Chapter 14ICSE Class 10 Maths100% Free

Equation of a Line — Important Questions

13 hand-picked ICSE Class 10 Maths important questions for Equation of a Line, each with a full model answer — the formats and topics most likely to appear in your board exam.

13
Questions
6
Question types
32
Total marks
₹0
With answers
Quick answer

High-yield ICSE Equation of a Line questions use the slope m=y2y1x2x1m=\dfrac{y_2-y_1}{x_2-x_1}, the forms y=mx+cy=mx+c and yy1=m(xx1)y-y_1=m(x-x_1), and the conditions for parallel lines (equal slopes) and perpendicular lines (product of slopes =1=-1). Finding a line through a point parallel or perpendicular to a given line is asked every year.

About Equation of a Line

In the ICSE Class 10 Maths chapter Equation of a Line you find the slope of a line, write its equation using the slope-intercept and point-slope forms, and use the conditions for parallel and perpendicular lines. You also read off gradients and intercepts and solve problems involving straight-line geometry.

Slope of a lineSlope-intercept form $y=mx+c$Point-slope and two-point formsParallel and perpendicular linesIntercepts on the axes

Key concepts & formulas

Slope

The slope of the line through (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2) is m=y2y1x2x1=tanθm=\dfrac{y_2-y_1}{x_2-x_1}=\tan\theta, where θ\theta is its inclination.

Forms of a line

Slope-intercept: y=mx+cy=mx+c; point-slope: yy1=m(xx1)y-y_1=m(x-x_1). These give the equation once the slope and a point (or intercept) are known.

Parallel and perpendicular

Parallel lines have equal slopes m1=m2m_1=m_2; perpendicular lines satisfy m1×m2=1m_1\times m_2=-1.

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Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The slope of the line joining (2,3)(2,3) and (4,7)(4,7) is:

  1. (a)

    22

  2. (b)

    12\tfrac12

  3. (c)

    2-2

  4. (d)

    44

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Answer: (a) 22.

m=7342=42=2.m=\dfrac{7-3}{4-2}=\dfrac{4}{2}=2.

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Q2MCQEasy1 mark

The slope of the line 2x+3y=62x+3y=6 is:

  1. (a)

    23-\tfrac23

  2. (b)

    23\tfrac23

  3. (c)

    32-\tfrac32

  4. (d)

    22

Show model answer

Answer: (a) 23-\tfrac23.

3y=2x+6y=23x+23y=-2x+6\Rightarrow y=-\dfrac23x+2, so the slope is 23.-\dfrac23.

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Q3MCQModerate1 mark

The equation of the line through (0,4)(0,4) parallel to 3xy=53x-y=5 is:

  1. (a)

    y=3x+4y=3x+4

  2. (b)

    y=3x+4y=-3x+4

  3. (c)

    y=13x+4y=\tfrac13x+4

  4. (d)

    y=3x4y=3x-4

Show model answer

Answer: (a) y=3x+4y=3x+4.

3xy=5y=3x53x-y=5\Rightarrow y=3x-5, slope 33. A parallel line has slope 33 and yy-intercept 44: y=3x+4.y=3x+4.

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Q4MCQHOTS1 mark

The slope of a line perpendicular to the line joining (1,2)(1,2) and (3,8)(3,8) is:

  1. (a)

    13-\tfrac13

  2. (b)

    33

  3. (c)

    13\tfrac13

  4. (d)

    3-3

Show model answer

Answer: (a) 13-\tfrac13.

Slope of the join =8231=3=\dfrac{8-2}{3-1}=3. Perpendicular slope =13.=-\dfrac1{3}.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): The lines y=2x+3y=2x+3 and 2xy+7=02x-y+7=0 are parallel.

Reason (R): Two lines are parallel when their slopes are equal.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

Show model answer

Answer: (a) The first line has slope 22; the second is y=2x+7y=2x+7, slope 22. Equal slopes mean the lines are parallel, so R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Find the slope and the yy-intercept of the line 4x2y+6=04x-2y+6=0.

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2y=4x+6y=2x+3.2y=4x+6\Rightarrow y=2x+3.

Slope =2=2 and yy-intercept =3.=3.

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Q7Very ShortEasy2 marks

Find the equation of the line passing through (2,3)(2,-3) with slope 4-4.

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yy1=m(xx1)y-y_1=m(x-x_1): y(3)=4(x2)y+3=4x+84x+y5=0.y-(-3)=-4(x-2)\Rightarrow y+3=-4x+8\Rightarrow 4x+y-5=0.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

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

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Slope m=2413=24=12.m=\dfrac{2-4}{-1-3}=\dfrac{-2}{-4}=\dfrac12.

y4=12(x3)2y8=x3x2y+5=0.y-4=\dfrac12(x-3)\Rightarrow 2y-8=x-3\Rightarrow x-2y+5=0.

