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

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 .

**NCERT Solutions for Class 11 Maths **

**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**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.

ar ABCD = ar ΔABC +arΔADC

The area of the triangle whose vertices are (x_{1},y_{1}),(x_{2},y_{2}) and (x_{3},y_{3})

1/2[x_{1}(y_{2}-y_{3}) +x_{2}(y_{3}-y_{4})+x_{3}(y_{2}-y_{3})]

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.

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(x _{1},y_{1}) and Q(x_{2},y_{2}) 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.

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

Here y_{1}=y_{2}

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

Here x_{1}=x_{2}

**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(x_{1},y_{1}) and Q(x_{2},y_{2})

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 (x_{1},y_{1}) and (x_{2},y_{2}) = (y_{2}-y_{1})/ (x_{2}-x_{1})

The slope of AB

The slope of BC

The slope of AC

The condition of the two lines whose slopes are m_{1}and m_{2}

m_{1} m_{2}=-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

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(x_{1},y_{1}) , B(x_{2},y_{2}) and C(x_{3},y_{3}) 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 m_{1 }and m_{2}is 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 (x _{1}, y_{1}) and (h, k). If slope of the line is m, show that k -y_{1} = m (h -x_{1}_{ }).**

Ans. The equation of the line with the slope m that passes through the point (**x _{1}**

**,y**

_{1}**)**is given as

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

The line is passing through (x_{1}, y_{1}) and (h, k)

m = (k-y_{1})/( h-x_{1})

k-y_{1}= m ( h-x_{1}), 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 System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

Chapter 7-Triangles | Chapter 15-Probability |

Chapter 8- Quadrilateral |

**NCERT Solutions of class 9 science **

**CBSE Class 9-Question paper of science 2020 with solutions**

**CBSE Class 9-Sample paper of science**

**CBSE Class 9-Unsolved question paper of science 2019**

**NCERT Solutions of class 10 maths**

**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**

**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-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**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 Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |