NCERT Solutions for Class 11 Maths Exercise 10.1 Chapter 10 Straight Lines - Future Study Point

NCERT Solutions for Class 11 Maths Exercise 10.1 Chapter 10 Straight Lines

Exercise 10.1 class 11 maths

NCERT Solutions for Class 11 Maths Exercise 10.1 Chapter 10 Straight Lines

Exercise 10.1 class 11 maths

NCERT Solutions for Class 11 Maths Exercise 10.1 Chapter 10 Straight Lines are created here for helping the class 11 maths students for the preparation of maths subjects in the CBSE board exam. NCERT Solutions for Class 11 Maths Exercise 10.1 Chapter 10 Straight Lines will help you in clearing your maths concept on coordinate geometry of straight lines which is required to solve other exercises of chapter 10 straight lines.NCERT Solutions for Class 11 Maths Exercise 10.1 Chapter 10 Straight Lines are solved by a step by step method so every student can understand the solutions .

Exercise 10.3 -Straight Lines

NCERT Solutions for Class 11 Maths  

NCERT solutions of class 11 maths

Chapter 1-SetsChapter 9-Sequences and Series
Chapter 2- Relations and functionsChapter 10- Straight Lines
Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbersChapter 13- Limits and Derivatives
Chapter 6- Linear InequalitiesChapter 14-Mathematical Reasoning
Chapter 7- Permutations and CombinationsChapter 15- Statistics
Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

 

Q1. Draw a quadrilateral in a cartesian plane,whose vertices are (-4,5),(0,7),(5,-5) and (-4,-2). Also, find its area.

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Ans.

exercise 10.1 class 11 maths

ar ABCD = ar ΔABC +arΔADC

The area of the triangle whose vertices are (x1,y1),(x2,y2) and (x3,y3)

1/2[x1(y2-y3) +x2(y3-y4)+x3(y2-y3)]

ar ΔABC = -29

Area can’t be – so area of ΔABC = 29 units

arΔ ADC = 1/2(75 -12) =63/2 = 31.5

Therefore area of ABCD = 29 +31.5 = 60.5 sq.unit

Q2.The base of an equilateral triangle with side 2a lies along the y-axis such that the midpoint of the base is at the origin . Find vertices of the triangle.

Ans.Q2 exercise 10.1 maths class 11

Let the vertices of the triangle are A,B and C where BC is the base of the triangle

AB =BC = AC (Since ΔABC is an equilateral triangle)

BC = 2a (Given)

The midpoint of AC is origin O, therefore coordinates of O are (0,0)

As seen in the diagram OB = OC = a

Therefore vertices of B are (-a,0) and of C are (a,0)

In ΔABC , AC = 2a

AO ⊥BC, so OA = √(AC²- OC²) =√[(2a)²- a²] = √(4a²-a²) = a√3

Therefore vertices of  A are (a√3,0) since the vertex of A is on Y-axis so it may be on the negative Y-axis, therefore coordinates of A are (±a√3,0)

Q3.Find the distance between P(x1,y1) and Q(x2,y2) when (i) PQ is parallel to the Y-axis (ii) PQ is parallel to the x-axis.

Ans. The points P and Q are given in both cases when PQ is parallel to X-axis and PQ is parallel to Y-axis.

Question 3 exercise 10.1 class 11 maths

 

 

 

 

 

 

The distance between P and Q when PQ is parallel to X axis

Here y1=y2

The distance between P and Q when PQ is parallel to X axis

Here x1=x2

Q5. Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, – 4) and B (8, 0).

Ans. The mid point of the line segment joining the points P (0, – 4) and B (8, 0) = (0 +8)/2, (-4 +0)/2 = (4,-2)

It is given that the line is passing through the origin (0,0) and (4,-2)

The slope(m) of the line segment joining the points P(x1,y1) and Q(x2,y2)

is given as follows

Therefore the required slope of the line is -1/2

See the video for the solutions of Q1 to Q5 of Exercise 10.1 Straight Lines


Q6.Without using the Pythagoras theorem, show that the points (4,4),(3,5), and (-1,-1) are the vertices of a right-angled triangle.

