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**NCERT Solutions for Class 11 Maths Chapter 10: Straight Lines Exercise 10.3-** In the previous exercises, you have already learned about the basic two-dimensional geometry slope of the line, conditions for parallelism and perpendicularity of lines, angles between two lines, collinearity of three points, equations of lines in various forms, etc. In the NCERT syllabus for Class 11 Maths chapter 10 exercise 10.3, you will learn about the general equation of the line in various forms like slope-intercept form, intercept form, normal form, etc. Also, the distance of a point from a line and the distance between two parallel lines are covered in exercise 10.3 Class 11 Maths. The Class 11 Maths chapter 10 exercise 10.3 is a very important exercise from the CBSE exam point of view.

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- NCERT Solutions for Class 11 Maths Chapter 10– Straight Lines Exercise 10.3
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- Key Features of Exercise 10.3 Class 11 Maths Solutions
- NCERT Solutions of Class 11 Subject Wise
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You must try to solve NCERT book Class 11th Maths chapter 10 exercise 10.3 problems by yourself. Most of the problems in the as well as exercise 10.3 Class 11 Maths are very basic and straightforward which you will be able to solve easily. You can refer to the Class 11 Maths chapter 10 exercise 10.3 solutions, meticulously crafted by subject experts at Careers360. These solutions are presented in detail and in easy-to-understand language. Additionally, PDF versions of the solutions are available, allowing students to access them at any time without incurring any charges. If you are looking for NCERT Solutions, click on the given link to get NCERT solutions at one place.

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****As per the CBSE Syllabus for 2023-24, this chapter has been renumbered as Chapter 9.**

Question:1(i) Reduce the following equations into slope - intercept form and find their slopes and the - intercepts.

Answer:

Given equation is

we can rewrite it as

-(i)

Now, we know that the Slope-intercept form of the line is

-(ii)

Where m is the slope and C is some constant

On comparing equation (i) with equation (ii)

we will get

and

Therefore, slope and y-intercept are respectively

Question:1(ii) Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts.

Answer:

Given equation is

we can rewrite it as

-(i)

Now, we know that the Slope-intercept form of line is

-(ii)

Where m is the slope and C is some constant

On comparing equation (i) with equation (ii)

we will get

and

Therefore, slope and y-intercept are respectively

Question:1(iii) Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts.

Answer:

Given equation is

-(i)

Now, we know that the Slope-intercept form of the line is

-(ii)

Where m is the slope and C is some constant

On comparing equation (i) with equation (ii)

we will get

and

Therefore, slope and y-intercept are respectively

Question:2(i) Reduce the following equations into intercept form and find their intercepts on the axes.

Answer:

Given equation is

we can rewrite it as

-(i)

Now, we know that the intercept form of line is

-(ii)

Where a and b are intercepts on x and y axis respectively

On comparing equation (i) and (ii)

we will get

a = 4 and b = 6

Therefore, intercepts on x and y axis are 4 and 6 respectively

Question:2(ii)Reduce the following equations into intercept form and find their intercepts on the axes.

Answer:

Given equation is

we can rewrite it as

-(i)

Now, we know that the intercept form of line is

-(ii)

Where a and b are intercepts on x and y axis respectively

On comparing equation (i) and (ii)

we will get

and

Therefore, intercepts on x and y axis are and -2 respectively

Question:2(iii) Reduce the following equations into intercept form and find their intercepts on the axes.

Answer:

Given equation is

we can rewrite it as

Therefore, intercepts on y-axis are

and there is no intercept on x-axis

Answer:

Given equation is

we can rewrite it as

Coefficient of x is -1 and y is

Therefore,

Now, Divide both the sides by 2

we will get

we can rewrite it as

Now, we know that the normal form of the line is

Where is the angle between perpendicular and the positive x-axis and p is the perpendicular distance from the origin

On comparing equation (i) and (ii)

we wiil get

Therefore, the angle between perpendicular and the positive x-axis and perpendicular distance from the origin is respectively

Answer:

Given equation is

we can rewrite it as

Coefficient of x is 0 and y is 1

Therefore,

Now, Divide both the sides by 1

we will get

we can rewrite it as

Now, we know that normal form of line is

Where is the angle between perpendicular and the positive x-axis and p is the perpendicular distance from the origin

On comparing equation (i) and (ii)

we wiil get

Therefore, the angle between perpendicular and the positive x-axis and perpendicular distance from the origin is respectively

Answer:

Given equation is

Coefficient of x is 1 and y is -1

Therefore,

Now, Divide both the sides by

we wiil get

we can rewrite it as

Now, we know that normal form of line is

Where is the angle between perpendicular and the positive x-axis and p is the perpendicular distance from the origin

On compairing equation (i) and (ii)

we wiil get

Therefore, the angle between perpendicular and the positive x-axis and perpendicular distance from the origin is respectively

Question:4 Find the distance of the point from the line .

