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NCERT Solutions For Class 11 Maths Chapter 8 Straight Lines Exercise 8.1 - 2025-26

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Maths Class 11 Chapter 8 Questions and Answers - Free PDF Download

In NCERT Solutions for Class 11 Maths Ch 8 Straight Lines Ex 8.1, you’ll unlock the basics of straight lines in coordinate geometry. This exercise helps you understand important ideas like slope, different ways to write the equation of a line, and how lines relate to each other. If you ever feel confused about finding the slope or writing equations for lines, these step-by-step solutions from Vedantu make things much easier.

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You’ll get to practice problems that show you how to find the distance between points, check if points are collinear, and even spot if a triangle is right-angled—without using the Pythagoras theorem! If you need the full syllabus, you can always check the Class 11 Maths Syllabus anytime for clarity on what to study.


All these solutions are available as a free PDF so you can revise anytime and check your answers. For more help on similar problems, you can access Class 11 Maths NCERT Solutions and prepare smarter for your exams.


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Access NCERT Solutions for Maths Class 11 Chapter 8 - Straight Lines

Exercise 8.1

1. Draw a quadrilateral in the Cartesian plane, whose vertices are $\left( { - 4,5} \right),\left( {0,7} \right),\left( {5, - 5} \right)$ and $\left( { - 4,2} \right)$. Also, find its area.

Ans: Let ABCD be the given quadrilateral with vertices $A\left( { - 4,5} \right),B\left( {0,7} \right),C\left( {5, - 5} \right)$ and $D\left( { - 4,2} \right).$

Then, by plotting A, B, C and D on the Cartesian plane and joining AB, BC, CD and DA, the given quadrilateral can be drawn as


Quadrilateral ABCD on cartesian plane


To find the area of quadrilateral ABCD, we draw one diagonal, say AC.

Accordingly, area (ABCD) $ = (\Delta ABC) + $ area$(\Delta ACD)$

We know that the area of a triangle whose vertices are $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$ and $\left( {{x_3},{y_3}} \right)$ is \[\frac{1}{2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|\]

Therefore, area of $\Delta ABC$

\[\begin{array}{l}= \frac{1}{2}\left| { - 4\left( {7 + 5} \right) + 0\left( { - 5 - 5} \right) + 5\left( {5 - 7} \right)} \right|uni{t^2}\\= \frac{1}{2}\left| { - 4\left( {12} \right) + 5\left( { - 2} \right)} \right|uni{t^2}\\= \frac{1}{2}\left| { - 48 - 10} \right|uni{t^2}\\ = \frac{1}{2}\left| { - 58} \right|uni{t^2}\\ = \frac{1}{2} \times 58uni{t^2}\\= 29uni{t^2}\end{array}\]

Area of $\Delta ACD$

\[\begin{array}{l}= \frac{1}{2}\left| { - 4\left( { - 5 + 2} \right) + 5\left( { - 2 - 5} \right) + \left( { - 4} \right)\left( {5 + 5} \right)} \right|uni{t^2}\\= \frac{1}{2}\left| { - 4\left( { - 3} \right) + 5\left( { - 7} \right) + \left( { - 4} \right)\left( {5 + 5} \right)} \right|uni{t^2}\\= \frac{1}{2}\left| { - 4\left( { - 3} \right) + 5\left( { - 7} \right) + \left( { - 4} \right)\left( {10} \right)} \right|uni{t^2}\\= \frac{1}{2}\left| {12 - 35 - 40} \right|uni{t^2}\\ = \frac{1}{2}\left| { - 63} \right|uni{t^2}\\= \frac{{63}}{2}uni{t^2}\end{array}\]

Thus, area (ABCD) $ = \left( {29 + \frac{{63}}{2}} \right)uni{t^2} = \frac{{58 + 36}}{2}uni{t^2} = \frac{{121}}{2}uni{t^2}$


2. The base of an equilateral triangle with side 2a lies along the y-axis such that the midpoint of the base is at the origin. Find vertices of the triangle.

Ans: Let ABC be the given equilateral triangle with side 2a. 

Accordingly, AB = BC = CA = 2a 

Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin. i.e., BO = OC = a, where O is the origin. 

Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are (0, -a). 

It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular. 

Hence, vertex A lies on the y-axis.


Equilateral triangle with base 2a


On applying Pythagoras theorem to $\Delta AOC$, we’ll get

\[\begin{array}{l}{\left( {AC} \right)^2} = {\left( {OA} \right)^2} + {\left( {OC} \right)^2}\\ \Rightarrow {\left( {2a} \right)^2} = {\left( {OA} \right)^2} + {a^2}\\\Rightarrow 4{a^2} - {a^2} = {\left( {OA} \right)^2}\\\Rightarrow {\left( {OA} \right)^2} = 3{a^2}\\\Rightarrow \left( {OA} \right) = \sqrt 3 a\end{array}\]

\[\therefore \] Coordinates of point A \[ = \left( { \pm \sqrt 3 a,0} \right)\]

Thus, the vertices of the given equilateral triangle are \[\left( {0,a} \right),\left( {0, - a} \right)\]and\[\left( {\sqrt 3 a,0} \right)\]or \[\left( {0,a} \right),\left( {0, - a} \right)\] and \[\left( { - \sqrt 3 a,0} \right)\]


3. Find the distance between P$\left( {{x_1},{y_1}} \right)$ and Q $\left( {{x_2},{y_2}} \right)$ when (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

Ans: The given points are P$\left( {{x_1},{y_1}} \right)$ and Q $\left( {{x_2},{y_2}} \right)$

  1. When PQ is parallel to y-axis, $\left( {{x_1} = {x_2}} \right)$

In this case, distance between P and Q \[ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{({y_2} - {y_1})}^2}} \]

\[\begin{array}{l}= \sqrt {{{({y_2} - {y_1})}^2}} \\= \left| {{y_2} - {y_1}} \right|\end{array}\]

  1. When PQ is parallel to x-axis, $\left( {{y_1} = {y_2}} \right)$

In this case, distance between P and Q \[ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{({y_2} - {y_1})}^2}} \]\[\begin{array}{l}= \sqrt {{{({x_2} - {x_1})}^2}} \\ = \left| {{x_2} - {x_1}} \right|\end{array}\]


4. Find a point on the x-axis, which is equidistant from the points $\left( {7,6} \right)$and$\left( {3,4} \right).$ 

Ans: Let $\left( {a,0} \right)$ be the point on the x-axis that is equidistance from the points $\left( {7,6} \right)$ and $\left( {3,4} \right).$

Accordingly, $\sqrt {{{\left( {7 - a} \right)}^2} + {{\left( {6 - 0} \right)}^2}}  = \sqrt {{{\left( {3 - a} \right)}^2} + {{\left( {4 - 0} \right)}^2}} $

\[\begin{array}{l}\Rightarrow \sqrt {49 + {a^2} - 14a + 36}  = \sqrt {9 + {a^2} - 6a + 16} \\\Rightarrow \sqrt {{a^2} - 14a + 85}  = \sqrt {{a^2} - 6a + 25} \end{array}\]

On squaring both sides, we’ll get

\[\begin{array}{l} \Rightarrow {a^2} - 14a + 85 = {a^2} - 6a + 25\\ \Rightarrow  - 14a + 6a = 25 - 85\\\Rightarrow  - 8a =  - 60\\ \Rightarrow a = \frac{{60}}{8} = \frac{{15}}{2}\end{array}\]

Thus, the required point on the x-axis is $\left( {\frac{{15}}{2},0} \right)$


5. Find the slope of a line, which passes through the origin, and the mid-point of the segment joining the points P $\left( {0, - 4} \right)$ and B $\left( {8,0} \right)$. 

Ans: The coordinates of the mid-point of the line segment joining the points P $\left( {0, - 4} \right)$ and B $\left( {8,0} \right)$ are  $\left( {\frac{{0 + 8}}{2},\frac{{ - 4 + 0}}{2}} \right) = \left( {4, - 2} \right)$

It is known that the slope (m) of a non-vertical line passing through the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$is given by \[m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}},{x_2} \ne {x_1}\]

Therefore, the slope of the line passing through $\left( {0,0} \right)$ and $\left( {4, - 2} \right)$ is $\frac{{ - 2 - 0}}{{4 - 0}} =  - \frac{2}{4} =  - \frac{1}{2}$

Hence, the required slope of the line is$ - \frac{1}{2}$.


