Class 11 Maths Chapter 10 Exercise 10.1

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

NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.1 on Straight Lines are designed to assist students in preparing for their CBSE board exams. These solutions provide a detailed, step-by-step explanation of the concepts related to the coordinate geometry of straight lines. Mastering these concepts is essential for tackling other exercises in Chapter 10 of the Class 11 Maths curriculum.

Our NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.1 are aimed at helping students understand the key principles of straight lines in an easy-to-follow manner. By studying these solutions, you will build a solid foundation in coordinate geometry, which will be crucial for solving more advanced problems in this chapter.

You can now download the NCERT Solutions PDF for Class 10 Maths Chapter 3 Exercise 3.1 on Pair of Linear Equations in Two Variables. This resource is crafted to help you easily understand and solve problems involving two linear equations with two variables, ensuring a better grasp of the concepts and enhancing your performance in exams.

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

Solution.

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)]

$ar\Delta ABC=\frac{1}{2}\left [ -4\left ( 7+5 \right ) +0\left ( -5-5 \right )+5\left ( 5-7 \right )\right ]$

$ar\Delta ABC=\frac{1}{2}\left [ -48 -10\right ]$

ar ΔABC = -29

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

$ar\Delta ADC=\frac{1}{2}\left [ -4\left ( -2+5 \right ) -4\left ( -5-5 \right )+5\left ( 5+2 \right )\right ]$

$ar\Delta ADC=\frac{1}{2}\left [ -12 +40+35\right ]$

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.

Solution.

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.

Solution. 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

$PQ =\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2} -y_{1}\right )^{2}}$

Here y₁=y₂

$PQ =\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{1} -y_{1}\right )^{2}}$

$PQ =\sqrt{\left ( x_{2}-x_{1} \right )^{2}}$

$PQ =\sqrt{\left ( x_{2}-x_{1} \right )^{2}}=\pm \left ( x_{2}-x_{1} \right )$

$PQ =\left | x_{2} -x_{1}\right |$

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

Here x₁=x₂

$PQ =\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2} -y_{1}\right )^{2}}$

$PQ =\sqrt{\left ( x_{1}-x_{1} \right )^{2}+\left ( y_{2} -y_{1}\right )^{2}}$

$PQ =\sqrt{\left ( y_{2} -y_{1}\right )^{2}}$

$PQ =\pm \left ( y_{2}-y_{1} \right )$

$PQ =\left | y_{2}-y_{1} \right |$

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).

Solution. 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

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\frac{-2-0}{4-0}$

$m=\frac{-2}{4}$

$m=-\frac{1}{2}$

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

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.

Solution. 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

$=\frac{5-4}{3-4}=-\frac{1}{1}=-1$

The slope of BC

$=\frac{-1-5}{-1-3}=-\frac{-6}{-4}=\frac{3}{2}$

The slope of AC

$=\frac{-1-4}{-1-4}=-\frac{-5}{-5}=1$

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.

Solution. 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.

Solution. 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

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y_{3}-y_{2}}{x_{3}-x_{2}}$

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

$\frac{1+1}{2-x}=\frac{5-1}{4-2}$

$\frac{2}{2-x}=2$

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.

Solution. 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

$=\frac{0+1}{4+2}=\frac{1}{6}$

Slope of BC

$=\frac{3-0}{3-4}=\frac{3}{-1}=-3$

Slope of DC

$=\frac{2-3}{-3-3}=\frac{-1}{-6}=\frac{1}{6}$

Slope of AD

$=\frac{2+1}{-3+2}=\frac{3}{-1}=-3$

Since slope of AB = Slope of DC

∴AB ∥ DC

Since slope of AD = Slope of BC

∴AD ∥ BC

Hence ABCD is a parallelogram

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

Solution. 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°

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

Solution. 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.

Solution. 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 m₁and m₂is given as

$tan\theta =\left | \frac{m_{2}-m_{1}}{1+m_{1}m_{2}} \right |$

$tan\theta =\left | \frac{m-2m}{1+2m^{2}} \right |$

$\frac{1}{3} =\left | \frac{-m}{1+2m^{2}} \right |$

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).

Solution. 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

Solution. 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?

Solution. 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

We hope that Class 11 Maths Chapter 10 – Straight Lines Exercise 10.1 is now clearer to you. If you still have any questions or need further clarification about this exercise, don’t hesitate to ask in the comments. We’re here to help you succeed!

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