Ans.The given vertices of the triangle are (4,4),(3,5), and (-1,-1)

Let the triangle is ΔABC, where A(4,4),B(3,5) and C(-1,-1)

The slope of the line joining the points (x1,y1) and (x2,y2) = (y2-y1)/ (x2-x1)

The slope of AB

The slope of BC

The slope of AC

The condition of the two lines whose slopes are m1and m2

m1 m2=-1

The slope of AC ×The slope of AB =1×-1 = -1

Therefore AC ⊥AB and BC is the hypotenuse of the right triangle

Q7.Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

Ans. The line described in the question is as shown below

 

Q7 ex.10.1 class 11 maths

 

 

 

 

 

 

The line makes an angle of 30° with the positive direction of y-axis  anticlockwise also makes an angle (90° +30°=120°) with the positive direction of x-axis anticlockwise

Hence the slope of the line is = tan120°

m= tan(π -60°)

m = -tan 60°= -√3

Therefore slope of the given line is -√3

Q8.Find the value of x for which the points (x, – 1), (2, 1) and (4, 5) are collinear.

Ans.Three points A(x1,y1) , B(x2,y2) and C(x3,y3)  are collinear when the slope of the line segment AB = Slope of the line segment BC

Here the points given to us are A(x, – 1), B(2, 1), and (4, 5)

4- 2x = 2

2x = 2⇒ x = 1

∴ The required value of x is 1

Q9.Without using the distance formula, show that points (– 2, – 1), (4, 0), (3, 3), and (–3, 2) are the vertices of a parallelogram.

Ans. Let the points given to us are named as A(– 2, – 1), B (4, 0), C (3, 3) and D(–3, 2)

In a parallelogram opposite sides are parallel

Therefore slope of two pairs of line segments among AB,BC,DC and AD must be equal to each other

Slope of AB

Slope of BC

Slope of DC

Slope of AD

Since slope of AB = Slope of DC

∴AB ∥ DC

Since slope of AD = Slope of BC

∴AD ∥ BC

Hence ABCD is a parallelogram

See the video for the solutions of the Q6 -Q9

Q10. Find the angle between the x-axis and the line joining the points (3, -1) and (4, -2).

Ans. Let he x axis makes an angle θ with the line joining two points (3,-1) and (4,-2)

For this let’s find the slope of the joining the given points i.e m = tan θ

The slope of m = (-2 +1)/(4 -3) = -1/1 = -1

tan θ = -1

tan θ = -tan 45°

tan θ = tan (180° -45°) [ Since tan θ is negative,so it lies on second quadrant]

tan θ = tan 135°

θ = 135°

Therefore the angle between the x axis and the given line is 135°

Q11.The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the lines.

Ans. Let the slope of a line is m then slope of another line is 2m

We are given the tangent of the angle(θ )between them ,tan θ = 1/3

The angle between the two lines with slope m1 and m2is given as

Taking positve sign

1/3 = (-m/(1 +2m²)

1 + 2m² =-3m

2m² +3m +1 =0

2m² + 2m +m +1=0

2m(m +1) +1(m+1) =0

(m +1) (2m +1) =0

m =-1, m =-1/2

If slope of one line is -1 then slope of another line is -2 and if slope of one line is -1/2 then slope of another line is -1

Taking negative sign

1/3 = -(-m/(1 +2m²)

1 + 2m² =3m

2m² -3m +1 =0

2m² – 2m -m +1=0

2m(m -1) -1(m-1) =0

(m -1) (2m -1) =0

m =1, m =1/2

If slope of one line is 1 then slope of another line is 2 and if slope of one line is 1/2 then slope of another line is 1

Q12.A line passes through (x1, y1) and (h, k). If slope of the line is m, show that k -y1 = m (h -x1 ).

Ans. The equation of the line with the slope m that passes through the point (x1,y1) is given as

y -y1 = m (x -x1 )

The line is passing through (x1, y1) and (h, k)

m = (k-y1)/( h-x1)

k-y1= m ( h-x1), Hence proved

Q13. If three points (h, 0), (a, b) and (0, k) lie on a line, show that a/h + b/k = 1 

Ans.Let the given point are A(h,0), B(a,b) and C(0,k) lie on a line

For the points A(h,0), B(a,b) and C(0,k) to be collinear

Slope of AB = Slope of AC

The slope of AB = (b-0)/(a -h) = b/(a-h)

Slope AC = (k-0)/(0-h) = k/-h

b/(a-h) =-k/h

-k( a-h) =bh

-ka +kh = bh

-ka -bh = -kh

ka + bh = kh

Dividing it by kh

ka/kh +bh/kh = 1

a/h + b/k = 1

Q14.Consider the following population and year graph (Fig 10.10), find the slope of the line AB and using it, find what will be the population in the year 2010?

