# NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines: In earlier classes, you have studied 2D coordinate geometry. This chapter is a continuation of the coordinate geometry to study the simplest geometric figure – a straight line. The word ‘straight’ means without a bend or not curved. A straight line is a line which is not bent or curved. In this article, you get NCERT solutions for class 11 maths chapter 10 straight lines. A straight line is the simplest figure in the geometry but it is the most important concept of geometry. Important topics like definition of the straight line, the slope of the line, collinearity between two points, the angle between two points, horizontal lines, vertical lines, general equation of a line, conditions for being parallel or perpendicular lines, the distance of a point from a line are covered in this chapter. In this chapter, there are 3 exercises with 52 questions. All these questions are explained in solutions of NCERT for class 11 maths chapter 10 straight lines. This chapter is very important for CBSE class 11 final examination as well as in various competitive exams like JEEmains, JEEAdvanced, BITSAT etc. There are 24 questions are given in a miscellaneous exercise. You should solve all the NCERT problems including examples and miscellaneous exercise to get command on this chapter. You can take help of CBSE NCERT solutions for class 11 maths chapter 10 straight lines which are prepared in a detailed manner. Check all NCERT solutions from class 6 to 12 at a single place to learn science and mathsThere are three exercises and a miscellaneous exercise in this chapter which are explained below.

Exercise:10.1

Exercise:10.2

Exercise:10.3

Miscellaneous Exercise

## Topics of NCERT Grade 11 Maths Chapter-10 Straight Lines

10.1 Introduction

10.2 Slope of a Line

10.3 Various Forms of the Equation of a Line

10.4 General Equation of a Line

10.5 Distance of a Point From a Line

## NCERT solutions for class 11 maths chapter 10 straight lines-Exercise: 10.1 Area of ABCD = Area of ABC + Area of ACD
Now, we know that the area of a triangle with vertices   is given by

Therefore,
Area of triangle ABC
Similarly,
Area of triangle ACD
Now,
Area of ABCD = Area of ABC + Area of ACD it is given that it is an equilateral triangle and length of all sides is 2a
The base of the triangle lies on y-axis such origin is the midpoint
Therefore,
Coordinates of point  A and B are   respectively
Now,
Apply Pythagoras theorem in triangle AOC

Therefore, coordinates of vertices of the triangle are

Question:3(i) Find the distance between and    when :

PQ is parallel to the -axis.

When PQ is parallel to the y-axis
then, x coordinates are equal i.e.
Now, we know that the distance between two points is given by

Now, in this case
Therefore,

Therefore, the distance between and    when  PQ is parallel to y-axis  is

Question:3(ii) Find the distance between    and    when :

When PQ is parallel to the x-axis
then, x coordinates are equal i.e.
Now, we know that the distance between two points is given by

Now, in this case
Therefore,

Therefore, the distance between and    when  PQ is parallel to the x-axis  is

Point is on the x-axis, therefore, y coordinate is 0
Let's assume the point is (x, 0)
Now, it is given that the given point (x, 0) is equidistance from point  (7, 6) and (3, 4)
We know that
Distance between two points is given by

Now,

and

Now, according to the given condition

Squaring both the sides

Therefore, the point is

Mid-point of the line joining  the points and  . is

It is given that line also passes through origin which means passes through the point (0, 0)
Now, we have two points on the line so we can now find the slope of a line by using formula

Therefore, the slope of the line is

It is given that  point A(4,4) , B(3,5) and C(-1,-1) are the vertices of a right-angled triangle
Now,
We know that the distance between two points is given by

Length of AB
Length of BC
Length of AC
Now, we know that  Pythagoras theorem is

Is clear that

Hence proved

It is given that the line makes an angle of   with the positive direction of  -axis measured anticlockwise
Now, we know that

line makes an angle of   with the positive direction of  -axis
Therefore, the angle made by line with the positive x-axis is =
Now,

Therefore, the slope of the line is

Point is collinear which means they lie on the same line by this we can say that their slopes are equal
Given points are A(x,-1) , B(2,1) and C(4,5)

Now,
The slope of AB = Slope of BC

Therefore, the value of x is 1

Given points are    and
We know the pair of the opposite side are parallel to each other in a parallelogram
Which means their slopes are also equal

The slope of AB =

The slope of BC =

The slope of CD =

We can clearly see that
The slope of AB = Slope of CD               (which means they are parallel)
and
The slope of BC = Slope of AD               (which means they are parallel)
Hence pair of opposite sides are parallel to each other
Therefore, we can say that points    and   are  the vertices of a parallelogram

We know that

So, we need to find the slope of line joining points (3,-1) and (4,-2)
Now,

Therefore, angle made by line with positive x-axis when measure in anti-clockwise direction is

Let   are the slopes of lines and  is the angle between them
Then, we know that

It is given that        and

Now,

Now,

Now,

According to which value of
Therefore,

Given that  A line passes through    and    and slope of the line is m
Now,

Hence proved

Points    and    lie on a line so by this we can say that their slopes are also equal
We know that

Slope of AB =

Slope of AC =
Now,
Slope of AB = slope of AC

Now divide both the sides by hk

Hence proved Given point A(1985,92) and B(1995,97)
Now, we know that

Therefore, the slope of line AB is
Now, the equation of the line passing through the point (1985,92) and with slope =    is given by

Now, in the year 2010 the population is

Therefore, the population in the year 2010 is 104.5 crore

## Question:1 Find the equation of the line which satisfy the given conditions:

Write the equations for the  -and  -axes.