6. Without using the Pythagoras theorem, show that the points $\left( {4,4} \right),\left( {3,5} \right)$ and $\left( { - 1, - 1} \right)$ are vertices of a right angled triangle.

Ans: The vertices of the given triangle are A$\left( {4,4} \right),$ B$\left( {3,5} \right)$and C$\left( { - 1, - 1} \right)$. 

It is known that the slope (m) of a non-vertical line passing through the $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is given by \[m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}},{x_2} \ne {x_1}\]

$\therefore $Slope of AB $\left( {{m_1}} \right)$$ = \frac{{5 - 4}}{{3 - 4}} =  - 1$

Slope of BC $\left( {{m_2}} \right)$ $ = \frac{{ - 1 - 5}}{{ - 1 - 3}} = \frac{{ - 6}}{{ - 4}} = \frac{3}{2}$

Slope of CA $\left( {{m_3}} \right)$ $ = \frac{{4 + 1}}{{4 + 1}} = \frac{5}{5} = 1$

It is observed that \[\left( {{m_1}{m_3}} \right) =  - 1\] 

This shows that line segments AB and CA are perpendicular to each other i.e., the given triangle is right-angled at A$\left( {4,4} \right)$. 

Thus, the points $\left( {4,4} \right),$$\left( {3,5} \right)$ and $\left( { - 1, - 1} \right)$ the vertices of a right-angled triangle.


7. Find the slope of the line, which makes an angle of $30^\circ $ with the positive direction of y-axis measured anticlockwise. 

Ans: If a line makes an angle of $30^\circ $ with positive direction of the y-axis measured anticlockwise, then the angle made by the line with the positive direction of the x-axis measured anticlockwise is $90^\circ  + 30^\circ  = 120^\circ $


Slope of the line with angle 30 degree to positive Y


Thus, the slope of the given line is \[\tan 120^\circ  = \tan \left( {180^\circ  - 60^\circ } \right) =  - \tan 60^\circ  =  - \sqrt 3 \]


8. Without using distance formula, show that points $\left( { - 2, - 1} \right),\left( {4,0} \right),\left( {3,3} \right)$and $\left( { - 3,2} \right)$ are vertices of a parallelogram. 

Ans: Let points $\left( { - 2, - 1} \right),\left( {4,0} \right),\left( {3,3} \right)$and $\left( { - 3,2} \right)$ be respectively denoted by A, B, C, and D. 


Parallelogram ABCD


Slopes of AB = $\frac{{0 + 1}}{{4 + 2}} = \frac{1}{6}$ 

Slopes of CD = $\frac{{2 - 3}}{{ - 3 - 3}} = \frac{{ - 1}}{{ - 6}} = \frac{1}{6}$

$ \Rightarrow $Slope of AB = Slope of CD 

$ \Rightarrow $AB and CD are parallel to each other. 

Now, slope of BC = $\frac{{3 - 0}}{{3 - 4}} = \frac{3}{{ - 1}} =  - 3$

Slope of AD = $\frac{{2 + 1}}{{ - 3 + 2}} = \frac{3}{{ - 1}} =  - 3$

$ \Rightarrow $Slope of BC = Slope of AD 

$ \Rightarrow $BC and AD are parallel to each other. 

Therefore, both pairs of opposite side of quadrilateral ABCD are parallel. Hence, ABCD is a parallelogram. 

Thus, points $\left( { - 2, - 1} \right),\left( {4,0} \right),\left( {3,3} \right)$and $\left( { - 3,2} \right)$ are the vertices of a parallelogram. 


9.  Find the angle between the x-axis and the line joining the points $\left( {3, - 1} \right)$ and $\left( {4, - 2} \right)$. 

Ans: The slope of the line joining the points $\left( {3, - 1} \right)$and $\left( {4, - 2} \right)$is $m = \frac{{ - 2 - \left( { - 1} \right)}}{{4 - 3}} =  - 2 + 1 =  - 1$

Now, the inclination ( θ ) of the line joining the points $\left( {3, - 1} \right)$and $\left( {4, - 2} \right)$is given by $\tan \theta  =  - 1$

\[ \Rightarrow \theta  = \left( {90^\circ  + 45^\circ } \right) = 135^\circ \] 

Thus, the angle between the x-axis and the line joining the points $\left( {3, - 1} \right)$and $\left( {4, - 2} \right)$is \[135^\circ \] 


10. The slope of a line is double of the slope of another line. If tangent of the angle between them is \[\frac{1}{3}\], find the slope of the lines. 

