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Edited By Ravindra Pindel | Updated on Sep 12, 2022 06:05 PM IST

**NCERT Exemplar Class 11 Maths solutions chapter 10** are prepared by our experts for helping students to understand and grasp the very basics of coordinate geometry. NCERT Exemplar solutions for Class 11 Maths chapter 10 focuses on the representation of a straight line in coordinate geometry. This is done through the determination of the slope of the straight line with reference to the previously studied concept of algebra and geometry.

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

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- NCERT Exemplar Class 11 Maths Solutions Chapter 10: Exercise: 10.3
- More About NCERT Exemplar Class 11 Maths Chapter 10
- Topics and Subtopics in NCERT Exemplar Class 11 Maths Solutions Chapter 10
- What will the students learn from NCERT Exemplar Class 11 Maths Solutions Chapter 10?
- NCERT Solutions for Class 11 Mathematics Chapters
- Important Topics To Cover in NCERT Exemplar Class 11 Maths Solutions Chapter 10

NCERT Exemplar Class 11 Maths solutions chapter 10 provide detailed and simple explanations for NCERT problems that help students in their competitive exams as well. The solutions are drafted very carefully by a detailed study of the concepts and exam pattern of CBSE, keeping the important topics in mind for better understanding and preparation.

**Also, check -** NCERT Solutions for Class 11

Question:1

Answer:

Equation of line in intercept form=

Given that

x+y=a……(1)

If equation (1) passes through (1,-2) we get 1+(-2) =a

1-2=a

a=-1 Putting the value of'a'in equation 1we get x+y=-1

Equation of straight line is x+y+1=0

Question:2

Answer:

Given points are A(5,2) B(2,3) and C(3,-1).

Slopeof the line joining points B and C =

It is given that line passing through the point A is perpendicular to BC

m_{1}*m_{2}= -1

-4*m_{2}=-1

m_{2}=1/4

Equation of line passing through point A

Equation of line: y-y_{1 }= m(x-x_{1)}

y-2=1/4 (x-5)

4y-8=x-5

x-5-4y+8=0

Equation of straight line passing through point A is x-4y+3=0

Question:3

Find the angle between the lines and

Answer:

Given equations are

................(1)

and

............(2)

In equation 1 the slope is as it is in the form of y=mx+b and in equation 2 it is

Let θ be the angle between the given m_{1} and m_{2} two lines

Putting the values of m_{1 }and m_{2 }in above equation we get

θ=60^{0}

Question:4

Answer:

Equation of line in intercept form =

Given that a+b=14 b=14-a

So equation of line

........................(1)

If equation 1 passes through (3,4) then

42-3a+4a-14a+a^{2}=0

a^{2}-13a+42=0

a^{2}-7a-6a+42=0

(a-6)(a-7)=0

a=6 or a=7

If a=6 then 6+b=14 b=8

If a=7 the 7+b=14 b=7

If a=6 and b=8 then equation of line is

4x+3y=24

If a=7 and b=7 then equation of line

x+y=7

Question:5

Find the points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10.

Answer:

Let x_{1}, y_{1 }be any point lying inthe equation x+y=4

x_{1}+y_{1}=4…......(1)

Distance of point x_{1},y_{1}from the equation 4x+3y=10

4x_{1}+3y_{1}-10=±5

4x_{1}+3y_{1}-10=5 or 4x_{1}+3y_{1}-10=-5

4x_{1}+3y_{1}=(15)...….(2) or 4x_{1}+3y_{1}=5….....(3)

From equation 1 we have y_{1}=4-x_{1}….. (4)

Putting the values of y_{1}in equation 2 we get 4x_{1}+3(4-x_{1})=15

4x_{1}+12-3x_{1}=15

x_{1}=15-12 =3

Putting the value of x_{1}in equation 4 we get y_{1}=1

Now, 4x1+34-x1=5

4x_{1}+12-3x_{1}=5

x_{1}=5-12 x_{1}=-7

Putting the value in equation 4 we get y_{1}=4--7=4+7=11

Hence, the required points on the given line are (3,1) and (-7,11)

Question:6

Show that the tangent of an angle between the lines and is

Answer:

Equation of line in intercept form ............(i) and ..............(ii)

y=mx+b Slope of equation 1 is

Similarly for equation 2 ,

Since, the above equation is in y=mx+b form

Slope of the equation 2 is

Let θ be the angle between the given two lines

Putting the values of m_{1 }and m_{2 }in above equation we get

Question:7

Find the equation of lines passing through (1, 2) and making angle 30° with y-axis.

Answer:

Given that line passes through (1,2) making an angle of 30^{0 }with y axis.

Angle made by line with x axis is 60^{0} Slope of the line, m=

Equation of line passing through x_{1},y_{1 }and having slope 'm' is y-y_{1}=m(x-x_{1)}

Here, (x_{1}, y_{1})=(1,2) and m=

Question:8

Answer:

. Given 2x+y=5…. (1) and x+3y=-8….(2)

Firstly, we find the point of intersection of both equations we get and

Hence, the point of intersection is

Now the slope of the equation 3x+4y=7 m=

Then the equation of the line passing through the point having slope is

y-y_{1}=m(x-x_{1})

Question:9

Answer:

Given equation is ax+by+8=0 or ax+by= -8

Now dividing by-8 to both sides

So the intercepts are and

Now, the second equation which is given is 2x-3y+6=0 or 2x-3y=-6

Dividing by-6 on both sides

So, the intercepts are-3 and 2

Now, according to the question and

and

Question:10

Answer:

. Let a and b be the intercepts on the given line

Coordinates of A and B are (a,0)and (0,b) respectiively.

Using the section formula we find the value of a and b

and

and

and

Coordinates of A and B are and

Equation of line AB

* *Hence, trequired equation is 8x-5y+60=0

Question:11

Answer:

Given that length of the perpendicular from the origin is 4 units and line makes an angle with positive direction of x-axis

∠BAX=120

∠BA0=180-120=60

∠MAO=60

Now in triangle AMO ∠MAO+∠AOM+∠OMA=180^{0 }

60^{0}+θ+ 90^{0}=180^{0 }

θ=30^{0} ∠AOM= 30^{0 }

x cosθ+ysinθ=p

xcos 30^{0}+ysin30^{0}=4

Question:12

Answer:

Equation of hypotenuse is 3x+4y=4 and opposite vertex is (2,2)

Slope of the equation of hypotenuse is

Now let the slope of AC be m

Putting the values of m_{1}and m_{2}in the equation

4m+3=4-3m

4m+3m=4-3

7m=1 m=

OR

4m+3=-4+3m

4m-3m=-4-3

m= -7

If m=1/7

equation of AC is y-y_{1}=m(x-x_{1})

y-2=1/7(x-2)

7y-14=x-2

x-7y-2+14=0

x-7y+12=0

If m= -7 then equation of AC is y-2=-(7)(x-2)

y-2=-7x+14

7x+y=16

The required equations are x-7y+12=0 and 7x+y=16

Question:13

Answer:

Let ABC be an equilateral triangle, BC is base, and altitude from A on BC meets at mid-point D.

Given:Equation of the base BC is x+y=2

Distance of point (x_{1},y_{1}) from the equation Ax+By+C=0

Now, length of perpendicular from vertex A(2,-1) to the line x+y=2

Squaring both the sides, we get

Question:14

Answer:

Let the variable line be ax+by=1

Length of perpendicular from (2,0) to the line ax+by-1=0

Now perpendicular distance from B(0,2) =

Now, perpendicular distance from C(1,1)=

The algebraic sum of the perpendicular from the given points (2,0), (0,2)

and (1,1) to this line is zero.

d_{1}+d_{2}+d_{3}=0

2a-1+2b-1+a+b-1=0

3a+3b-3=0

a+b-1=0

a+b=1

So, the equation ax+by=1 represents a family of straight lines passing

through a fixed point .

Comparing equation ax+by=1 and a+b=1, we get x=1 and y=1.

Coordinates of fixed point is (1,1)

Question:15

Answer:

Let the given line x+y=4 and the required line'l'intersect at B (a,b)

Slope of line'l'is

Given that

So, by distance formula for point A(1,2) and B(a,b) we get d

Point B(a,b)also satisfies x+y=4

a+b=4; b=4-a

Putting the value of b in equation (i )we get 3a^{2}+3(4-a)^{2}-6a-12(4-a)+13=0

3a^{2}+48+3a^{2}-24a-6a-48+12a+13=0

6a^{2}-18a+13=0

Using the formula

Putting the value of a in the equation we get

Now putting the value of a and b in equation

θ= 60^{0}-45^{0}=15^{0}

Similarly, taking

θ= 60^{0}+45^{0}=105^{0}

Question:16

Answer:

Equation of line in intercept form=xa+yb=1

where a and b are intercepts on the axes

Given that

Then,

This shows that the line is passing through the fixed point that is (k,k)

Question:17

Answer:

Let the line cut the x axis at (a,0) and Y axis at (0,b) Therefore

It is given that the point (-4,3) divides the line internally in 5: 3 ratio. Hence applying the formula for internal division of line segment, we get and

Hence Using slope intercept form we get

Question:18

Answer:

Given two lines are: x-y+1=0 and 2x-3y+5=0

Solving these two equations gives us points of intersection we get y=3 and x=2

(x,y)=(2,3)

Let m be the slope of the required line

Then, equation of the line is y-3=mx-2

y-3=mx-2m

mx-y-2m+3=0….(1)

Since, the perpendicular distance from the point 3,2to the line is75then

Squaring both the sides, we get

49m^{2}+1=25(m+1)^{2}

49m^{2}+49=25m^{2}+25+50m

25m^{2}+25+50m-49m^{2}-49=0

-24m^{2}+50m-24=0

-12m^{2}+25m-12=0

Factorising, we get (3m-4)(4m-3)=0

3m-4=0 or 4m-3=0

3m=4 or 4m=3

m=4/3 or 3/4

Putting the value of m=4/3 in equation 1 we get 4x/3-y-2(4/3)+3=0

4x/3-y-8/3+3=0

4x/3-y=-1/3

4x-3y+1=0

Putting the value of m=3/4 in the equation we get 3x/4-y-2(3/4)+3=0

3x/4-y-3/2+3=0

3x/4-y+3/2=0

3x-4y+6=0

Hence, the required equation are 4x-3y+1=0 and 3x-4y+6=0

Question:19

Answer:

Let the coordinates of a moving point P be (a,b)

It is given that the sum of the distance from the axes to the point is always 1 |x|+|y|=1

±x±y=1 -x-y=1, x+y=1, -x+y=1 and x-y=1

Hence, these equations gives us the locus of the point P which is a square.

Question:20

Answer:

Given lines are

If x≥0, then ..............(i)

If x<0 then .............(ii)

On adding both the equations we get

2y=4

y=2

Putting the value of y=2 in equation (ii), we get

Point of intersection of given lines is (0,2 )

Now we find the slopes of given lines Slope of equation (i) is

Comparing the above equation with y=mx+b, we get

and we know that

θ= 60^{0}

Slope of equation (ii) is , we get

θ=180^{0}- 60^{0}= 120^{0}

In ACB,

Hence, the coordinates of the foot of perpendicular

Question:21

Answer:

Equation of line in intercept form=

Since, p is the length of perpendicular drawn from the origin to the gievn line p=

Squaring both the sides, we have

Since, a^{2,}b^{2 }and p^{2 }are in AP

2p^{2}=a^{2}+b^{2 }

From equation (i) and (ii) we get

Hence, proved.

Question:22

A line cutting off intercept – 3 from the y-axis and the tangent at angle to the x-axis is 3/5, its equation is

A. 5y – 3x + 15 = 0

B. 3y – 5x + 15 = 0

C. 5y – 3x – 15 = 0

D. None of these

Answer:

. Given that tanθ=3/5

We know that slope of a line, m=tanθ Slope of line, m=3/5

Since, the lines cut off intercepts -3 on y axis so the line is passing

through the point (0,-3)

So, the equation of line is y-y_{1}=m(x-x_{1})

y-(-3)=3/5(x-0)

y+3=3x/5

5y+15=3x

5y-3x+15=0

Hence, the correct option is (a)

Question:23

Slope of a line which cuts off intercepts of equal lengths on the axes is

A. – 1

B. – 0

C. 2

D.

Answer:

Equation of line in intercept form=

Given that a=b ;

x+y=a……(1)

y=-x+a

y=(-1)x+a

Since, the above equation is in y=mx+b form

So, the slope of the line is (-1) .

Question:24

The equation of the straight line passing through the point (3, 2) and perpendicular to the line y = x is

A. x – y = 5

B. x + y = 5

C. x + y = 1

D. x – y = 1

Answer:

Given that straight line passing through the point (3,2) and is perpendicular to

line y=x

Let the equation of line L is (y-y_{1}) =m(x-x_{1})

Since, L is passing through the point (3,2)

y-2=m(x-3)..…(i)

Now, given equation is y=x

Comparing it with y=mx+b

we get slope of the equation as 1.

It is also given that line L and y=x are perpendicular to each other.

We know that when two lies are perpendicular then m_{1}*m_{2}= -1 m*1=-1

m=-1

Putting this value of m in equation (i) we get y-2=-(1)(x-3)

y-2=-x+3

x+y=5

Hence, the correct option is b

Question:25

The equation of the line passing through the point (1, 2) and perpendicular to the line x + y + 1 = 0 is

A. y – x + 1 = 0

B. y – x – 1 = 0

C. y – x + 2 = 0

D. y – x – 2 = 0

Answer:

Given that line passing through the point (1,2) and perpendicular to the line

x+y+1=0

Let the equation of line L is x-y+k=0…..(i)

Since, L is passing through the point (1,2)

1-2+k=0

k=1

Putting this value of k in equation )i) we get x-y+1=0 or y-x-1=0

Hence, the correct option is b

Question:26

The tangent of angle between the lines whose intercepts on the axes are a, – b and b, – a, respectively, is

A.

B.

C.

D. None of these

Answer:

First equation of line in intercept form=

bx-ay=ab……. (i)

Let the second equation of line having intercepts on the axes b, -a is

ax-by=ab…..(ii)

Now we find the slope in the first equation bx-ay=ab

ay=bx-ab

Slope of the equation

Now, we find the slope of equation (ii) ax-by=ab

by=ax-ab

Slope of the equation

Let θ be the angle between the given two lines

Putting the values of m1 and m2 in above equation, we get

is the required angle.

Hence, the correct option is (c)

Question:27

If the line passes through the points (2, –3) and (4, –5), then (a, b) is

A. (1, 1)

B. (– 1, 1)

C. (1, – 1)

D. (– 1, –1)

Answer:

Given points are (2, -3) and (4, -5)

Firstly the equation of line is found

We know that the equation of line when two points are given is y-y_{1}=

Putting the values we get

y+3= -1( x-2)

y+3= -x+2

x+y=2-3

x+y=-1

in intercept form

Comparing the equaton with intercept form of equation that is

the value of a=-1 and b=-1

Hence, the correct option is (d)

Question:28

Answer:

(a) Given lines are:

and

Solving these lines, we get point of intersection as

therefore Distance of this point from the line

Question:29

The equations of the lines which pass through the point (3, –2) and are inclined at 60° to the line is

A. y + 2 = 0,

B. x – 2 = 0,

C.

D. None of these

Answer:

Given equation is and θ= 60^{0 }

Slope of the equation

Slope of the equation

Let slope of the required line be m_{2}

Then

Putting the values in the above equation we gettan

Taking+sign we get

Taking -ve sign we get

m_{2}=0

Equation of line passing through (3,-2) with slope is y-y_{1}=m(x-x_{1 })

and equation of line passing through (3, -2) with slope 0 is y-y_{1}=m(x-x_{1})

y-(-2)=0(x-3)

y+2=0

Hence, the required equations are and y+2=0

Hence, the correct option is (a)

Question:30

The equations of the lines passing through the point (1, 0) and at a distance from the origin, are

A.

B.

C.

D. None of these.

Answer:

Let the equation of any line passing through the point (1,0) is y-y_{1}=m(x-x_{1})

y-0=m(x-1)

y=mx-m

mx-m-y=0….........(i)

Distance from the origin of the line is Distance of point (x_{1},y_{1}) from the equation Ax+By+C=0

Squaring both the sides we get

3(m^{2}+1)=4m^{2 }

3m^{2}+3=4m^{2}

4m^{2}-3m^{2}=3

m^{2}=3

Putting the value of m= in equation (i) we get x-y-=0

Now, putting the value of m= - in the same equation we get-x-y+=0

Hence, the correct option is (a)

Question:31

The distance between the lines y = mx + c_{1} and y = mx + c_{2} is

A.

