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Class 11 Maths Chapter 10 Exercise 10.3 - Straight Lines NCERT Solutions
The NCERT Solutions for Class 11 Maths Chapter 10, Exercise 10.3, focuses on the concept of Straight Lines. This exercise is crucial for students preparing for the Class 11 Maths CBSE exam. Each solution is thoroughly explained by experts to ensure students fully understand the concepts covered in this exercise.
In Class 11 Maths Chapter 10 Exercise 10.3, you will learn about parallel and perpendicular lines, the distance of a point from a line, and the general and normal form of a straight line. The questions in this exercise are mostly formula-based, requiring students to apply standard equations with various parameters to solve them.
These solutions will help you grasp the fundamentals of straight lines and improve your problem-solving skills for the exam.
Q1. Reduce the following equations into slope intercept form and find their slopes and the y intercepts.
(i) x + 7y =0
(ii) 6x + 3y -5 = 0
(iii) y=0
Solution. The slope-intercept form of the equation is
y = mx + c
Where m is the slope of the line and c is the y-intercept
(i) The given equation is
x + 7y =0
Rearranging the given equation into the standard form
7y = -x
Comparing it with the standard equation
m = -1/7, c = 0
Therefore slope of the equation is -1/7 and y intercept is 0
Q3.Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y -12 = 0
(ii) 4x -3y =6
(iii) 3y +2 = 0
Solution. Equation of the line in the intecept form is given below
Where a and b are the x-intercept on the x-axis and y-intercept on the y axis
3x + 2y -12 = 0
3x +2y=12
Dividing the equation by 2
Comparing with the standard equation,we get a =4 and b=6
Hence x-intercept of the equation is 4 and y-intercept is 6
Q3. Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between the perpendicular and the positive x-axis.
(i) x -√3 y +8 = 0
Solution. The normal form of the equation Ax +By + C = 0 is x cosω +y sinω =p
Rearranging the given equation
-x +√3 y = 8( since C>0)
Where A =-1, B= √3, C = 8, so √(A² +B²) =√(1 +3) = 2
Since sinω is positive and cosω is negative, so ω lies in second quadrant
cosω =- cos 60° = cos(180°-60°) = cos 120°⇒ω =120°
also sinω =sin60° = sin(180°-60°)= sin 120°⇒ω=120°
Putting the values of ω and p in the normal form
x cos120° +y sin120° =4
The perpendicular distance from the origin is 4 units and the angle between the perpendicular and the positive x-axis is 120°
Q4. Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2)
Solution. The given line is 12(x + 6) = 5(y – 2)
Rearranging the line in the general form Ax + By + C = 0
12x + 72 = 5y -10
12x – 5y + 82 = 0, where A = 12, B = -5 and C = 82
Perpendicular distance(d) of the point (x1,y1) from the line Ax + By + C = 0
∴Distance of the point (–1, 1) from the given line 12x – 5y + 82 = 0
Hence the distance of the point (-1,1) from the given line is 5 units
Q5.Find the points on the x-axis, whose distances from the line x/3 + y/4 = 1 are 4 units.
Ans. Let the points on the x axis is (a,0) whose distance from the line is 4
The given line is x/3 + y/4 = 1
Rewriting the given line into the general form Ax +By +C =0
4x + 3y -12 = 0
Perpendicular distance(d) of the point (x1,y1) from the line Ax + By + C = 0
Where d = 4, x1= a,y1=0
4a -12 = 20 or 4a -12 = -20
4a = 32 or 4a = -8
a = 8 or a = -2
Therefore the points on the x axis is (8,0) or (-2,0) whose distances from the given line is 4 units
Q6. Find the distance between parallel lines
(i) 15x +8y -34 = 0 and 15x + 8y +31 = 0
Solution. The distance(d) between two parallel lines Ax + By+ C1=0 and Ax + By + C2=0 is given by
∴ Distance between two parallel lines 15x +8y -34 = 0 and 15x + 8y +31 = 0, where C1
= -34, C2= 31, A =15 and B = 8
Therefore the distance between two given lines is 65/17
Q7. Find equation of the line parallel to the line 3x − 4y + 2 = 0 and passing through the point (-2, 3).
Solution. The given line is 3x − 4y + 2 = 0 which is passing through the point (-2,3)
Rearranging the given line
4y = 3x +2
Comparing with y = mx + c,where m is the slope of line and c is the x intercept
m = 3/4 and c = 2
The equation of a line passing through the point (x1,y1) is given as follows
(y – y1) = m(x – x1)
The equation of the line passing through (-2,3)
(y – 3) = m(x +2)
Two parallel lines have the same slope
4y -12 = 3x + 6
3x – 4y +18 = 0
Therefore the equation of the line parallel to given line is 3x – 4y +18 = 0
Q8. Find the equation of the line perpendicular to the line x – 7y + 5 = 0 and having x-intercept 3.
