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NCERT Solutions for Class 11 Maths Chapter 10 : Straight Lines Miscellaneous Exercise- In this chapter, you have already learned about the slope of the line, parallel lines, perpendicular lines, angles between two lines, colinearity of three points, equations of a line in various forms, the distance of a point from a line, etc. In the Class 11 Maths Chapter 10 miscellaneous exercise solutions, you will get mixed equations from all the concepts given in this Class 11 NCERT syllabus chapter. If you are done with previous exercises, you can start solving problems from the miscellaneous exercise.
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Miscellaneous exercise chapter 10 Class 11 is considered to be tougher than the other exercises of this chapter, so you may not able to solve them by yourself at first. NCERT solutions for Class 11 Maths chapter 10 miscellaneous exercise are here to help you to solve them very easily. Questions from the miscellaneous exercises are not much asked in the CBSE exam but these exercises are very important for engineering entrance exams like JEE Main, SRMJEE, etc. The class 11 maths ch 10 miscellaneous exercise solutions are meticulously crafted by highly qualified subject experts from Careers360. These solutions offer detailed and comprehensive explanations, ensuring a thorough understanding of the concepts. Additionally, the availability of a PDF version allows students the flexibility to access the class 11 chapter 10 maths miscellaneous solutions offline, anytime and anywhere, free of charge. This resource is designed to facilitate convenient and effective learning for students studying Straight Lines in Class 11 Mathematics. If you are looking for NCERT Solutions, you can click on the above link where you will NCERT solutions from Classes 6 to 12 at one place.
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**In the CBSE Syllabus for 2023-24, this miscellaneous exercise class 11 chapter 10 has been renumbered as Chapter 9.
Question:1(a) Find the values of for which the line is
Parallel to the x-axis.
Answer:
Given equation of line is
and equation of x-axis is
it is given that these two lines are parallel to each other
Therefore, their slopes are equal
Slope of is ,
and
Slope of line is ,
Now,
Therefore, value of k is 3
Question:1(b) Find the values of for which the line is
Parallel to the y-axis.
Answer:
Given equation of line is
and equation of y-axis is
it is given that these two lines are parallel to each other
Therefore, their slopes are equal
Slope of is ,
and
Slope of line is ,
Now,
Therefore, value of k is
Question:1(c) Find the values of k for which the line is Passing through the origin.
Answer:
Given equation of line is
It is given that it passes through origin (0,0)
Therefore,
Therefore, value of k is
Question:2 Find the values of and , if the equation is the normal form of the line .
Answer:
The normal form of the line is
Given the equation of lines is
First, we need to convert it into normal form. So, divide both the sides by
On comparing both
we will get
Therefore, the answer is
Answer:
Let the intercepts on x and y-axis are a and b respectively
It is given that
Now, when
and when
We know that the intercept form of the line is
Case (i) when a = 3 and b = -2
Case (ii) when a = -2 and b = 3
Therefore, equations of lines are
Question:4 What are the points on the -axis whose distance from the line is units.
Answer:
Given the equation of the line is
we can rewrite it as
Let's take point on y-axis is
It is given that the distance of the point from line is 4 units
Now,
In this problem
Case (i)
Therefore, the point is -(i)
Case (ii)
Therefore, the point is -(ii)
Therefore, points on the -axis whose distance from the line is units are and
Question:5 Find perpendicular distance from the origin to the line joining the points and .
Answer:
Equation of line passing through the points and is
Now, distance from origin(0,0) is
Answer:
Point of intersection of the lines and
It is given that this line is parallel to y - axis i.e. which means their slopes are equal
Slope of is ,
Let the Slope of line passing through point is m
Then,
Now, equation of line passing through point and with slope is
Therefore, equation of line is
Answer:
given equation of line is
we can rewrite it as
Slope of line ,
Let the Slope of perpendicular line is m
Now, the ponit of intersection of and is
Equation of line passing through point and with slope is
Therefore, equation of line is
Question:8 Find the area of the triangle formed by the lines and .
Answer:
Given equations of lines are
The point if intersection of (i) and (ii) is (0,0)
The point if intersection of (ii) and (iii) is (k,-k)
The point if intersection of (i) and (iii) is (k,k)
Therefore, the vertices of triangle formed by three lines are
Now, we know that area of triangle whose vertices are is
Therefore, area of triangle is
Question:9 Find the value of so that the three lines and may intersect at one point.
Answer:
Point of intersection of lines and is
Now, must satisfy equation
Therefore,
Therefore, the value of p is
Question:10 If three lines whose equations are and are concurrent, then show that .
Answer:
Concurrent lines means they all intersect at the same point
Now, given equation of lines are
Point of intersection of equation (i) and (ii)
Now, lines are concurrent which means point also satisfy equation (iii)
Therefore,
Hence proved
Question:11 Find the equation of the lines through the point which make an angle of with the line .
Answer:
Given the equation of the line is
The slope of line ,
Let the slope of the other line is,
Now, it is given that both the lines make an angle with each other
Therefore,
Now,
Case (i)
Equation of line passing through the point and with slope
Case (ii)
Equation of line passing through the point and with slope 3 is
Therefore, equations of lines are and
Answer:
Point of intersection of the lines and is
We know that the intercept form of the line is
It is given that line make equal intercepts on x and y axis
Therefore,
a = b
Now, the equation reduces to
-(i)
It passes through point
Therefore,
Put the value of a in equation (i)
we will get
Therefore, equation of line is
Question:13 Show that the equation of the line passing through the origin and making an angle with the line is .