B.

C.

D. 0

Answer:

Given equations are y=mx+c_{1}….(i) and y=mx+c_{2}…….(ii)

Firstly, we find the slope of both the equations

Since, both of them have the same slope they are parallel lines.

We know that distance between two parallel lines Ax+By+C_{1}=0 and Ax+By+C_{2}=0

Hence, the correct option is b.

Question:32

The coordinates of the foot of perpendiculars from the point (2, 3) on the line y = 3x + 4 is given by

A.

B.

C.

D.

Answer:

Given equations are y=3x+4….(i)

Comparing this equation with y=mx+b form , the slope of the equation is 3.

Let the equation of any line passing through the point (2,3) is y-y_{1}=m(x-x_{1})

y-3=m(x-2)……(ii)

Given that equation (i) is perpendicular to equation (ii)

And we know that, if two lines are perpendicular then m_{1}m_{2}= -1

3*m_{2}=-1

m_{2}=-1/3 which is the slope of the required line

Putting the value of slope in equation ii we get y-3=-1/3(x-2)

3y-9=-x+2

x+3y-9-2=0

x+3y-11=0……(iii)

Now we have to find the coordinates of foot of the perpendicular

Solving equation (i) and (iii) we get x+3(3x+4)-11=0

x+9x+12-11=0

10x+1=0

x=-1/10

Putting the value of x in equation i , we get y=3(-1/10)+4

y= -3/10+4

y=37/10

So the required coordinates are (-1/10,37/10)

Hence, the correct option is (b)

Question:33

If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is (3, 2), then the equation of the line will be

A. 2x + 3y = 12

B. 3x + 2y = 12

C. 4x – 3y = 6

D. 5x – 2y = 10

Answer:

Let the given line meets the axes at A(a,0) and B(0,b)

Given that (3,2) is the midpoint

a=6 and

b=4

Intercept form of the line AB is

Putting the value of a and b in above equation, we get

2x+3y=12

Hence, the correct option is (a)

Question:34

Equation of the line passing through (1, 2) and parallel to the line y = 3x – 1 is

A. y + 2 = x + 1

B. y + 2 = 3 (x + 1)

C. y – 2 = 3 (x – 1)

D. y – 2 = x – 1

Answer:

Given equation of the line is y=3x-1

Now we find the slope of the above equation by comparing it with y=mx+b form

So m=3

Now, we find the equation of line passing through the point (1,2) and parallel to the given line

with slope=3

y-y_{1}=m(x-x_{1})

y-2=3(x-1)

Hence, the correct option is (c)

Question:35

Equations of diagonals of the square formed by the lines x = 0, y = 0, x = 1 and y = 1 are

A. y = x, y + x = 1

B. y = x, x + y = 2

C. 2y = x, y + x = 1/3

D. y = 2x, y + 2x = 1

Answer:

It is given that the lines x=0, y=0 , x=1 and y=1 form a square of side 1 unit

Let us form a square OABC having corners O(0,0) from the given lines

with A(1,0), B(1,1) and C(0,1)

Now we have to find the equation of the diagonal AC

Equation of a line is found out by

y=-1(x-1)

y=-x+1

x+y=1

Equation of diagonal OB is

y=x

y=x

Hence, the correct option is (a)

Question: 36

For specifying a straight line, how many geometrical parameters should be known?

A. 1

B. 2

C. 4

D. 3

Answer:

Equation of straight line in intercept form=

where a and b are the intercepts on the axis.

In intercept form we need 2 parameters a and b to specify a straight line

Slope-Intercept Form y=mx+c where m=tanθ

and θ is the angle made with positive x axis and c is the intercept on y axis

So, we need two parameters 'm' and 'c' to specify a straight line

Hence, the correct option is b

Question:37

The point (4, 1) undergoes the following two successive transformations:

(i) Reflection about the line y = x

(ii) Translation through a distance 2 units along the positive x-axis. Then the final coordinates of the point are

A. (4, 3)

B. (3, 4)

C. (1, 4)

D.

Answer:

Let Q(x,y) be the reflection of P(4,1) about the line y=x, then

midpoint of PQ

which lies on y=x

4+x=1+y

x-y+3=0……(i)

Now, we find the slope of given equation y=x

Since this equation is in y=mx+b form

So the slope=m=1

Slope of

Since, PQ is perpendicular to y=x

And when two lines rae perpendicular then m_{1}m_{2}=-1

y-1=-(x-4)

x+y-5=0…..(ii)

On adding equation i and equation ii we get x-y+3+x+y-5=0

2x-2=0

x-1=0

x=1

Putting the value of x=1 in equation i we get 1-y+3=0 -y+4=0 y=4

It is given that translation through a distance of 2 units along the positive x axis takes place

The point after translaton is (1+2,4)=(3 ,4)

Hence, the correct option is (b)

Question:38

A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is

A. (1, –1)

B. (1, 1)

C. (0, 0)

D. (0, 1)

Answer:

Given equations are 4x+3+10=0 …..(i)

5x-12y+26=0…......(ii)

and 7x+24y-50=0

Let p,qbe the point which is equidistant from the givenlines

Now, we find the distance of (p,q) from the given lines

Distance of point (x_{1},y_{1}) from the equation Ax+By+C=0

Distance of (p,q) from equation i is

Distance of (p,q) from equation ii is

Distance of (p,q) from equation iii is

Given that (p,q) is equidistant from the given lines =

On putting the value of (p,q) as (0,0) we get

Hence, the correct option is c

Question:39

A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is

A. 1/3

B. 2/3

C. 1

D. 4/3

Answer:

Given line is 3x+y=3 It can be re written as y=-3x+3

Comparing it with the y=mx+b form of equation we get the slope as m= -3

So slope of the perpendicular line will be1/3

The line passes through (2,2) and has a slope1/3 is

3y-6=x-2

3y=x+4

So, the y intercept is

Hence, the correct option is (d)

Question:40

The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is

A. 1 : 2

B. 3 : 7

C. 2 : 3

D. 2 : 5

Answer:

Given lines are 3x+4y+5=0…....(i)

3x+4y-5=0…....(ii)

and 3x+4y+2=0….......(iii)

Since the coefficient of x and y are same equation i, ii and iii are parallel to each other

We know that in case of i and iii

distance between two parallel lines is

Similarly in case of ii and iii distance between two parallel lines is

Ratio between the distance is

Hence, the correct option is b

Question:41

One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is

A. (–1, –1)

B. (2, 2)

C. (–2, –2)

D. (2, –2)

Answer:

Let ABC be an equilateral triangle with vertex A (a,b)

Let AD be perpendicular to BC and let (p,q) be the coordinates of D

Given that the centroid P lies at the origin (0,0)

We know that, the centroid of a triangle divides the median in the ratio 1:2

Now, using the section formula, we get

and

a+2p=0 and b+2q=0…A…

a+2p=b+2q

2p-2q=b-a …i

It is given that BC=x+y-2=0

Since, the above equation passes through p,q

p+q-2=0

Now we find the slope of line AP

Equation of line BC is x+y-2=0

y= -x+2

y=-(1)x+2

Since the above equation is in y=mx+b form

So, slope of line BC is m_{BC}=-1

Since both lines rae perpendicular

b=a

Putting this value in equation (i )we get 2p-q=b-b=0

p=q

Now putting this value in equation (i) we get p+q-2=0

2p=2

p=1

q=1

Putting the value of p and q in equation A , we get a+2*1=0 and b+2*1=0

a=-2 and b=-2

So, the coordinates of vertex A (a,b) is (-2,-2)

Hence, the correct option is c

Question:42

Given that a,b,c are in AP

2b=a+c

a-2b+c=0….i

Now comparing equation iwith the given equation ax+by+c=0 we get x=1, y=-2

So, the line will pass through (1,-2 )

Question:43

Answer:

Equation of straight line in intercept form= where a and b are intercepts

Given that a=b

x+y=a….(i)

If equation ipasses through the point (1, -2) we get 1+(-2)=a

1-2=a

a=-1

Putting the value of a in equation i we get x+y=-1

x+y+1=0

Question:44

Answer:

Given equation is x-2y=3

x-3=2y

The slope of the equation can be found by comparing with y=mx+b form

So,

We have to find an equation which is passing through the point (3,2)

A line passing through the point x_{1},y_{1} has an equation y-y_{1}=m(x-x_{1})

So, here x_{1}=3 and y_{1}=2

y-2=m(x-3)....…(ii)

Now, it is given that the angle between the given two lines is 45^{0 }^{ }

Putting the values of m_{1}and m_{2} in above equation we get tan45^{0 = }

2m-1=2+m or-(2m-1)=2+m

2m-m=2+1 or -2m+1-m=2

m=3 or-3m=1 or m=-1/3

Putting the value of m=3 in equation (ii) we get y-2=3(x-3)

y-2=3x-9

3x-y-9+2=0

3x-y-7=0

Putting the value of m=-1/3 in equation (ii) we get y-2= -1/3(x-3)

3(y-2)=3-x

3y-6=3-x

x+3y-6-3=0

x+3y-9=0

Question:45

Answer:

Given line 3x-4y-8=0 and given points are (3,4) and (2,-6)

For point (3,4)

3(3)-4(4)-8=9-16-8

=9-24= -15<0

Forpoint (2, -6) 3(2)-4(-6)-8 =6+24-8

=30-8=22>0

So the points (3,4)and (2,-6) are situated on the opposite sides of 3x-4y-8=0

Question:46

Answer:

Given point is (3, -2) and equation of line is 5x-12y=3

Let (p,q) be any moving point

Distance between them (p,q) and (3,-2)

(d_{1})^{2}=(p-3)^{2}+(q+2)^{2}

Now, distance of the point (p,q) from the given line 5x-12y-3=0

=

Taking numerical values only, we have (p-3)^{2}+(q-2)^{2}=

13[(p-3)^{2}+(q+2)^{2}]=5p-12q-3

13[p^{2}+9-6p+q^{2}+4+4q]=5p-12q-3

On solving we get 13p^{2}+13q^{2}-83p+64q+172=0

A point moves so that square of its distance from the point (3,-2) is numerically equal to its distance from the line 5x-12y=3 .

The equation of its locus is 13p^{2}+13q^{2}-83p+64q+172=0

Question:47

Answer:

Given equation of the line is x sinθ+y cosθ=p…..(i)

Let Ph,kbe the midpoint of the given line where it meets the two axis at a,0and 0,b.

Since (a,0) lies on equation (i) then asinθ+0=p a= …..(ii)

(0,b) also lies on the equation i then 0+bcosθ=p

……(iii)

Since, P (h,k) is the midpoint of the given line

2h=a and

2k=b

Putting the value of a=2h in equation (ii) we get

…..(iv)

Putting the value of b=2k in equation (ii) we get

….(v)

Squaring and adding equation (iv) and (v), we get

or

or 4x^{2}y^{2}=p^{2}y^{2}+p^{2}x^{2}

or 4x^{2}y^{2}=p^{2}(x^{2}+y^{2}) is the locus of the mid-points of the portion of the line intercepted between the axes

Question:48

Answer:

Let ABC be a triangle with vertices A(x_{1},y_{1}), B (x_{2},y_{2}) and C (x_{3}, y_{3}) where x_{i}, y_{i}, i=1,2,3 are integers

Then area of

Since, xi and yi all are integers but is a rational number.