Answer:

Given the equation of the line is

we can rewrite it as

Now, we know that

where A and B are the coefficients of x and y and C is some constant and is point from which we need to find the distance

In this problem A = 12 , B = -5 , c = 82 and = (-1 , 1)

Therefore,

Therefore, the distance of the point from the line is 5 units

Question:5 Find the points on the x-axis, whose distances from the line are units.

Answer:

Given equation of line is

we can rewrite it as

Now, we know that

In this problem A = 4 , B = 3 C = -12 and d = 4

point is on x-axis therefore = (x ,0)

Now,

Now if x > 3

Then,

Therefore, point is (8,0)

and if x < 3

Then,

Therefore, point is (-2,0)

Therefore, the points on the x-axis, whose distances from the line are units are (8 , 0) and (-2 , 0)

Question:6(i) Find the distance between parallel lines and .

Answer:

Given equations of lines are

and

it is given that these lines are parallel

Therefore,

Now,

Therefore, the distance between two lines is

Question:6(ii) Find the distance between parallel lines and

Answer:

Given equations of lines are

and

it is given that these lines are parallel

Therefore,

Now,

Therefore, the distance between two lines is

Question:7 Find equation of the line parallel to the line and passing through the point .

Answer:

It is given that line is parallel to line which implies that the slopes of both the lines are equal

we can rewrite it as

The slope of line =

Now, the equation of the line passing through the point and with slope is

Therefore, the equation of the line is

Question:8 Find equation of the line perpendicular to the line and having intercept .

Answer:

It is given that line is perpendicular to the line

we can rewrite it as

Slope of line ( m' ) =

Now,

The slope of the line is

Now, the equation of the line with intercept i.e. (3, 0) and with slope -7 is

Question:9 Find angles between the lines and .

Answer:

Given equation of lines are

and

Slope of line is,

And

Slope of line is ,

Now, if is the angle between the lines

Then,

Therefore, the angle between the lines is

Question:10 The line through the points and intersects the line at right angle. Find the value of .

Answer:

Line passing through points ( h ,3) and (4 ,1)

Therefore,Slope of the line is

This line intersects the line at right angle

Therefore, the Slope of both the lines are negative times inverse of each other

Slope of line ,

Now,

Therefore, the value of h is

Question:11 Prove that the line through the point and parallel to the line is

Answer:

It is given that line is parallel to the line

Therefore, their slopes are equal

The slope of line ,

Let the slope of other line be m

Then,

Now, the equation of the line passing through the point and with slope is

Hence proved

Answer:

Let the slope of two lines are respectively

It is given the lines intersects each other at an angle of and slope of the line is 2

Now,

Now, the equation of line passing through point (2 ,3) and with slope is

-(i)

Similarly,

Now , equation of line passing through point (2 ,3) and with slope is

-(ii)

Therefore, equation of line is or

Question:13 Find the equation of the right bisector of the line segment joining the points and .

Answer:

Right bisector means perpendicular line which divides the line segment into two equal parts

Now, lines are perpendicular which means their slopes are negative times inverse of each other

Slope of line passing through points and is

Therefore, Slope of bisector line is

Now, let (h , k) be the point of intersection of two lines

It is given that point (h,k) divides the line segment joining point and into two equal part which means it is the mid point

Therefore,

Now, equation of line passing through point (1,3) and with slope -2 is

Therefore, equation of line is

Question:14 Find the coordinates of the foot of perpendicular from the point to the line .

Answer:

Let suppose the foot of perpendicular is

We can say that line passing through the point is perpendicular to the line

Now,

The slope of the line is ,

And

The slope of the line passing through the point is,

lines are perpendicular

Therefore,

Now, the point also lies on the line

Therefore,

On solving equation (i) and (ii)

we will get

Therefore,

Question:15 The perpendicular from the origin to the line meets it at the point . Find the values of and .