Solution. The given line is x – 7y + 5 = 0
Rewriting the equation in the form of y = mx +c, where m is the slope and c is the x-intercept
7y = x +5
Let’s find out the equation of the line which is perpendicular to the given line and has x-intercept 3
Two lines with the slopes m1 and m2 are perpendicular to each other when
m1 m2 = -1
The slope of given line is m1 =1/7
∴m2 = -1/m1 = -7
The equation of the line with slope m and passing through the points (x1,y1)
(y – y1) = m(x – x1)
We are given x intercept is 3 that means the line is passing through (3,0)
Putting the value m=m2 =-7 and x1 = 3 and y1= 0
(y -0) = -7(x – 3)
y = -7x + 21
7x +y-21 = 0
Hence required equation of the line is 7x +y-21 = 0
Q9.Find angles between the lines √3x + y = 1 and x + √3y = 1.
Solution. The given lines are √3x + y = 1 and x +√3y = 1
Rearranging the given lines into the slope intercept form, y = mx + c
So, y = -√3x + 1.. (i) and
y = -1/√3x + 1/√3x.. (ii)
Slope of line (i) is m1 = -√3, while the slope of line (ii) is m2 = -1/√3
Let θ be the angle between two lines
So,
Taking positive sign
tan θ= – 1/√3 = – tan 30° = tan (180° – 30°) = tan 150°
θ = 150 °
Taking negative sign
tan θ= – (-1/√3) = 1/√3 = tan 30°
θ = 30°
Hence the angle between the lines is either 30° or 150°
Q10.The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y -19 = 0. At right angle. Find the value of h.
Solution. The slope of the line segment joining the points (h,3) and (4,1),
m1=(1-3)/(4-h)= -2/(4-h)
The given line is 7x – 9y -19 = 0.
-9y = -7x +19
y = -7/(-9)x +19/(-9)
y = (7/9)x -19/9
m2= 7/9( Since y = mx +c)
The angle between the both lines is 90°
The angle between the two lines which have slopes m1and m2
The condition of two lines to be perpendicular to each other
m1 m2=-1
(-2/(4-h) (7/9) =-1
-14 = -9(4-h)
14 = 36 -9h
9h = 22
h = 22/9
Q11.Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x -x1) + B (y – y1) = 0.
Solution. The given line is Ax + By + C =0
We know the slopes of two parallel lines are equal to each other
Therefore writting the given equation into the slope intercept form, y= mx +c
By = -Ax -C
y = (-A/B)x – C/B
m = -A/B
The equation of the line passing from (x1, y1) with the slope m is given as
(y- y1) = m(x – x1 )
(y- y1) = (-A/B) (x – x1 )
B(y- y1) = -A(x- x1)
A(y- y1) + B(x- x1) =0,Hence proved
Q12.Two lines passing through the point (2, 3) intersects each other at an angle of 60o. If slope of one line is 2, find equation of the other line.
Solution. Let the slope other line is m
The angle between the lines is given as 60°
Taking positive sign
√3 = (m-2)/(1 +2m)
√3 (1 +2m) = m- 2
√3 + 2√3 m = m-2
(2√3 -1)m =-2 -√3
m= -(2 +√3)/(2√3 -1)
For this value of m and passing through the point (2,3), the equation of line is
y -3 =[ -(2 +√3)/(2√3 -1)](x -2)
y(2√3 -1) = -(2 +√3)(x -2)
y(2√3 -1) = -(2 +√3)x+ 2(2 +√3)
y(2√3 -1) +(2 +√3)x- 2(2 +√3) =0
Similarly taking the negative value and evaluating the value of m from the equation
1/√3 = -(m-2)/(1 +2m), we can get another equation of the line
We hope that Class 11 Maths Chapter 10 – Straight Lines Exercise 10.3 is now clearer to you. If you still have any questions or need further clarification about this exercise, don’t hesitate to ask in the comments. We’re here to help you succeed!
NCERT Solutions of Class 11 Maths
| Chapter 1-Sets | Chapter 9-Sequences and Series |
| Chapter 2- Relations and functions | Chapter 10- Straight Lines |
| Chapter 3- Trigonometry | Chapter 11-Conic Sections |
| Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |
| Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |
| Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |
| Chapter 7- Permutations and Combinations | Chapter 15- Statistics |
| Chapter 8- Binomial Theorem | Chapter 16- Probability |