So, the result comes out to be a rational number. i.e . Area of ABC=a rational number

Suppose, ABC be an equilateral triangle, then area of ABC is=

It is given that vertices are integral coordinates, it means the value of coordinates is in whole

number. Therefore, the value of (AB)^{2} is also an integer.

(positive integer)

But, is an irrational number

Area of triangle ABC=an ir-rational number

This contradicts the fact that the area is a rational number

Hence, the given statement is true.

Question:49

Answer:

Given points are A(-2,1), B (0,5) and C(-1,2)

There are two ways to find that given points are collinear or not .

The first is if 3 points are collinear, then slope of any two pairs of points will be equal.

Second way is that if the value of area of triangle formed by the 3 points is zero, then the points

are collinear.

Slope of AB

Slope of BC

Slope of CA is

Since, the slopes are different.

So, the given points are not collinear

Hence, the given statement is False

Question:50

Answer:

Let the equation of line y=mx+c….(i)

So, slope of the above equation is m

Given equation of line is x secθ+y cosecθ=a sin 2θ.

Since the slope of the equation is m'=

Given that equation (i) is perpendicular to x secθ+y cosec θ=a sin 2θ

m*m'= -1

Putting the value of m in equation (i)we get

ysecθ=x cosec θ+c secθ

x cosec θ-y secθ=k….(ii)

If equation (ii) passes through the point (a cos^{3}θ, a sin^{3}θ)

(a cos^{3}θ )cosecθ-(a sin^{3}θ) secθ=k

a[(cos^{2}θ-sin^{2}θ)(cos^{2}θ+sin^{2}θ)]=x cosθ-ysinθ

a[cos^{2}θ-sin^{2}θ]=xcosθ-ysinθ

a[cos2θ]=xcos θ-y sinθ

The given statement is FALSE

Question:51

Answer:

Given equations are x+2y-10=0….(i) and 2x+y+5=0…..(ii)

The point of intersection is obtained by solving them together i.e.

If the given line 5x+4y=0 passes through the point then

0=0

So, the given line passes through the point of intersection of the given lines.

Hence, the given statement is True.

Question:52

Answer:

Let ABC be an equilateral triangle with vertex (2,3) and the equation of the opposite side is x+y=2

In the case of equilateral triangle θ=60

Let the slope of line AB is m and the slope of the given equation x+y=2 is m_{2}=-1

We know that

Putting the values of m_{1} and m_{2} in the above equation, we get

Rationalizing both the equations we get

So, the slope of line AB is

Thus, the equations of the other two lines joining the point (2, 3) are

Hence, the given statement is True.

Question:53

Answer:

Given two lines are 4x+y-1=0…(i ) and 7x-3y-35=0…..(ii)

Now, point of intersection of these lines can be find out as x=2 and y=-7

To find the equation of the line joining the point (3,5) and (2,-7)

y-5=12x-3

y-5=12x-36

12x-y-31=0……iv

Now, the distance of equation (iv) from the point (0,0) is d=

Now the distance of

equation iv from the point (8,34) is

Hence, the equation of line 12x-y-31=0 is equidistant from (0,0)and (8,34)

Hence, the given statement is True.

Question:54

Answer:

Equation of line ……i

Equation of line passing through the origin and perpendicular to the given line ….ii

Now the foot of perpendicular from origin on the line (i) is the point of intersection of lines (i) and ii

So, to find its locus we have to eliminate the variable a and b

Squaring and adding both the equations we get

x^{2}+y^{2}=c^{2}

Hence, the given statement is true.

Question:55

Answer:

Given that ax+2y+1=0 bx+3y+1=0 and cx+4y+1=0 are concurrent .

For the lines to be concurrent

Now expanding along first column we get

a[3-4]-b[2-4]+c[2-3]=0

-a+2b-c=0

2b=a+c and we know that if a, b ,c are in AP then

This means given lines are in AP not in GP

Hence, the given statement is False.

Question:56

Answer:

Given points are (3,-4), (-2,6), (-3,6) and (9,-18)

Now we find the slope since the lines are perpendicular, the product of the slopes is -1 i.e. m_{1}m_{2}= -1

Slope of the line joining the points (3,-4) and (-2,6)

Here, x_{1}=3, x_{2}=-2 , y_{1}=-4 and y_{2}=6

Now, slope of the line joining the points -(3,6) and (9,-18)

Here, x_{1}=-3, x_{2}=9, y_{1}=6 and y_{2}=-18

m_{1}=m_{2}=-2

and m_{1}m_{2}= -2*(-2)=-4 ≠-1

So, the lines are parallel and not perpendicular

Hence, the given statement is False

Question: 57

Match the questions given under Column C1 with their appropriate answers given under the Column C2.