Answer:

We can say that line passing through point is perpendicular to line

Now,

The slope of the line passing through the point is ,

lines are perpendicular

Therefore,

- (i)

Now, the point also lies on the line

Therefore,

Therefore, the value of m and C is respectively

Question:16 If and are the lengths of perpendiculars from the origin to the lines and , respectively, prove that .

Answer:

Given equations of lines are and

We can rewrite the equation as

Now, we know that

In equation

Similarly,

in the equation

Now,

Hence proved

Question:17 In the triangle with vertices , and , find the equation and length of altitude from the vertex .

Answer:

Let suppose foot of perpendicular is

We can say that line passing through point is perpendicular to line passing through point

Now,

Slope of line passing through point is ,

And

Slope of line passing through point is ,

lines are perpendicular

Therefore,

Now, equation of line passing through point (2 ,3) and slope with 1

-(i)

Now, equation line passing through point is

Now, perpendicular distance of (2,3) from the is

-(ii)

Therefore, equation and length of the line is and respectively

Answer:

we know that intercept form of line is

we know that

In this problem

On squaring both the sides

we will get

Hence proved

Class 11th Maths chapter 10 exercise 10.3 consists of questions related to changing the equations of lines in different forms, finding the slope of the line given the equation of the line, the distance between two parallel lines, the distance between a point and a line, equation of the line passing through a point and parallel to a line, coordinates of the foot of the perpendicular from the point to a line, etc.

**Also Read| **Straight Lines Class 11 Notes

Most of the important concepts related to straight-line are covered in exercise 10.3 Class 11 Maths.

Although this ex 10.3 class 11 is a bit lengthy, you must try to solve all the problems of Class 11 Maths chapter 10 exercise 10.3 by yourself.

If you are not able to solve NCERT problems at first, Class 11 Maths chapter 10 exercise 10.3 solutions are here to help you.

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Download EBook**Complete Exercise Coverage:**11th class maths exercise 10.3 answers encompass all exercises 10.3 problems in the Class 11 Mathematics textbook.**Step-by-Step Approach:**Ex 10.3 class 11 solutions are presented in a clear, step-by-step format to aid understanding and application of problem-solving methods.**Clarity and Precision:**Written with clarity and precision, ensuring easy comprehension of mathematical concepts and techniques.**Proper Mathematical Notation:**Utilizes appropriate mathematical notations and terminology, fostering fluency in mathematical language.**Conceptual Understanding:**Class 11 maths ex 10.3 solutions focus on deepening understanding rather than memorization, encouraging critical thinking.**Free Accessibility:**Class 11 ex 10.3 solutions are typically available free of charge, providing an accessible resource for self-study.**Supplementary Learning Tool:**Class 11 Maths Chapter 10 exercise 10.3 Solutions serve as supplementary resources to complement classroom instruction and support exam preparation.**Homework and Practice:**Students can use solutions to cross-verify work, practice problem-solving, and enhance overall math skills.

**Happy learning!!!**

1. Find the slope of line 3y + x = 1 ?

Given line 3y + x = 1

3y = -x + 1

y = -x/3 + 1/3

Compare with y = mx + c

Slope of the line (m) = -1/3

2. Write the equation of line passing through (1,0) and slope is 1 ?

Equation of line with slope 1 => y = x + c

Line pass through (1,0)

0 = 1 + c

c = -1

Equation of line => y = x - 1

3. Write equation of line passing through (0,0) and parallel to line y = 2x + 3 ?

Equation of line parallel to line (y = 2x + 3 ) => y = 2x + c

Line pass through (0,0) => 0 = 0 + c => c = 0

Equation of line => y = 2x

4. Find the slope of line 2y = 3x -1 ?

Given line 2y = 3x -1

y = 3x/2 - 1/2

Compare with y = mx + c

Slope (m) = 3/2

5. What is the weightage of straight line in the JEE Main ?

Chapter straight lines has 6.6% weightage in the JEE Main exam.

6. Do I need to buy NCERT solutions book for Class 11 Chemistry ?

No, You don't need to buy any NCERT solutions book for any class. Here you will get NCERT Solutions for Class 11 Chemistry.

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