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Edited By Ramraj Saini | Updated on Sep 23, 2023 06:43 PM IST

**NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines **are provided here.** **In earlier classes, you have studied 2D coordinate geometry. This NCERT Book chapter is a continuation of the coordinate geometry to study the simplest geometric figure – a straight line. A straight line is a line which is not bent or curved. In this article, you get straight lines class 11 NCERT solutions. These NCERT Solutions are prepared by experts keeping in mind CBSE latest syllabus 2023. The Important topics included in the chapter 10 class 11 maths are 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. Also, students can practice NCERT solutions for class 11 to command class 11th concepts which are foundation for class 12th concepts.

**JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | **

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This Story also Contains

- Straight Lines Class 11 Questions And Answers
- Straight Lines Class 11 Questions And Answers PDF Free Download
- Straight Lines Class 11 Solutions - Important Formulae
- Straight Lines Class 11 NCERT Solutions (Intext Questions and Exercise)
- Class 11 maths chapter 10 NCERT solutions - Topics
- NCERT solutions for class 11 mathematics - Chapter Wise
- Key Feature of class 11 maths chapter 10 NCERT solutions
- NCERT solutions for class 11 - Subject wise

In this NCERT Book ch 10 maths class 11, there are 3 exercises with 52 questions. All these questions are explained in comprehensively by Careers360 experts. 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.

**Also read:**

**Distance Formula:**

The distance between two points A(x_{1}, y_{1}) and B (x_{2}, y_{2}) is given by:

AB = √((x

_{2}-x_{1})^{2}+ (y_{2}-y_{1})^{2})

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download EBook**Distance of a Point from the Origin:**

The distance of a point A(x, y) from the origin O(0, 0) is given by:

OA = √(x

^{2}+ y^{2})

**Section Formula for Internal Division:**

The coordinates of the point which divides the line segment joining (x_{1}, y_{1}) and (x_{2}, y_{2}) in the ratio m:n internally is:

((mx

_{2}+ nx_{1})/(m + n), (my_{2}+ ny_{1})/(m + n))

**Section Formula for External Division:**

The coordinates of the point which divides the line segment joining (x_{1}, y_{1}) and (x_{2}, y_{2}) in the ratio m:n externally is:

((mx

_{2}- nx_{1})/(m - n), (my_{2}- ny_{1})/(m - n))

**Midpoint Formula:**

The midpoint of the line segment joining (x_{1}, y_{1}) and (x_{2}, y_{2}) is:

((x

_{1}+ x_{2})/2, (y_{1}+ y_{2})/2)

**Coordinates of Centroid of a Triangle:**

For a triangle with vertices (x_{1}, y_{1}), (x_{2}, y_{2}), and (x_{3}, y_{3}), the coordinates of the centroid are:

((x

_{1}+ x_{2}+ x_{3})/3, (y_{1}+ y_{2}+ y_{3})/3)

**Area of a Triangle:**

The area of a triangle with vertices (x_{1}, y_{1}), (x_{2}, y_{2}), and (x_{3}, y_{3}) is given by:

(1/2) |x

_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})|

**Collinearity of Points:**

If the points (x_{1}, y_{1}), (x_{2}, y_{2}), and (x_{3}, y_{3}) are collinear, then:

x

_{1}(y_{2}– y_{3}) + x_{2}(y_{3}– y_{1}) + x_{3}(y_{1}– y_{2}) = 0

**Slope or Gradient of a Line:**

The slope (m) of a line passing through points P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) is given by:

m = (y

_{2}- y_{1})/(x_{2}- x_{1})

**Angle between Two Lines:**

The angle (θ) between two lines with slopes m_{1} and m_{2} is given by:

tan θ = |(m

_{2}- m_{1})/(1 + m_{1}m_{2})|

**Point of Intersection of Two Lines:**

Let the equations of lines be a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0. The point of intersection is:

((b

_{1}c_{2}- b_{2}c_{1})/(a_{1}b_{2}- a_{2}b_{1}), (a_{2}c_{1}- a_{1}c_{2})/(a_{1}b_{2}- a_{2}b_{1}))