Column C | Column C |

a) The coordinates of the points P and Q on | i) (3,1), (-7,11) |

b) The coordinates of the points on | ii) -1/3, 11/3, 4/3, 7/3 |

c) The coordinates of the points on the line joining A | iii) 1, 12/5,-3,16/5 |

Answer:

Let P (x_{1},y_{1}) be any point on the given line x+5y=13

x_{1}+5y_{1}=13

5y_{1}=13-x_{1}….(i)

Distance of the point P(x_{1},y_{1})from the equation 12x-5y+26=0

2=|x_{1}+1|

2= ±(x_{1}+1)

So x_{1}=1….(ii) or x_{1}=-3…. (iii)

Putting the value in equation (i) we get 5y_{1}=13-1=12

Putting the value of x_{1}=-3 in the same equation we get 5y_{1}=13-(-3)=16

Hence, the required points on the given line areand -3,165

Hence, (a)-(iii)

bLet P x_{1}, y_{1} be any point lying in the equation x+y=4

x_{1}+y_{1}=4…. i

Now, the distance of the point from the equation is

4x_{1}+3y_{1}-10= ±5

either 4x_{1}+3y_{1}-10=5 or 4x_{1}+3y_{1}-10=-5

4x_{1}+3y_{1}=15 ….(ii) or 4x_{1}+3y_{1}=5…..(iii)

From equation i we have y_{1}=4-x_{1}….(iv)

Putting the value of y_{1} in equation ii we get 4x_{1}+34-x_{1}=15

4x_{1}+12-3x_{1}=15

x_{1}=3 Putting the value of x_{1 }in equation (iv) we get y_{1}=4-3=1

Putting the value of y_{1 }in equation iii we get 4x_{1}+34-x_{1}=5

4x_{1}+12-3x_{1}=5

x_{1}=5-12=-7

Putting the value of x_{1} in equation (iv) , we get y_{1}=4-(-7)

y_{1}=4+7=11

Hence, the required points on the given line are (3,1) and (-7,11)

Hence, b-i

c Given that AP=PQ=QB and given points are A(-2,5) and B(3,1)

Firstly, we find the slope of the line joining the points (-2,5) and (3,1)

Slope of line joining two points=

Now equation of line passing through the point (-2, 5) y-5=-4/5[x-(-2)]

5y-25=-4(x+2)

4x+5y-17=0 Let P (x_{1,}y_{1}) and Q (x_{2},y_{2}) be any two points on the AB

P(x_{1},y_{1}) divides the line AB in the ratio 1:2

Now, Q (x_{2},y_{2}) is the midpoint of PB

Hence, the coordinates of Q (x_{2},y_{2}) is (4/3,7/3)

Hence, c-ii

Question:58

The value of the λ, if the lines (2x + 3y + 4) + λ (6x – y + 12) = 0 are

Column C | Column C |

a) parallel to y-axis is | i) |

b) perpendicular to 7x+y-4=0 | ii) |

c) Passes through (1,2) is | iii) |

d) parallel to x-axis is | iv) |

Answer:

a ) Given equation is (2x+3y+4)+ λ(6x-y+12)=0

2x+3y+4+6λx-λy+12λ=0

(2+6λ)x+(3-λ)y+4+12λ=0…. (i)

If equation i is parallel to y-axis, then 3-λ=0

λ=3

Hence, iv.

b) Given equation is 2x+3y+4+λ6x-y+12=0

2x+3y+4+6λx-λy+12λ=0

(3-λ)y= -4-12λ-(2+6λ)x

Since, the above equation is in y=mx+b form

So the slope of equation (i) is

Now the second equation is 7x+y-4=0…..(ii)

y=-7x+4

So, the slope of equation (ii) is m_{2}=-7

Now equation i is perpendicular to equation (ii)

m_{1}m_{2}=-1

(2+6λ)*7=-(3-λ)

On solving we get λ=-17/41

Hence, (b)-(iii)

c) Given equation is (2x+3y+4)+λ(6x-y+12)=0

If the above equation passes through the point (1,2) then [2*1+3*2+4]+λ[6*1-2+12]=0

2+6+4+λ(6+10)=0

12+16λ=0

12=-16λ

λ=-12/16=-3/4

Hence, (c)-(i)

d) Given equation is (2x+3y+4)+λ(6x-y+12)=0

2x+3y+4+6λx-λy+12λ=0

(2+6λ)x+(3-λ)y+4+12λ=0…..(i)

If equation (i) is parallel to x axis, then 2+6λ=0 6λ=-2

λ=-1/3

(d)-(ii)

Question:59

The equation of the line through the intersection of the lines 2x – 3y = 0 and 4x – 5y = 2 and

Column C | Column C |

a) through the point (2,1) is | i) 2x-y =4 |

b)perpendicular to the line x+2y+1=0 is | ii)x+y-5=0 |

c)parallel to the line 3x+4y+5=0 is | iii)x-y-1=0 |

d) equally inclined to the axes is | iv) 3x-4y-1=0 |

Answer:

.a) Given equations are 2x-3y=0….(i) and 4x-5y=2….(ii )

Equation of line passing through equation iand (ii), we get 2x-3y+λ4x-5y-2=0….(iii)

If the above equation passes through the point (2,1) we get (2*2-3*1)+λ(4*2-5*1-2)=0

(4-3)+λ(8-5-2)=0

1+λ=0

λ=-1

Putting the value of λ in equation iii we get (2x-3y)+(-1)(4x-5y-2)=0

2x-3y-4x+5y+2=0

-2x+2y+2=0

x-y-1=0

Hence, (a)-(iii)

b) Given equations are 2x-3y=0….(i) and 4x-5y=2….(ii)

Equation of line passing through equation i and ii, we get 2x-3y+λ4x-5y-2=0…(iii)

2x-3y+4λx-5λy-2λ=0

x(2+4λ)-y(3+5λ)-2λ=0

-y(3+5λ)= -(2+4λ)+2λ

Slope of equation iii is m_{1}=

Now, we find the slope of thegiven line x+2y+1=0…..(iv)

2y=-x-1 y=-1/2x+(-1/2)

So slope of equation (iv) is m_{2}=(-1/2)

We know that, if two lines are perpendicular to each other then the product of their slopes is equal to-1

So, m_{1}*m_{2}=-1

2+4λ=2*(3+5λ)

