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NCERT Solutions for Class 11 Maths Chapter 10: Straight Lines - Exercise 10.1

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NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.1 (Ex 10.1) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 10 Straight Lines Exercise 10.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 11

Subject:

Class 11 Maths

Chapter Name:

Chapter 10 - Straight Lines

Exercise:

Exercise - 10.1

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

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Competitive Exams after 12th Science

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.1

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.Find the value of x for which the points \[\left( {x, - 1} \right)\], $\left( {2,1} \right)$ and $\left( {4,5} \right)$ are collinear.

Ans: If points A\[\left( {x, - 1} \right)\], B$\left( {2,1} \right)$, and C$\left( {4,5} \right)$ are collinear, then 

Slope of AB = Slope of BC 

\[\begin{array}{l} \Rightarrow \frac{{1 - \left( { - 1} \right)}}{{2 - x}} = \frac{{5 - 1}}{{4 - 2}}\\\Rightarrow \frac{{1 + 1}}{{2 - x}} = \frac{4}{2}\\ \Rightarrow \frac{2}{{2 - x}} = 2\\ \Rightarrow 2 = 4 - 2x\\\Rightarrow 2x = 2\\ \Rightarrow x = 1\end{array}\]

Thus, the required value of x is 1.


9. 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. 


10.  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 \[135^\circ \] 


11. 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 \[1\]


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


13. If three points $\left( {h,0} \right),$$\left( {a,b} \right)$and$\left( {0,k} \right)$, lie on a line, show that$\frac{a}{h} + \frac{b}{k} = 1$. 

Ans: If the points A$\left( {h,0} \right),$B$\left( {a,b} \right)$ and C$\left( {0,k} \right)$ lie on a line, then 

Slope of AB = Slope of BC

$\frac{{b - 0}}{{a - h}} = \frac{{k - b}}{{0 - a}}$

$\begin{array}{l}\Rightarrow \frac{b}{{a - h}} = \frac{{k - b}}{a}\\\Rightarrow  - ab = \left( {k - b} \right)\left( {a - h} \right)\\\Rightarrow  - ab = ka - kh - ab + bh\\ \Rightarrow ka + bh = kh\end{array}$

On dividing both sides by kh, we’ll get

\[\begin{array}{l}\frac{{ka}}{{kh}} + \frac{{bh}}{{kh}} = \frac{{kh}}{{kh}}\\\Rightarrow \frac{a}{h} + \frac{b}{k} = 1\end{array}\] 

Hence, $\frac{a}{h} + \frac{b}{k} = 1$


14. Consider the given population and year graph. Find the slope of the line AB and using it, find what will be the population in the year 2010? 


Square and Rectangle with same perimeter, Plot measurement, Shape of the garden, Food pieces of different shapes


Ans: Since line AB passes through points A $\left( {1985,92} \right)$and B $\left( {1995,97} \right)$, 

its slope is \[\frac{{97 - 92}}{{1995 - 1985}} = \frac{5}{{10}} = \frac{1}{2}\]

Let y be the population in the year 2010. 

Then, according to the given graph, line AB must pass through point C $\left( {2010,y} \right)$. $\therefore $Slope of AB = Slope of BC 

\[\begin{array}{l} \Rightarrow \frac{1}{2} = \frac{{y - 97}}{{2010 - 1995}}\\ \Rightarrow \frac{1}{2} = \frac{{y - 97}}{{15}}\\ \Rightarrow \frac{{15}}{2} = y - 97\\ \Rightarrow y - 97= 7.5\\\Rightarrow y = 97 + 7.5\\\Rightarrow y = 104.5\end{array}\]

Thus, the slope of line AB is \[\frac{1}{2}\], while in the year 2010, the population will be 104.5 crores.


NCERT Solutions for Class 11 Maths Chapters

 

NCERT Solution Class 11 Maths of Chapter 10 All Exercises

Chapter 10 - Straight Lines Exercises in PDF Format

Exercise 10.1

14 Questions & Solutions

Exercise 10.2

20 Questions & Solutions

Exercise 10.3

18 Questions & Solutions

Miscellaneous Exercise

24 Questions & Solutions


All Topics of NCERT Class 11 Maths for Exercise 10.1

The topics and subtopics covered under exercise 10.1 Maths Class 11 NCERT are given below.

Section 10.1 is related to the Introduction to the Straight Lines.

Section 10.2 is the most important part which is related to finding the slope of the line. This section is further divided into 4 sub-sections.

Section 10.2.1: To find the slope of a line when two points are given along with their coordinates

Section 10.2.2: We will find the slope of a line when certain conditions for perpendicularity and parallelism of lines are given

Section 10.2.3 To find the slope of a line when the angle between the two lines is given

Section 10.2.4  Conditions of collinearity of three points


NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.1

Opting for the NCERT solutions for Ex 10.1 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 10.1 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.


Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Subject Straight Lines textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 10 Exercise 10.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.


Besides these NCERT solutions for Class 11 Maths Chapter 10 Exercise 10.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.


Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 10 Exercise 10.1 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

FAQs on NCERT Solutions for Class 11 Maths Chapter 10: Straight Lines - Exercise 10.1

1. What is meant by the slope of a line?

The slope of a line is the measure of its steepness. In coordinate geometry, the letter m is used to denote the slope of a line. The direction of a line can be determined by its slope. Also, it is a measure of the change in the y coordinate for every unit change in the x coordinate, along the straight line. There are several formulas to find the slope of a given line. There are four types of slopes, namely, zero, positive, negative, and undefined. The basic formula for calculating the slope of a line is m = Δy/Δx, where Δy represents the change in the y coordinate and Δx represents the change in the x coordinate.

2. How to determine the slope of a straight line?

There are three forms of equations that can be used to find the slope of straight lines. These are standard form, point-slope form, and slope-intercept form.


Slope - Intercept Form: The slope m is used in the representation of the standard equation of a straight line, which is, y = mx + c. This type of equation is called the slope-intercept equation, wherein, ‘c’ is the y-intercept of the straight line.  


Point - Slope Form: The general form of equation used to define the slope of a straight line is, m = Δy/Δx. This equation can be elaborated as m = [y - y1]/[x - x1]. This form of the slope equation is known as the point-slope form. In this equation, (x1,y1) represents any point on the given straight line and m is the slope of the straight line. 


Standard Form: Ax + By = C This is the standard form of the equation that is used to determine the slope of a straight line. In this equation, A, B, and C are constants.

3. How will you show that a group of three given points are the vertices of a right-angled triangle, without using Pythagoras Theorem?

If any two straight lines are perpendicular to each other, then the product of their slopes is always equal to -1. For example, if (3,5), (4,4) and (-1,-1) are the vertices of a right-angled triangle, then the product of the slopes of any of the two straight lines formed with these points will be -1. Here, the straight line formed between the points (4,4) and (3,5) is -1, and the slope of the straight line formed between the points (4,4) and (-1,-1) is 1. The product of these two slopes is -1, therefore, these two lines are perpendicular to each other. The triangle formed by (3,5), (4,4), and (-1,-1) is right-angled at (4,4).

4. Why should you refer to the NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines (Ex 10.1) Exercise 10.1?

Practicing all the sums from the Class 11 NCERT maths book is an important part of the preparation for your Maths exam. These NCERT solutions available on Vedantu are prepared by our subject matter experts in a stepwise manner. You can download the PDF of these NCERT solutions for free and refer to it for clearing your doubts. Chapter 10 of NCERT Class 11 Maths covers various concepts of Straight Lines. The topics included in this chapter are,

  1. Introduction to Straight Lines

  2. The Slope of a Line

  • Collinearity of three points

  • To find the slope of a line when the coordinates of any two points lying on the line are given

  • The angle between two lines

  • Slopes of parallel lines

  • Slopes of perpendicular lines

5. Find the value of x for which the points (x, – 1), (2, 1), and (4, 5) are collinear, according to Exercise 10.1 of Chapter 10 of Class 11 Maths?

The slope of AB = Slope of BC if the points (x, – 1), (2, 1), and (4, 5) are collinear.

Then (1+1)/(2-x) = (5-1)/(4-2)

2/(2-x) = 4/2

2/(2-x) = 2

2 = 2(2-x)

2 = 4 – 2x

2x = 4 – 2

2x = 2

x = 2/2

= 1

6. What is the benefit of Vedantu’s solutions for Exercise 10.1 of Chapter 10 of Class 11 Maths?

Vedantu's in-house professionals have meticulously addressed the exercise's problems/questions while adhering to all CBSE requirements. Students of Class 11 who are thoroughly familiar with all of the topics presented in the Subject Straight Lines textbook and all of the problems presented in the exercises can easily achieve the best possible score on the final exam. Students may readily grasp the pattern of questions asked in the test from this Chapter with the help of NCERT Solutions for Exercise 10.1 of Chapter 10 of Class 11 Maths and the Chapter's marks weightage for them to be prepared for the end-term exam. These solutions are available free of cost on the Vedantu website and the Vedantu app.

7. What all things are included in Exercise 10.1 of Chapter 10 of Class 11 Maths?

Students should measure geometrical shapes and compute their area or perimeter using measures like side lengths and angles formed by neighboring sides in NCERT Solutions for Exercise 10.1 of Chapter 10 of Class 11 Maths. This activity provides us with a new tool for learning and measuring geometrical forms through the questions. Coordinate geometry is this new tool. As a result, rather than having the length of sides measured, we now have the geographic coordinates of a particular point about a fixed point termed the 'Origin.'

8. How can students solve Exercise 10.1 of Chapter 10 of Class 11 Maths?

NCERT's recommendations Exercise 10.1 of Chapter 10 of Class 11 Maths Straight Lines is about navigating the coordinate geometry environment. The essential pillars of coordinate geometry are the two perpendicular axes known as the X and Y axes. To better understand the difficulties, pupils need to have a good sense of direction. There are also important considerations about determining the slope of a line in this exercise. The tasks can ask youngsters to determine if two lines are perpendicular or parallel to one another and the angle between two intersecting lines, using this method.

9. How can students find a slope in Exercise 10.1 of Chapter 10 of Class 11 Maths?

NCERT's recommendations Exercise 10.1 of Chapter 10 of Class 11 Maths encourages students to juggle several inputs of knowledge regarding a point plotted between the X and Y axes. Because this assignment involves the usage of numerous formulas, students should create a formula chart to help them understand the concepts. After completing these problems correctly, students will calculate the slope of a straight line and interpret the results graphically.