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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. How do you find the slope of a line using different forms of equations as per NCERT Class 11 Maths Chapter 8?

According to the NCERT solutions for the 2025-26 syllabus, the method to find the slope (m) depends on the information given:

  • Slope-Intercept Form (y = mx + c): The slope is the coefficient 'm' of x.
  • Point-Slope Form (y - y₁) = m(x - x₁): The slope is directly given as 'm'.
  • Two-Point Form: If two points (x₁, y₁) and (x₂, y₂) are known, the correct method is to calculate the slope using the formula m = (y₂ - y₁) / (x₂ - x₁).
  • General Equation (Ax + By + C = 0): The slope is calculated as m = -A/B.

2. What is the correct method to prove that three given points are the vertices of a right-angled triangle using the concept of slopes from Chapter 8?

To solve this as per the NCERT methodology, you should not use the Pythagoras theorem. Instead, follow these steps:

  • Let the vertices be A, B, and C. Calculate the slopes of the three lines formed by these points: slope of AB (m₁), slope of BC (m₂), and slope of AC (m₃).
  • Check if the product of the slopes of any two lines is equal to -1.
  • If m₁ × m₂ = -1, or m₂ × m₃ = -1, or m₁ × m₃ = -1, it proves that two lines are perpendicular.
  • The presence of a perpendicular pair of lines confirms that the vertices form a right-angled triangle.

3. How do you solve problems on the collinearity of three points using the slope formula in NCERT Chapter 8?

The step-by-step NCERT solution to check if three points A, B, and C are collinear is based on the principle that collinear points lie on the same straight line and thus must have the same slope. The correct method is:

  • Calculate the slope of the line segment AB.
  • Calculate the slope of the line segment BC.
  • If the slope of AB is equal to the slope of BC, the points A, B, and C are collinear. This is a fundamental technique used in many problems in the chapter.

4. What is the approach for solving questions in the Miscellaneous Exercise of Chapter 8, Straight Lines?

The Miscellaneous Exercise in Chapter 8 requires a comprehensive understanding of all concepts. To solve these problems correctly:

  • First, ensure you have mastered the individual concepts like slope calculation, different forms of line equations (point-slope, slope-intercept, intercept form), angle between two lines, and distance formulas.
  • These problems often combine two or more concepts. For example, a question might require you to find the equation of a line and then calculate its distance from a point.
  • The NCERT Solutions demonstrate how to break down these complex problems into smaller, manageable steps, which is key to finding the correct answer.

5. What is a common mistake made when finding the angle between two lines, and how do NCERT Solutions help prevent it?

A common mistake is forgetting to use the absolute value in the formula for the angle (θ) between two lines. Students often calculate tan θ = (m₂ - m₁)/(1 + m₁m₂), which may result in a negative value and only give the obtuse angle. The NCERT Solutions for the 2025-26 session consistently use the correct formula tan θ = |(m₂ - m₁)/(1 + m₁m₂)|. This ensures you always find the acute angle between the lines, which is the standard convention unless specified otherwise.

6. Why is the concept of slope (m = tan θ) so fundamental to solving most problems in Chapter 8, Straight Lines?

The concept of slope is fundamental because it numerically defines a line's two most critical properties: its steepness and direction. This single value 'm' is the key to:

  • Determining the relationship between lines (parallel if m₁=m₂, perpendicular if m₁m₂=-1).
  • Calculating the angle of inclination with the x-axis.
  • Deriving the various forms of a line's equation, as almost every form (like point-slope and slope-intercept) directly uses the slope.
  • Solving problems related to collinearity and geometric shapes.

7. When solving NCERT problems, how do you decide which form of a linear equation is the most efficient to use?

Choosing the right form saves time and prevents errors. The NCERT Solutions implicitly guide this choice:

  • Use Point-Slope Form [ (y - y₁) = m(x - x₁) ] when you know one point on the line and its slope.
  • Use Slope-Intercept Form [ y = mx + c ] when the slope and the y-intercept are known.
  • Use Two-Point Form [ y - y₁ = (y₂-y₁)/(x₂-x₁) * (x - x₁) ] when the coordinates of two points on the line are given.
  • Use Intercept Form [ x/a + y/b = 1 ] when the x-intercept (a) and y-intercept (b) are known. This is the most direct method for such problems.

8. How does the solution for finding the distance of a point from a line build upon concepts from earlier exercises in Chapter 8?

The formula for the distance of a point from a line is an application that synthesizes earlier concepts. The solution process demonstrates this connection:

  • To use the distance formula, you first need the equation of the line in the general form (Ax + By + C = 0).
  • Deriving this equation often requires using foundational skills from the chapter, such as calculating the slope (m) from given points or conditions.
  • Therefore, finding the distance is not an isolated skill; it's a multi-step problem that relies on your ability to first define the line using the basic principles of slope and linear equations taught at the start of the chapter.