**Distance of a Point from a Line:**

The perpendicular distance (d) of a point P(x_{1}, y_{1}) from the line Ax + By + C = 0 is given by:

d = |(Ax

_{1}+ By_{1}+ C)/√(A^{2}+ B^{2})|

**Distance Between Two Parallel Lines:**

The distance (d) between two parallel lines y = mx + c_{1} and y = mx + c_{2} is given by:

d = |c

_{1}- c_{2}|/√(1 + m^{2})

**Different Forms of Equation of a Line:**

Various forms of the equation of a line include:

Normal form: x cos α + y sin α = p

Intercept form: x/a + y/b = 1

Slope-intercept form: y = mx + c

One-point slope form: y - y

_{1}= m(x - x_{1})

Two-point form: y - y_{1} = ((y_{2} - y_{1})/(x_{2} - x_{1}))(x - x_{1})

Free download **NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines **for CBSE Exam.

**Straight lines class 11 solutions - Exercise: 10.1 **

**Question:1 ** Draw a quadrilateral in the Cartesian plane, whose vertices are and . Also, find its area.

** Answer: **

Area of ABCD = Area of ABC + Area of ACDNow, we know that the area of a triangle with vertices is given byTherefore,Area of triangle ABC Similarly,Area of triangle ACD Now,Area of ABCD = Area of ABC + Area of ACD

** Answer: **

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 :

** Answer: **

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 :

** Answer: **

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

** Question:4 ** Find a point on the x-axis, which is equidistant from the points and .

** Answer: **

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

** Answer: **

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

** Answer: **

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

** Answer: **

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

** Question:8 ** Find the value of for which the points and are collinear.

** Answer: **

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

** Question:9 ** Without using the distance formula, show that points and are the vertices of a parallelogram.

** Answer: **

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 =

The slope of AD

=

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

** Question:10 ** Find the angle between the x-axis and the line joining the points and .

** Answer: **

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

** Answer: **

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,

** Question:12 ** A line passes through and . If slope of the line is , show that .

** Answer: **

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

Now,

Hence proved

** Question:13 ** If three points and lie on a line, show that

** Answer: **

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

** Answer: **

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

** Straight lines class 11 solutions - Exercise: 10.2 **

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

Write the equations for the -and -axes.

** Answer: **

Equation of x-axis is y = 0

and

Equation of y-axis is x = 0

** Question:2 ** Find the equation of the line which satisfy the given conditions:

Passing through the point with slope .

** Answer: **

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

** Question:3 ** Find the equation of the line which satisfy the given conditions:

** Answer: **

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

** Question:4 ** Find the equation of the line which satisfy the given conditions:

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

** Answer: **

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

** Question:5 ** Find the equation of the line which satisfy the given conditions:

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

** Answer: **

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

** Question:6 ** Find the equation of the line which satisfy the given conditions:

** Answer: **

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

** Question:7 ** Find the equation of the line which satisfy the given conditions:

Passing through the points and .

** Answer: **

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

** Question:8 ** Find the equation of the line which satisfy the given conditions:

** Answer: **

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

** Question:9 ** The vertices of are and . Find equation of the median through the vertex .

** Answer: **

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

** Question:10 ** Find the equation of the line passing through and perpendicular to the line through the points and .

** Answer: **

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

** Answer: **

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

** Answer: **

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

** Question:13 ** Find equation of the line passing through the point and cutting off intercepts on the axes whose sum is .