Equation of x-axis is y = 0
and
Equation of y-axis is x = 0

Passing through the point    with slope  .

We know that , equation of line passing through point  and with slope m is given by

Now,  equation of line passing through point (-4,3) and with slope  is

Therefore, equation of the line  is

Passing through  with slope .

We know that the equation of the line passing through the point  and with slope m is given by

Now, the equation of the line passing through the point (0,0) and with slope m is

Therefore, the equation of the line  is

Passing through    and inclined with the x-axis at an angle of .

We know that the equation of the line passing through the point  and with slope m is given by

we know that

where  is angle made by line with positive x-axis measure in the anti-clockwise direction

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

Therefore, the equation of the line  is

Intersecting the -axis at a distance of  units to the left of origin with slope

We know that the equation of the line passing through the point  and with slope m is given by

Line Intersecting the -axis at a distance of  units to the left of origin which means the point is (-3,0)
Now, the equation of the line passing through the point (-3,0) and with slope -2  is

Therefore, the equation of the line  is

Intersecting the -axis at a distance of units above the origin and making an angle of   with positive direction of the x-axis.

We know that , equation of line passing through point  and with slope m is given by

Line Intersecting the y-axis at a distance of 2 units above the origin which means point is (0,2)
we know that

Now, the equation of the line passing through the point (0,2) and with slope   is

Therefore, the equation of the line  is

Passing through the points    and .

We know that , equation of line passing through point  and with slope m is given by

Now, it is given that line passes throught point (-1 ,1) and (2 , -4)

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

Perpendicular distance from the origin is units and the angle made by the  perpendicular with the positive -axis is

It is given that length of perpendicular is 5 units  and  angle made by the  perpendicular with the positive -axis is
Therefore, equation of line is

In this case p = 5 and

Therefore, equation of the line  is The vertices of   are    and
Let m be RM b the median through vertex R
Coordinates of M (x, y ) =
Now, slope of line RM

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

equation of line passing through point (0 , 2) and with slope  is

Therefore, equation of median is

It is given that the line passing through    and perpendicular to the line through the points    and
Let the slope of the line passing through the point (-3,5) is m and
Slope of line  passing through points (2,5) and (-3,6)

Now this line is perpendicular to line passing through point (-3,5)
Therefore,

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

equation of line passing through point (-3 , 5) and with slope  5 is

Therefore, equation of line  is

Co-ordinates of point which divide line segment joining the points    and    in the ratio  is

Let the slope of the perpendicular line is m
And Slope of  line segment joining the points    and   is

Now, slope of perpendicular line is

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

equation of line passing through point  and with slope  is

Therefore, equation of line  is

Let (a, b) are the intercept on x and y-axis respectively
Then, the equation of the line is given by

Intercepts are equal which means a = b

Now, it is given that line passes through the point (2,3)
Therefore,

therefore, equation of the line is

Let (a, b) are the intercept on x and y axis respectively
Then, the equation of line is given by

It is given that
a + b = 9
b = 9 - a
Now,

It is given that line passes through point (2 ,2)
So,

case (i)  a = 6  b = 3

case (ii)   a = 3 , b = 6

Therefore, equation of line is 2x + y = 6 , x + 2y = 6

We know that

Now, equation of line passing through point (0 , 2) and with slope  is

Therefore, equation of line is                   -(i)

Now, It is given that line crossing the -axis at a distance of units below the origin which means coordinates are  (0 ,-2)
This line is parallel to above line which means slope of both the lines are equal
Now, equation of line passing through point (0 , -2) and with slope  is

Therefore, equation of line is

Let the slope of the line is m
and slope of a perpendicular line is which passes through the origin (0, 0) and (-2, 9) is

Now, the slope of the line is

Now, the equation of line passes through the point (-2, 9) and with slope   is

Therefore, the equation of the line is

It is given that
If  then
and  If    then
Now, if assume C along x-axis and L along y-axis
Then, we will get coordinates of two points (20 , 124.942)  and (110 , 125.134)
Now, the relation between C and L is given by equation