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Q9Short AnswerModerate3 marks

Find the equation of the line passing through (2,3)(-2,3) and perpendicular to the line 2x3y=62x-3y=6.

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2x3y=6y=23x22x-3y=6\Rightarrow y=\dfrac23x-2, slope 23\dfrac23.

Perpendicular slope =32.=-\dfrac32.

y3=32(x+2)2y6=3x63x+2y=0.y-3=-\dfrac32(x+2)\Rightarrow 2y-6=-3x-6\Rightarrow 3x+2y=0.

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Q10Short AnswerHOTS3 marks

A line passes through the point (4,3)(4,3) and makes equal intercepts on the two axes. Find its equation.

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Equal intercepts mean the line is xa+ya=1\dfrac{x}{a}+\dfrac{y}{a}=1, i.e. x+y=a.x+y=a.

It passes through (4,3)(4,3): 4+3=aa=7.4+3=a\Rightarrow a=7.

The equation is x+y=7.x+y=7.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

Points A(1,4)A(1,4) and B(3,0)B(3,0) are given. (i) Find the equation of line ABAB. (ii) Find the equation of the perpendicular bisector of ABAB. (iii) Verify whether the point (0,1)(0,1) lies on this perpendicular bisector.

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(i) Slope of AB=0431=2.AB=\dfrac{0-4}{3-1}=-2. Using A(1,4)A(1,4): y4=2(x1)2x+y=6.y-4=-2(x-1)\Rightarrow 2x+y=6.

(ii) Mid-point of AB=(1+32,4+02)=(2,2).AB=\left(\dfrac{1+3}{2},\dfrac{4+0}{2}\right)=(2,2). Perpendicular slope =12.=\dfrac12.

y2=12(x2)2y4=x2x2y+2=0.y-2=\dfrac12(x-2)\Rightarrow 2y-4=x-2\Rightarrow x-2y+2=0.

(iii) Put (0,1)(0,1): 02(1)+2=00-2(1)+2=0, which holds, so (0,1)(0,1) lies on the perpendicular bisector.

ICSE Class 10 Maths — Equation of a Line: Points A(1,4) and B(3,0) are given. (i) Find the equation of line AB. (ii) Find the equation of the perpendicular bisector of AB. (iii) Ve
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Q12Long AnswerHOTS5 marks

A line LL passes through the point P(2,3)P(-2,3) and is parallel to the line joining A(2,1)A(2,-1) and B(6,3)B(6,3). (i) Find the equation of LL. (ii) Find the coordinates of the points where LL cuts the xx-axis and the yy-axis. (iii) Find the point where LL meets the line 2x+y=42x+y=4.

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(i) Slope of AB=3(1)62=44=1.AB=\dfrac{3-(-1)}{6-2}=\dfrac{4}{4}=1. Line LL has slope 11 through P(2,3)P(-2,3): y3=1(x+2)y=x+5.y-3=1(x+2)\Rightarrow y=x+5.

(ii) xx-axis (y=0y=0): 0=x+5x=50=x+5\Rightarrow x=-5, point (5,0).(-5,0). yy-axis (x=0x=0): y=5y=5, point (0,5).(0,5).

(iii) Solve y=x+5y=x+5 and 2x+y=42x+y=4: 2x+(x+5)=43x=1x=13, y=x+5=143.2x+(x+5)=4\Rightarrow 3x=-1\Rightarrow x=-\dfrac13,\ y=x+5=\dfrac{14}{3}. The point is (13,143).\left(-\dfrac13,\dfrac{14}{3}\right).

ICSE Class 10 Maths — Equation of a Line: A line L passes through the point P(-2,3) and is parallel to the line joining A(2,-1) and B(6,3). (i) Find the equation of L. (ii) Find th
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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A straight stretch of road on a map passes through the points A(0,2)A(0,2) and B(4,10)B(4,10).

(i) Find the slope of the road.

(ii) Find the equation of the road.

(iii) Does the point (3,8)(3,8) lie on this road?

(iv) Find the equation of a road parallel to it that passes through (0,1)(0,-1).

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(i) Slope =10240=84=2.=\dfrac{10-2}{4-0}=\dfrac{8}{4}=2.

(ii) The yy-intercept is 22, so y=2x+2.y=2x+2.

(iii) At x=3x=3: y=2(3)+2=8y=2(3)+2=8, so (3,8)(3,8) does lie on the road.

(iv) A parallel road has slope 22; through (0,1)(0,-1) its equation is y=2x1.y=2x-1.

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