** Answer: **

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

** Answer: **

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

** Question:15 ** The perpendicular from the origin to a line meets it at the point , find the equation of the line.

** Answer: **

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

** Answer: **

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

** Answer: **

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

** Question:18 ** is the mid-point of a line segment between axes. Show that equation of the line is .

** Answer: **

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

** Question:19 ** Point divides a line segment between the axes in the ratio . Find equation of the line.

** Answer: **

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

** Question:20 ** By using the concept of equation of a line, prove that the three points and are collinear.

** Answer: **

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

** Straight lines class 11 NCERT solutions - Exercise: 10.3 **

** 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

** Straight lines NCERT solutions - Miscellaneous Exercise **

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

** 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

** 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

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

**If you are interested in class 11 maths chapter 10 question answer of exercises then these are listed below.**

- Straight Lines Class 11 exercises 10.1
- Straight Lines Class 11 exercises 10.2
- Straight Lines Class 11 exercises 10.3
- Straight Lines Class 11 miscellaneous exercises

chapter-1 | Sets |

chapter-2 | Relations and Functions |

chapter-3 | Trigonometric Functions |

chapter-4 | Principle of Mathematical Induction |

chapter-5 | Complex Numbers and Quadratic equations |

chapter-6 | Linear Inequalities |

chapter-7 | Permutation and Combinations |

chapter-8 | Binomial Theorem |

chapter-9 | Sequences and Series |

chapter-10 | Straight Lines |

chapter-11 | Conic Section |

chapter-12 | Introduction to Three Dimensional Geometry |

chapter-13 | Limits and Derivatives |

chapter-14 | Mathematical Reasoning |

chapter-15 | Statistics |

chapter-16 | Probability |

**Clear and Concise Explanations**: The NCERT Solution of ch 10 maths class 11 provide clear and concise explanations for all the concepts covered in the chapter. This makes it easier for students to understand and retain the information.

**Step-by-Step Solutions:** Each problem in the NCERT Solutions of maths chapter 10 class 11 is solved step-by-step, making it easy for students to follow the logic and understand the solution.

**Revision Notes:** The class 11 maths chapter 10 question answer also include revision notes that summarize the key concepts covered in the chapter. This makes it easier for students to revise the chapter before exams.

NCERT solutions for class 11 biology |

NCERT solutions for class 11 maths |

NCERT solutions for class 11 chemistry |

NCERT solutions for Class 11 physics |

**Important**** Formulas **

- 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 NCERT solutions for class 11 maths chapter 10 straight lines.

** Happy Reading !!! **

1. What are the important topics of the chapter Straight Lines ?

The slope of a line, various forms of the equation of a line, the general equation of a line, the distance of a point from a line are the important chapters of this chapter. Important topics are enumerated in NCERT syllabus.

2. Mention the topics and subtopics covered in NCERT solution of class 11th maths chapter 10.

The contents of straight lines class 11 NCERT solutions are structured as follows:

10.1 Introduction

10.2 Slope of a Line

- 10.2.1 Calculation of slope when coordinates of any two points on the line are given
- 10.2.2 Criteria for parallelism and perpendicularity of lines based on their slopes
- 10.2.3 Calculation of angle between two lines
- 10.2.4 Determination of collinearity of three points

10.3 Various Forms of the Equation of a Line

- 10.3.1 Discussion of horizontal and vertical lines
- 10.3.2 Derivation of point-slope form
- 10.3.3 Derivation of two-point form
- 10.3.4 Derivation of slope-intercept form
- 10.3.5 Derivation of intercept-form
- 10.3.6 Derivation of normal form

10.4 General Equation of a Line

- 10.4.1 Explanation of different forms of Ax + By + C = 0

10.5 Distance of a Point From a Line

- 10.5.1 Computation of distance between two parallel lines.

3. Explain the equation of a line in general form as per class 11 chapter 10 maths.

Class 11 maths ch 10 question answer can be expressed in various forms to represent a graphical line. Among these forms, the most prevalent one is the general equation, which is used to represent a line in two variables, namely, x and y, of the first degree. The general equation, Ax + By + C = 0, where A and B are non-zero constants and C is a constant that belongs to the set of real numbers, is commonly used. The variables x and y represent the respective axes' coordinates.

4. Where can I find the complete solutions of NCERT for class 11 maths ?

Here you will get the detailed NCERT solutions for class 11 maths by clicking on the link. Both careers360 official website and this article listed NCERT book solutions. Interested students refer them and also for easy they can study straight lines class 11 pdf both online and offline mode.

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