Which is the required relation

It is given that the owner of a milk store sell
980 litres milk each week at
and    litres of milk each week at
Now, if we assume the rate of milk as x-axis and Litres of milk as y-axis
Then, we will get coordinates of two points i.e.  (14, 980)  and   (16, 1220)
Now, the relation between  litres of milk and Rs/litres is given by equation

Now, at  he could sell

He could sell 1340 litres of milk each week at Now, let coordinates of point A is (0 , y) and of point B is (x , 0)
The,

Therefore, the coordinates of point A is (0 , 2b) and of point B is (2a , 0)
Now, slope of line passing through points (0,2b) and (2a,0) is

Now, equation of line passing through point (2a,0) and with slope    is

Hence proved Let the coordinates of Point A is (x,0) and of point B is (0,y)
It is given that point R(h , k) divides the line segment between the axes in the ratio
Therefore,
R(h , k)

Therefore, coordinates of point A is   and of point B is
Now, slope of line passing through points  and   is

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

Therefore, the equation of line is

Points are collinear means they lies on same line
Now,  given points are    and
Equation of line passing through point A and B is

Therefore, the equation of line passing through A and B is

Now, Equation of line passing through point B and C is

Therefore, Equation of line passing through point B and C is
When can clearly see that  Equation of line passing through point A nd B  and through B and C is the same
By this we can say that points   and   are collinear points

CBSE NCERT solutions for class 11 maths chapter 10 straight lines-Exercise: 10.3

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

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

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

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

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

Given equation is

we can rewrite it as

Therefore,  intercepts on y-axis are
and there is no intercept on x-axis

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

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

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

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

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  .

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

Given equations of lines are
and
it is given that these lines are parallel
Therefore,

Now,

Therefore, the distance between two lines is

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

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   .

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

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

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

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

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

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,

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

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 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

we know that intercept form of line is

we know that

In this problem

On squaring both the sides
we will get

Hence proved

NCERT solutions for class 11 maths chapter 10 straight lines-Miscellaneous Exercise

Question:1(a) Find the values of   for which the line   is

Parallel to the x-axis.

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.

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

Given equation of line is

It is given that it passes through origin (0,0)
Therefore,

Therefore, value of k is

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

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

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

Equation of line passing through the points     and   is

Now, distance from origin(0,0) is

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

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

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

Point of intersection of lines  and  is
Now,  must satisfy equation
Therefore,

Therefore, the value of p is

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

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

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

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

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 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

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 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 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

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

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

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 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

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

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

## NCERT solutions for class 11 mathematics

 chapter-1 NCERT solutions for class 11 maths chapter 1 Sets chapter-2 Solutions of NCERT for class 11 chapter 2 Relations and Functions chapter-3 CBSE NCERT solutions for class 11 chapter 3 Trigonometric Functions chapter-4 NCERT solutions for class 11 chapter 4 Principle of Mathematical Induction chapter-5 Solutions of NCERT for class 11 chapter 5 Complex Numbers and Quadratic equations chapter-6 CBSE NCERT solutions for class 11 maths chapter 6 Linear Inequalities chapter-7 NCERT solutions for class 11 maths chapter 7 Permutation and Combinations chapter-8 Solutions of NCERT for class 11 maths chapter 8 Binomial Theorem chapter-9 CBSE NCERT solutions for class 11 maths chapter 9 Sequences and Series chapter-10 NCERT solutions for class 11 maths chapter 10 Straight Lines chapter-11 Solutions of NCERT for class 11 maths chapter 11 Conic Section chapter-12 CBSE NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry chapter-13 NCERT solutions for class 11 maths chapter 13 Limits and Derivatives chapter-14 Solutions of NCERT for class 11 maths chapter 14 Mathematical Reasoning chapter-15 CBSE NCERT solutions for class 11 maths chapter 15 Statistics chapter-16 NCERT solutions for class 11 maths chapter 16 Probability

## NCERT solutions for class 11- Subject wise

 Solutions of NCERT for class 11 biology CBSE NCERT solutions for class 11 maths NCERT solutions for class 11 chemistry Solutions of NCERT for Class 11 physics

## Important points to remember from NCERT solutions for class 11 maths chapter 10 straight lines-

• If a non-vertical line passing through the points  and  then the slope(m) of the line is given by-

• The slope of the line which makes an angle with the positive x-axis is given by .
• The slope of the horizontal line is zero and the slope of the vertical line is undefined.
• An acute angle () between lines   and   with slopes  and  is given by-

• Two lines (with slopes  and ) are parallel if and only if their slopes () are equal.
• Two lines (with slopes  and ) are perpendicular if and only if the product of their slopes is –1 or .
• The equation of a line passing through the points and  is given by-

Tip- If you facing difficulties in the memorizing the formulas, you should write formula every time when you are solving the problem. You should try to solve every problem on your own and reading the solutions won't be much helpful. You can take help from CBSE NCERT solutions for class 11 maths chapter 10 straight lines.