Home > NCERT Solutions > NCERT Solutions for Class 11 Maths Chapter 8 Straight Lines Exercise 8.1

NCERT Solutions for Class 11 Maths Chapter 8 Straight Lines Exercise 8.1

NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.1 – Straight Lines helps students understand the concept of straight lines in coordinate geometry. This exercise introduces key concepts such as the slope of a line, the equation of a line in different forms, and the relationship between the slope and the intercepts. Students will learn to find the equation of a line when different data points are given, such as the slope and a point on the line, or two points on the line.

Prepared according to the latest CBSE Class 11 Maths syllabus, Exercise 8.1 focuses on finding the slope-intercept form, point-slope form, and two-point form of the equation of a straight line. It also includes problems that help students apply these formulas and concepts to solve real-life problems and practice interpreting the results geometrically.

NCERT Solutions for Class 11 Maths Chapter 8 Straight Lines Exercise 8.1

Straight Lines

Difficult

Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area.

Straight Lines

Difficult

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

Straight Lines

Medium

Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Straight Lines

Medium

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

Straight Lines

Difficult

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

Straight Lines

Difficult

\begin{array}{l} The slope of a line is double of the slope of another line. If tangent of the angle between them \\ is \frac{1}{3}, find the slopes of the lines. \end{array}

Straight Lines

Easy

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

Straight Lines

Medium

If three points (h, 0), (a, b) and (0, k) lie on a line, show that \frac{a}{h} + \frac{b}{k} =1.

Straight Lines

Difficult

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?

Straight Lines

Difficult

In Exercises 1 to 8, find the equation of the line which satisfy the given conditions:

1. Write the equations for the x-and y-axes.
2. $Passing through the point (- 4, 3) with slope \frac{1}{2} .$
3. Passing through (0, 0) with slope m.
4. $Passing through (2, 2 \sqrt{3}) and inclined with the x-axis at an angle of 75° .$
5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.
6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of the x-axis.
7. Passing through the points (–1, 1) and (2,– 4).
8. Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30°.

Straight Lines

Difficult

A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.

Straight Lines

Medium

Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

Straight Lines

Medium

\begin{array}{l} Find equation of the line through the point (0, 2) making an angle \frac{2π}{3} with the positive x-axis. Also, \\ find the equation of line parallel to it and crossing the y - axis at a distance of 2 units below the origin. \end{array}

Straight Lines

Medium

The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L= 125.134 when C = 110, express L in terms of C.

Straight Lines

Medium

The owner of a milk store finds that, he can sell 980 litres of milk each week at ₹14/litre and 1220 litres of milk each week at ₹16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹17/litre?

Straight Lines

Difficult

Find the equation of the line through the intersection of lines x + 2y – 3 = 0 and 4x – y + 7 = 0 and which is parallel to 5x + 4y – 20 = 0

NCERT Solutions for Class 11 Maths Chapter 8 Straight Lines Exercise 8.1

The solutions are explained in a step-by-step format so that students can easily grasp the method and approach, making it easier to solve problems related to straight lines in their exams.

1. Draw a quadrilateral in the Cartesian plane, whose vertices are (4, 5), (0, 7), (5, 5), and (4, 2). Also, find its area.

The quadrilateral is defined by points

$P(4, 5)$ $P (4, 5)$ ,

$Q(0, 7)$ $Q (0, 7)$ ,

$R(5, 5)$ $R (5, 5)$ , and

$S(4, 2)$ $S (4, 2)$ .
After plotting the points and joining them, you calculate the area of the quadrilateral by breaking it into two triangles:

$\triangle PQR$ $△ PQR$ and

$\triangle PRS$ $△ PRS$ .
The area of the quadrilateral is the sum of these areas.

The area of

$\triangle PQR$

$△ PQR$ and

$\triangle PRS$

$△ PRS$ is calculated using the formula for the area of a triangle with given vertices:

$\text{Area} = \frac{1}{2} | x_1 y_2 + x_2 y_3 + x_3 y_1 - y_1 x_2 - y_2 x_3 - y_3 x_1 |$

$Area = 21 ∣ x_{1} y_{2} + x_{2} y_{3} + x_{3} y_{1} - y_{1} x_{2} - y_{2} x_{3} - y_{3} x_{1} ∣$

After performing the calculation:

$\text{Area of PQRS} = 121 \, \text{units}^2$

$Area of PQRS = 121 units^{2}$

2. The base of an equilateral triangle with side

$2a$

$2 a$ lies along the y-axis such that the midpoint of the base is at the origin. Find the vertices of the triangle.