Ans: Let \[{m_1}\], \[m\]be the slopes of the two given lines such that \[{m_1} = 2m\]. 

We know that if \[\theta \] is the angle between the lines \[{l_1}\] and \[{l_2}\] with slopes m and then 

\[\tan \theta  = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|\]

It is given that the tangent of the angle between the two lines is \[\frac{1}{3}\].

\[\begin{array}{l}\therefore \frac{1}{3} = \left| {\frac{{m - 2m}}{{1 + \left( {2m} \right).m}}} \right|\\ \Rightarrow \frac{1}{3} = \left| {\frac{{ - m}}{{1 + 2{m^2}}}} \right|\\ \Rightarrow \frac{1}{3} = \frac{m}{{1 + 2{m^2}}}\end{array}\]

Now, Case I:

\[ \Rightarrow \frac{1}{3} = \frac{{ - m}}{{1 + 2{m^2}}}\]

\[\begin{array}{l}\Rightarrow 1 + 2{m^2} =  - 3m\\ \Rightarrow 2{m^2} + 3m + 1 = 0\\ \Rightarrow 2{m^2} + 2m + m + 1 = 0\end{array}\]

\[\begin{array}{l} \Rightarrow 2m\left( {m + 1} \right) + 1\left( {m + 1} \right) = 0\\ \Rightarrow \left( {m + 1} \right)\left( {2m + 1} \right) = 0\\ \Rightarrow m =  - 1 or m =  - \frac{1}{2}\end{array}\]

If \[m =  - 1\], then the slopes of the lines are \[ - 1\]and \[ - 2\]. 

If \[m =  - \frac{1}{2}\], then the slopes of the lines are \[ - \frac{1}{2}\] and \[ - 1\]

Now, Case II:

\[\frac{1}{3} = \frac{m}{{1 + 2{m^2}}}\]

\[\begin{array}{l}\Rightarrow 2{m^2} + 1 = 3m\\ \Rightarrow 2{m^2} - 3m + 1 = 0\\\Rightarrow 2{m^2} - 2m - m + 1 = 0\\ \Rightarrow 2m\left( {m - 1} \right) - 1\left( {m - 1} \right) = 0\\\Rightarrow \left( {m - 1} \right)\left( {2m - 1} \right) = 0\\ \Rightarrow m = 1 or m = \frac{1}{2}\end{array}\]

If \[m = 1\], then the slopes of the lines are \[1\]and \[2\]. 

If \[m = \frac{1}{2}\], then the slopes of the lines are \[\frac{1}{2}\] and \[1\]

Hence, the slopes of the lines are \[ - 1\] and \[ - 2\] or \[ - \frac{1}{2}\] and \[ - 1\] or \[1\]and \[2\]or \[\frac{1}{2}\] and \[1\]


11.  A line passes through $\left( {{x_1},{y_1}} \right)$ and $\left( {h,k} \right)$. If slope of the line is \[m\], show that $k - {y_1} = m\left( {h - {x_1}} \right)$

Ans: The slope of the line passing through $\left( {{x_1},{y_1}} \right)$ and $\left( {h,k} \right)$, is $\frac{{k - {y_1}}}{{h - {x_1}}}$. 

It is given that the slope of the line is \[m\]. 

\[\therefore \]$\frac{{k - {y_1}}}{{h - {x_1}}} = m$

$ \Rightarrow k - {y_1} = m\left( {h - {x_1}} \right)$

Hence, $k - {y_1} = m\left( {h - {x_1}} \right)$


Conclusion

Ex 8.1 Class 11 Maths NCERT Solutions has provided an introduction to the basic principles of straight lines in coordinate geometry. You have learned how to determine the slope of a line from two given points and how to express the equation of a line in various forms, such as slope-intercept form, point-slope form, and general form. By practising these problems, you have built a strong foundation in understanding and working with straight lines. This knowledge is essential for solving more complex geometric problems and will be beneficial in advanced mathematical studies.