2+4λ=6+10λ

-6λ=4

λ=-4/6= -2/3

Putting the value of λ in equation (iii) we get (2x-3y)+(-2/3)(4x-5y-2)=0

6x-9y-8x+10y+4=0

-2x+y+4=0

2x-y=4

b-i

c) Given equations are 2x-3y=0….(i) and 4x-5y=2…..(ii)

Equation of line passing through equation iand (ii), we get (2x-3y)+ λ(4x-5y-2)=0….iii

2x-3y+4λx-5λy-2λ=0

x(2+4λ)-y(3+5λ)-2λ=0

So slope of equation iii is m_{1}=

Now, we find the slope of the given line 3x-4y+5=0

3x+5=4y y=3/4x+5/4

So slope of equation is 3/4

If 2 lines are parallel then their slopes are also equal So

4(2+4λ)=3(3+5λ)

8+16λ=9+15λ

λ=1

Putting the values of λ in equaaion iii we get (2x-3y)+(4x-5y-2)=0

2x-3y+4x-5y-2=0

6x-8y-2=0

3x-4y-1=0

Hence, c-iv

d) Given equations are 2x-3y=0…(i) and 4x-5y=2….(ii)

Equation of line passing through equation (i) and (ii), we get (2x-3y)+λ(4x-5y-2)=0

x(2+4λ)-y(3+5λ)-2λ=0

-y(3+5λ)= -(2+4λ)+2λ

Slope of equation iii is

Since the equation is equally inclined with axes, it means that the line makes equal angles with both the

coordinate axes.

It will make an angle of 45^{0} or 135^{0}

m_{2}=tan45^{0} andtan135^{0}

= 1 and-1

2+4λ=-3+5λ or 2+4λ=3+5λ

So λ=-5/9 or-1

Putting the value in equation iii we get (2x-3y)+(-5/9)(4x-5y-)2=0

18x-27y-20x+25y+10=0

-2x-2y+10=0

x+y-5=0

If λ=-1, then the required equation is (2x-3y)+(-1)( 4x-5y-2)=0

2x-3y-4x+5y+2=0

-2x+2y+2=0

x-y-1=0

d-ii, iii

The students trying to ace their examinations can access NCERT Exemplar Class 11 Maths solutions chapter 10 PDF download from here. This will help them in getting a helpful approach towards the preparation of your exams that are fabricated by our experts through thorough study.

The NCERT Exemplar solutions for Class 11 Maths chapter 10 tries to aid students with properly examined solutions from the perspective of exams and its usage in various other fields.

Also read - NCERT Solutions for Class 11 Maths Chapter 10

- Introduction
- Slope of a line
- Slope of a line when coordinates of any two points on the line are given
- Conditions for parallelism and perpendicularity of lines in terms of their slopes
- Angle between two lines
- Collinearity of three points
- Various forms of the equation of the line
- Horizontal and vertical lines
- Point-slope form
- Two-point form
- Slope-intercept form
- Intercept-form
- Normal form
- General Equation of a line
- Different forms of Ax + By + C = 0
- Distance of a point from a line
- Distance between two parallel lines

The students will get a brief introduction of coordinate geometry in two dimensions along with the main focus on one of the simplest figures, which is a straight line. NCERT Exemplar solutions for Class 11 Maths chapter 10 covers different concepts relating to a straight line that will teach students about the calculation of angle between two lines, the distance between two parallel lines, determining the slope of a line and other forms of the equation of a line. A straight line is the simplest shape of geometry, but it is also the most important part as it acts as the base for the introduction of other shapes such as curves that will help students in the further lessons. The learners will also benefit from NCERT Exemplar Class 11 Maths solutions chapter 10 through its use and application in different fields, including physics, architecture, geometry, and much more.

**Also, check -**

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download EBookSome of the important topics for students to review from this chapter are:

The learners should revise all the basics and concepts of coordinate geometry learned previously for better understanding.

NCERT Exemplar Class 11 Maths chapter 10 solutions covers the calculation of inclination of the line along with gradient or slope of a line in two-dimensional planes.

The students should study all the solutions of NCERT books along with examples to get a better understanding of concepts.

The students should cover various forms of the equation of lines from an examination perspective.

**Check Chapter-Wise NCERT Solutions of Book**

Chapter-1 | |

Chapter-2 | |

Chapter-3 | |

Chapter-4 | |

Chapter-5 | |

Chapter-6 | |

Chapter-7 | |

Chapter-8 | |

Chapter-9 | |

Chapter-10 | Straight Lines |

Chapter-11 | |

Chapter-12 | |

Chapter-13 | |

Chapter-14 | |

Chapter-15 | |

Chapter-16 |

**Read more NCERT Solution subject wise -**

- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Physics
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology

**Also, read NCERT Notes subject wise -**

- NCERT Notes for Class 11 Maths
- NCERT Notes for Class 11 Physics
- NCERT Notes for Class 11 Chemistry
- NCERT Notes for Class 11 Biology

**Also Check NCERT Books and NCERT Syllabus here:**

1. Are these solutions helpful from an exam point of view?

Yes, NCERT Exemplar Class 11 Maths chapter 10 solutions are very important from the perspective of exams including Board exams as well as other competitive exams.

2. Is the chapter important for the Board examination?

Yes, NCERT Exemplar Class 11 Maths solutions chapter 10 are very important for the Board examination as it has the relevant weightage of marks along with its importance in other competitive exams.

3. What are some of the important topics from this chapter that have practical application in other fields?

The important topics to be covered in this chapter are the slope of a line, general equation of a line, the distance of a point from a line and various forms of the equation of a line in coordinate geometry.

Sep 06, 2024

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