The base of the triangle lies along the y-axis, with the midpoint at the origin.
The coordinates of points

$R(0, a)$ $R (0, a)$ and

$Q(0, -a)$ $Q (0, - a)$ are determined as per the problem.
The vertex

$P$ $P$ lies on the y-axis and is found using the Pythagorean theorem:

$OP = \sqrt{3}a$

$OP = 3 a$

Thus, the vertices of the equilateral triangle are

$(0, a)$ $(0, a)$ ,

$(0, -a)$ $(0, - a)$ , and

$(\sqrt{3}a, 0)$ $(3 a, 0)$ .

3. Find the distance between

$P(x_1, y_1)$

$P (x_{1}, y_{1})$ and

$Q(x_2, y_2)$

$Q (x_{2}, y_{2})$ when (i) PQ is parallel to the y-axis and (ii) PQ is parallel to the x-axis.

(i) When PQ is parallel to the y-axis, the x-coordinates are the same (

$x_1 = x_2$

$x_{1} = x_{2}$ ), so the distance is:

$\text{Distance} = |y_2 - y_1|$

$Distance = ∣ y_{2} - y_{1} ∣$

(ii) When PQ is parallel to the x-axis, the y-coordinates are the same (

$y_1 = y_2$

$y_{1} = y_{2}$ ), so the distance is:

$\text{Distance} = |x_2 - x_1|$

$Distance = ∣ x_{2} - x_{1} ∣$

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

$A(7, 6)$

$A (7, 6)$ and

$B(3, 4)$

$B (3, 4)$ .

Let the point

$C(a, 0)$ $C (a, 0)$ on the x-axis be equidistant from

$A(7, 6)$ $A (7, 6)$ and

$B(3, 4)$ $B (3, 4)$ .
The distances

$AC = BC$ $A C = BC$ are set equal, and by applying the distance formula, we solve for

$a$ $a$ .
The result is that

$a = 15/2$ $a = 15/2$ , so the point on the x-axis is

$(15/2, 0)$ $(15/2, 0)$ .

5. Find the slope of a line, which passes through the origin and the mid-point of the segment joining the points

$P(0, 4)$

$P (0, 4)$ and

$B(8, 0)$

$B (8, 0)$ .

The midpoint of the segment joining

$P(0, 4)$ $P (0, 4)$ and

$B(8, 0)$ $B (8, 0)$ is calculated as

$(4, 2)$ $(4, 2)$ .
The slope

$m$ $m$ of the line passing through the origin and the midpoint is:

$m = \frac{2 - 0}{4 - 0} = \frac{1}{2}$

$m = 4 - 02 - 0 = 21$

Thus, the slope is

$1/2$

$1/2$ .

6. Without using the Pythagoras theorem, show that the points

$(4, 4)$

$(4, 4)$ ,

$(3, 5)$

$(3, 5)$ , and

$(-1, -1)$

$(- 1, - 1)$ are vertices of a right-angled triangle.

The slopes of the lines joining the points are calculated:

$\text{Slope of PQ} = 1, \quad \text{Slope of QR} = -1, \quad \text{Slope of RP} = -1$

$Slope of PQ = 1, Slope of QR = - 1, Slope of RP = - 1$

Since two of the slopes are perpendicular (

$\text{Slope of PQ} = \text{Slope of RP} = 1$ $Slope of PQ = Slope of RP = 1$ ), the triangle is right-angled at

$P(4, 4)$ $P (4, 4)$ .
Therefore, the points

$(4, 4)$ $(4, 4)$ ,

$(3, 5)$ $(3, 5)$ , and

$(-1, -1)$ $(- 1, - 1)$ form a right-angled triangle.

7. Find the slope of the line, which makes an angle of

$30^\circ$

$3 0^{\circ}$ with the positive direction of the y-axis measured anticlockwise.

The angle between the line and the positive x-axis is

$120^\circ$ $12 0^{\circ}$ (since

$90^\circ + 30^\circ = 120^\circ$ $9 0^{\circ} + 3 0^{\circ} = 12 0^{\circ}$ ).
The slope of the line is:

$\text{Slope} = \tan(120^\circ) = -\sqrt{3}$

$Slope = tan (12 0^{\circ}) = - 3$

Thus, the slope of the line is

$-\sqrt{3}$

$- 3$ .

8. Find the value of

$x$

$x$ for which the points

$(x, 1)$

$(x, 1)$ ,

$(2, 1)$

$(2, 1)$ , and

$(4, 5)$

$(4, 5)$ are collinear.