 

 

Ans. Slope of the line AB = (97-92)/(1995-1985) = 5/10 = 1/2

Let the population in the year 2010 is a,in the graph it is represented by the point C along the line AB

So, the coordinates of the point C are (2010,a)

The slope of AB = Slope of BC

1/2 = (a-97)/(2010-1995)

1/2 = (a -97)/15

15/2 = a -97

7.5 = a -97

a = 97 + 7.5 = 104.5

Hence the slope of the line is 1/2 and the population in 2010 is 104.5

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NCERT Solutions of  Science and Maths for Class 9,10,11 and 12

NCERT Solutions of class 9 maths

Chapter 1- Number SystemChapter 9-Areas of parallelogram and triangles
Chapter 2-PolynomialChapter 10-Circles
Chapter 3- Coordinate GeometryChapter 11-Construction
Chapter 4- Linear equations in two variablesChapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s GeometryChapter 13-Surface Areas and Volumes
Chapter 6-Lines and AnglesChapter 14-Statistics
Chapter 7-TrianglesChapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions of class 9 science 

Chapter 1-Matter in our surroundingsChapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?Chapter 10- Gravitation
Chapter3- Atoms and MoleculesChapter 11- Work and Energy
Chapter 4-Structure of the AtomChapter 12- Sound
Chapter 5-Fundamental unit of lifeChapter 13-Why do we fall ill ?
Chapter 6- TissuesChapter 14- Natural Resources
Chapter 7- Diversity in living organismChapter 15-Improvement in food resources
Chapter 8- MotionLast years question papers & sample papers

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CBSE Class 9-Sample paper of science

CBSE Class 9-Unsolved question paper of science 2019

NCERT Solutions of class 10 maths

Chapter 1-Real numberChapter 9-Some application of Trigonometry
Chapter 2-PolynomialChapter 10-Circles
Chapter 3-Linear equationsChapter 11- Construction
Chapter 4- Quadratic equationsChapter 12-Area related to circle
Chapter 5-Arithmetic ProgressionChapter 13-Surface areas and Volume
Chapter 6-TriangleChapter 14-Statistics
Chapter 7- Co-ordinate geometryChapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

NCERT solutions of class 10 science

Chapter 1- Chemical reactions and equationsChapter 9- Heredity and Evolution
Chapter 2- Acid, Base and SaltChapter 10- Light reflection and refraction
Chapter 3- Metals and Non-MetalsChapter 11- Human eye and colorful world
Chapter 4- Carbon and its CompoundsChapter 12- Electricity
Chapter 5-Periodic classification of elementsChapter 13-Magnetic effect of electric current
Chapter 6- Life ProcessChapter 14-Sources of Energy
Chapter 7-Control and CoordinationChapter 15-Environment
Chapter 8- How do organisms reproduce?Chapter 16-Management of Natural Resources

Solutions of class 10 last years Science question papers

CBSE Class 10 – Question paper of science 2020 with solutions

CBSE class 10 -Latest sample paper of science

NCERT solutions of class 11 maths

Chapter 1-SetsChapter 9-Sequences and Series
Chapter 2- Relations and functionsChapter 10- Straight Lines
Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbersChapter 13- Limits and Derivatives
Chapter 6- Linear InequalitiesChapter 14-Mathematical Reasoning
Chapter 7- Permutations and CombinationsChapter 15- Statistics
Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT solutions of class 12 maths

Chapter 1-Relations and FunctionsChapter 9-Differential Equations
Chapter 2-Inverse Trigonometric FunctionsChapter 10-Vector Algebra
Chapter 3-MatricesChapter 11 – Three Dimensional Geometry
Chapter 4-DeterminantsChapter 12-Linear Programming
Chapter 5- Continuity and DifferentiabilityChapter 13-Probability
Chapter 6- Application of DerivationCBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

 

 

 

 

 

 

 

 

 

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