(12x+5y-2)$
$39x-52y+91 = -60x-25y+10$
$99x-27y+81=0 \Rightarrow 11x-3y+9=0$

Bisector through origin: Make constants positive (7>0, −2<0 → rewrite 2nd as $-12x-5y+2=0$). With both positive, (+) case contains origin: $21x+77y-101=0$. Check: $0+0-101\neq 0$... Let me verify by substituting origin $(0,0)$ into both: $21(0)+77(0)-101=-101<0$ and $11(0)-3(0)+9=9>0$. Bisector containing origin is $11x-3y+9=0$.

∴ Bisectors: 21x+77y−101=0 and 11x−3y+9=0. The bisector containing the origin is 11x−3y+9=0.

3

Prove that the lines $2x-3y+4=0$, $4x-y-2=0$, and $x+2y-3=0$ are concurrent and find the point of concurrency.

NEB 2077 [4 marks]

Solution

Step 1 — Solve first two:
$2x-3y=-4$ ... (i) $4x-y=2$ ... (ii)

From (ii): $y=4x-2$. Substitute into (i): $2x-3(4x-2)=-4 \Rightarrow 2x-12x+6=-4 \Rightarrow -10x=-10 \Rightarrow x=1$.
Then $y=4(1)-2=2$. Point: $(1,2)$.

Step 2 — Verify in third line:
$1+2(2)-3=1+4-3=2 \neq 0$...

(Recalculating: $x+2y-3=0$ at $(1,2)$: $1+4-3=2 \neq 0$. These three specific lines are NOT concurrent. For the proof type question, the approach is shown above — substitute the intersection into the third line and show it equals 0.)

∴ Method: solve first two lines, substitute intersection into third. If = 0, concurrent.

4

A line cuts the x-axis at $A$ and y-axis at $B$. If the midpoint of $AB$ is $(h, k)$, find the equation of the line in terms of h and k.

NEB 2076 [3 marks]

Solution

Let $A=(a,0)$ and $B=(0,b)$.

Midpoint $= \left(\frac{a}{2}, \frac{b}{2}\right) = (h,k)$

$\Rightarrow a = 2h$ and $b = 2k$

Using double intercept form:
$\frac{x}{2h} + \frac{y}{2k} = 1$

$$\frac{x}{h} + \frac{y}{k} = 2 \quad \Rightarrow \quad kx + hy = 2hk$$

∴ kx + hy = 2hk

Chapter Test — Full Mixed Paper

Attempt this test in 50 minutes without notes. This mirrors the NEB exam format.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

Slope of $x + y = 0$ is:

Correct!

Incorrect. Correct: −1

Solution

Rewrite $x+y=0$ as $y = -x$. Slope = $-1$.

∴ −1

2

The equation of the x-axis is:

Correct!

Incorrect. Correct: y = 0

Solution

The x-axis consists of all points where $y = 0$. Therefore its equation is $y = 0$.

∴ y = 0

3

Distance between $(0,0)$ and $(3,4)$ is:

Correct!

Incorrect. Correct: 5

Solution

$d = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9+16} = \sqrt{25} = 5$

∴ 5

2 Section B — Short Questions (2 Marks Each)

1

Find the slope of the line joining $(1,2)$ and $(3,5)$.

[2 marks]

Solution

$m = \frac{5-2}{3-1} = \frac{3}{2}$

∴ m = 3/2

2

Are $y = 2x+1$ and $y = 2x-5$ parallel?

[2 marks]

Solution

Both have slope $m = 2$. Since slopes are equal and the lines are distinct (different constants), they are parallel.

∴ Yes, parallel (both have slope 2)

3 Section C — Long Questions (4–5 Marks Each)

1

Find the equation of angle bisectors of $3x-4y+7=0$ and $12x+5y-2=0$.

[5 marks]

Solution

$$\frac{3x-4y+7}{5} = \pm\frac{12x+5y-2}{13}$$

(+): $39x-52y+91=60x+25y-10 \Rightarrow 21x+77y-101=0$

(−): $39x-52y+91=-60x-25y+10 \Rightarrow 99x-27y+81=0 \Rightarrow 11x-3y+9=0$

∴ 21x + 77y − 101 = 0 and 11x − 3y + 9 = 0

2

A line passes through $(3,4)$ such that its intercepts on the axes are equal. Find its equation.

[4 marks]

Solution

Case 1 — Equal nonzero intercepts (a = b):
Using $x/a + y/a = 1 \Rightarrow x+y=a$. Passes through $(3,4)$: $3+4=a \Rightarrow a=7$.
Equation: $x+y=7$.

Case 2 — Both intercepts zero (line through origin):
$y/x = 4/3 \Rightarrow 4x-3y=0$.

Two equations: $x+y=7$ and $4x-3y=0$.

∴ x + y = 7 OR 4x − 3y = 0

One-Page Revision Cheat Sheet

1

Distance: $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ — always positive.

2

Section (internal): $P = \left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\; \frac{m_1y_2+m_2y_1}{m_1+m_2}\right)$. External: replace + with −.

3

Centroid: Average of coordinates: $G = \left(\frac{\sum x}{3}, \frac{\sum y}{3}\right)$.

4

Area: $\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$. Zero → collinear.

5

Slope: $m = \tan\theta = (y_2-y_1)/(x_2-x_1)$. Vertical → undefined. Horizontal → 0.

6

Parallel: $m_1 = m_2$. Same ax+by, change constant.

7

Perpendicular: $m_1 m_2 = -1$. Swap a,b; negate one; change constant.

8

Point-slope: $y-y_1 = m(x-x_1)$. Two-point: substitute slope formula.

9

Intercept form: $x/a + y/b = 1$ — a is x-intercept, b is y-intercept.

10

Normal form: $x\cos\alpha + y\sin\alpha = p$ — p is perpendicular distance from origin.

11

Angle between lines: $\tan\theta = \pm(m_1-m_2)/(1+m_1m_2)$.

12

Perp distance: $d = |ax_1+by_1+c|/\sqrt{a^2+b^2}$ — memorise this cold.

13

Parallel lines distance: $d = |c_1-c_2|/\sqrt{a^2+b^2}$ — match coefficients first.

14

Angle bisectors: $(a_1x+b_1y+c_1)/\sqrt{a_1^2+b_1^2} = \pm(a_2x+b_2y+c_2)/\sqrt{a_2^2+b_2^2}$.

15

Bisector through origin: Make constants positive → use (+) case.