For the points to be collinear, the slope between the points

$(x, 1)$ $(x, 1)$ and

$(2, 1)$ $(2, 1)$ should be equal to the slope between

$(2, 1)$ $(2, 1)$ and

$(4, 5)$ $(4, 5)$ .
Solving for

$x$ $x$ , we get:

$x = 1$

9. Without using distance formula, show that points

$(-2, 1)$

$(- 2, 1)$ ,

$(4, 0)$

$(4, 0)$ ,

$(3, 3)$

$(3, 3)$ , and

$(-3, 2)$

$(- 3, 2)$ are vertices of a parallelogram.

The slopes of the line segments are checked:

$\text{Slope of PQ} = \text{Slope of RS}, \quad \text{Slope of QR} = \text{Slope of PS}$

$Slope of PQ = Slope of RS, Slope of QR = Slope of PS$

Since both pairs of opposite sides are parallel, the quadrilateral is a parallelogram.

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

$(3, 1)$

$(3, 1)$ and

$(4, 2)$

$(4, 2)$ .

The slope

$m$ $m$ of the line joining the points

$(3, 1)$ $(3, 1)$ and

$(4, 2)$ $(4, 2)$ is:

$m = -1$

$m = - 1$

The angle

$\theta$ $θ$ between the line and the x-axis is:

$\theta = 135^\circ$

$θ = 13 5^{\circ}$

11. A line passes through

$(x_1, y_1)$

$(x_{1}, y_{1})$ and

$(h, k)$

$(h, k)$ . Its slope is

$m$

$m$ . Show that

$k - y_1 = m(h - x_1)$

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

The slope of the line is defined as:

$m = \frac{k - y_1}{h - x_1}$

$m = h - x ^{1} k - y ^{1}$

Rearranging gives:

$k - y_1 = m(h - x_1)$

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

This confirms the desired result.

FAQs – Class 11 Maths Chapter 8 Exercise 8.1 Straight Lines

Q1. What is the focus of Exercise 8.1?
Exercise 8.1 focuses on understanding and applying the equations of a straight line in different forms, and calculating the slope and intercepts of a line.

Q2. What are the different forms of the equation of a straight line?
The main forms of the equation of a straight line are:

Slope-intercept form:

$y = mx + c$ $y = m x + c$
Where

$m$ $m$ is the slope of the line, and

$c$ $c$ is the y-intercept.
Point-slope form:

$y - y_1 = m(x - x_1)$ $y - y_{1} = m (x - x_{1})$
Where

$m$ $m$ is the slope, and

$(x_1, y_1)$ $(x_{1}, y_{1})$ is a point on the line.
Two-point form:

$\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$ $y ^{2} - y ^{1} y - y ^{1} = x ^{2} - x ^{1} x - x ^{1}$
Where

$(x_1, y_1)$ $(x_{1}, y_{1})$ and

$(x_2, y_2)$ $(x_{2}, y_{2})$ are two distinct points on the line.

Q3. How is the slope of a line calculated?
The slope of a line is calculated as:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = x ^{2} - x ^{1} y ^{2} - y ^{1}$

Where

$(x_1, y_1)$

$(x_{1}, y_{1})$ and

$(x_2, y_2)$

$(x_{2}, y_{2})$ are two distinct points on the line.

Q4. Why is Exercise 8.1 important for exams?
This exercise is crucial because straight lines form a foundational concept in coordinate geometry. The ability to work with different forms of the equation of a line and calculate slope and intercepts is fundamental for solving many advanced problems in mathematics.

Q5. How can students prepare effectively for Exercise 8.1?
Students should:

Understand the different forms of the equation of a straight line.
Practice problems on finding the equation of a line using the slope and given points.
Work on interpreting the geometric meaning of slope and intercepts.

NCERT Solutions for Class 11 Maths Chapter 8 Straight Lines Exercise 8.1

NCERT Solutions for Class 11 Maths Chapter 8 Straight Lines Exercise 8.1

NCERT Solutions for Class 11 Maths Chapter 8 Straight Lines Exercise 8.1

FAQs – Class 11 Maths Chapter 8 Exercise 8.1 Straight Lines

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NCERT Solutions for Class 11 Maths Chapter 8 Straight Lines Exercise 8.1

NCERT Solutions for Class 11 Maths Chapter 8 Straight Lines Exercise 8.1

NCERT Solutions for Class 11 Maths Chapter 8 Straight Lines Exercise 8.1

FAQs – Class 11 Maths Chapter 8 Exercise 8.1 Straight Lines

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