Class 11 Maths Chapter 8: Exercises Breakdown

Exercise

Number of Questions

Exercise 8.2

19 Questions & Solutions

Miscellaneous Exercise

23 Questions & Solutions


CBSE Class 11 Maths Chapter 8 Other Study Materials


Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



Important Related Links for CBSE Class 11 Maths

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FAQs on NCERT Solutions For Class 11 Maths Chapter 8 Straight Lines Exercise 8.1 - 2025-26

1. Is an Arithmetic Progression (AP) the only type of sequence with a pattern?

No, a sequence is any list of numbers following a specific rule, and an Arithmetic Progression (AP) is just one type among many. The misconception arises because APs are often the first type of sequence taught.




2. Do the NCERT Solutions for Class 11 Maths only give the final answers?

No, NCERT Solutions provide detailed, step-by-step explanations for every problem, not just the final result. This helps you understand the logic and method for solving all Sequences and Series Class 11 questions and answers, showing the exact application of formulas.


3. Is the common ratio 'r' in a Geometric Progression (GP) always a whole number?

No, the common ratio (r) in a GP can be any non-zero real number, including fractions, negative numbers, or irrational numbers. The confusion often comes from introductory examples that use simple integers like 2 or 3 for clarity.


For instance, the sequence 16, 8, 4, 2, ... is a valid GP where the common ratio is a fraction, r = 1/2. Similarly, the sequence 5, -10, 20, -40, ... is a GP with a negative common ratio, r = -2, causing the signs of the terms to alternate.


To find the common ratio correctly, always use the rule: r = aₙ / aₙ₋₁ (divide any term by its preceding term). This rule works for all types of common ratios.


The common ratio can be an integer, fraction, or negative value.


4. Can a sequence have a finite sum to infinity?

Yes, an infinite Geometric Progression (GP) can have a finite sum, but only if its common ratio 'r' is between -1 and 1 (i.e., |r| < 1). The sum of an infinite AP is never finite. This is a key concept in Class 11 Maths Chapter 8 question answer sets.


5. Is the Sequences and Series Class 11 NCERT PDF just a digital copy of the textbook?

No, the NCERT Solutions for Class 11 Maths Chapter 8 in PDF format is a separate solution manual, not the textbook itself. It is a companion resource designed to explain the answers to the textbook exercises.


This common misunderstanding leads students to look for theory in the solutions file. Instead, this PDF contains detailed, step-by-step workings for every exercise question in the NCERT chapter on Sequences and Series. It focuses on demonstrating *how* to arrive at the correct answer using the right formulas and methods.




6. Are the terms 'sequence' and 'series' interchangeable?

No, a sequence is a list of numbers arranged in a specific order (e.g., 2, 4, 6, 8), whereas a series is the sum of the terms of that sequence (e.g., 2 + 4 + 6 + 8). The key difference is between the ordered list and its calculated sum.


7. Do I need to be online to use the Sequences and Series ncert pdf solutions?

No, once you download the Sequences and series NCERT PDF solutions, you can access and use them completely offline. This makes them a reliable study tool without needing a constant internet connection.




8. Are the solutions in the PDF only for the difficult questions?

No, the Sequences and Series Class 11 questions and answers PDF provides solutions for every single question in the NCERT exercise, from the easiest to the most complex. This ensures complete coverage so you can check your work for any problem without gaps.


9. Is the Arithmetic Mean (A.M.) just the average of two numbers?

No, the Arithmetic Mean (A.M.) can be calculated for any number of terms, and the concept extends to inserting multiple A.M.s between two given numbers. The simple formula (a+b)/2 for two numbers is just the most basic case.


The main idea is that when you insert A.M.s between two numbers, you create a complete Arithmetic Progression. For example, to insert three A.M.s (A₁, A₂, A₃) between 3 and 19, you form the AP: 3, A₁, A₂, A₃, 19. Here, a = 3 and the 5th term is 19. Using the formula aₙ = a + (n-1)d, we get 19 = 3 + (5-1)d, which gives a common difference d = 4.




10. Must a sequence always be infinite?

No, a sequence can be either finite or infinite. A finite sequence has a defined last term (e.g., the first 10 terms of an AP), while an infinite sequence continues without end (e.g., the set of all natural numbers). The Sequences And Series NCERT PDF covers both types.


11. Are NCERT Solutions a shortcut to avoid practicing Sequences and Series?

No, the ncert solution Class 11 Maths Chapter 8 is a verification and learning tool, not a shortcut to skip practice. Using them effectively requires you to first attempt the problems on your own.