Answer:
Slope of line is m
Let the slope of other line is m'
It is given that both the line makes an angle with each other
Therefore,
Now, equation of line passing through origin (0,0) and with slope is
Hence proved
Question:14 In what ratio, the line joining and is divided by the line ?
Answer:
Equation of line joining and is
Now, point of intersection of lines and is
Now, let's suppose point divides the line segment joining and in
Then,
Therefore, the line joining and is divided by the line in ratio
Question:15 Find the distance of the line from the point along the line .
Answer:
point lies on line
Now, point of intersection of lines and is
Now, we know that the distance between two point is given by
Therefore, the distance of the line from the point along the line is
Answer:
Let be the point of intersection
it lies on line
Therefore,
Distance of point from is 3
Therefore,
Square both the sides and put value from equation (i)
When point is
and
When point is
Now, slope of line joining point and is
Therefore, line is parallel to x-axis -(i)
or
slope of line joining point and
Therefore, line is parallel to y-axis -(ii)
Therefore, line is parallel to x -axis or parallel to y-axis
Answer:
Slope of line OA and OB are negative times inverse of each other
Slope of line OA is ,
Slope of line OB is ,
Now,
Now, for a given value of m we get these equations
If
Question:18 Find the image of the point with respect to the line assuming the line to be a plane mirror.
Answer:
Let point is the image of point w.r.t. to line
line is perpendicular bisector of line joining points and
Slope of line ,
Slope of line joining points and is ,
Now,
Point of intersection is the midpoint of line joining points and
Therefore,
Point of intersection is
Point also satisfy the line
Therefore,
On solving equation (i) and (ii) we will get
Therefore, the image of the point with respect to the line is
Question:19 If the lines and are equally inclined to the line , find the value of .
Answer:
Given equation of lines are
Now, it is given that line (i) and (ii) are equally inclined to the line (iii)
Slope of line is ,
Slope of line is ,
Slope of line is ,
Now, we know that
Now,
and
It is given that
Therefore,
Now, if
Then,
Which is not possible
Now, if
Then,
Therefore, the value of m is
Answer:
Given the equation of line are
Now, perpendicular distances of a variable point from the lines are
Now, it is given that
Therefore,
Which is the equation of the line
Hence proved
Question:21 Find equation of the line which is equidistant from parallel lines and .
Answer:
Let's take the point which is equidistance from parallel lines and
Therefore,
It is that
Therefore,
Now, case (i)
Therefore, this case is not possible
Case (ii)
Therefore, the required equation of the line is
Answer:
From the figure above we can say that
The slope of line AC
Therefore,
Similarly,
The slope of line AB
Therefore,
Now, from equation (i) and (ii) we will get
Therefore, the coordinates of . is
Question:23 Prove that the product of the lengths of the perpendiculars drawn from the points and to the line is .
Answer:
Given equation id line is
We can rewrite it as
Now, the distance of the line from the point is given by
Similarly,
The distance of the line from the point is given by
Hence proved
Answer:
point of intersection of lines and (junction) is
Now, person reaches to path in least time when it follow the path perpendicular to it
Now,
Slope of line is ,
let the slope of line perpendicular to it is , m
Then,
Now, equation of line passing through point and with slope is
Therefore, the required equation of line is
As the name suggests miscellaneous exercise chapter 10 class 11 contains the mixed kind of questions from all the concepts you have learned in the previous exercises of this chapter. There are six solved examples given in the NCERT textbook before this exercise. You must go through these solved examples before solving this exercise. You must solve problems from this exercise to get conceptual clarity. Also, it will check your understanding of the concepts.
The Miscellaneous Exercise in NCERT Solutions for Class 11 Maths Chapter 10 - Straight Lines covers the following topics:
Mathematics, being a subject that benefits from consistent practice, requires a thorough understanding of the best problem-solving methods. By practising with the help of these NCERT Solutions for Class 11 Maths, students can enhance their problem-solving skills, ultimately leading to the potential to score high marks in examinations.
Comprehensive Coverage: Solutions cover all themiscellaneous exercise class 11 chapter 10, addressing various aspects of Straight Lines.
Step-by-Step Solutions: Each class 11 maths miscellaneous exercise chapter 10 problem is solved with a step-by-step approach, facilitating a clear and logical progression of concepts for Class 11 students.
Clarity in Language: The class 11 chapter 10 miscellaneous exercise solutions are presented in clear and concise language, making complex geometric concepts associated with Straight Lines more accessible for students.
PDF Version Access: The class 11 maths ch 10 miscellaneous exercise solutions are available in a downloadable PDF format, providing students with offline access for study anytime and anywhere.
Happy learning!!!
Equation of line parallel to y-axis => x = c
Line passing through the point (5,4) => 5 = c
Equation of line => x = 5
Equation of line parallel to x-axis => y = c
Line passing through the point (5,4) => 4 = c
Equation of line => y = 4
Line make angle of 45 degree with positive x-axis => slope of line = 1
Equation of line with slope 1 => y = x + c
Line pass through the (1,0) => 0 = 1 + c => c=-1
Equation of line => y = x -1
Given line 4y = 3x - 6
y = 3x/4 - 6/4
Compare with y = mx + c
Slope (m) = 3/4
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