Straight Lines Class 11 NCERT Maths Solutions – Chapter 9 PDF Free

Q: 20. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30o with positive direction of the x-axis.

20. The slope of the line is m = tan 30° = 1√3 And x-intercept, c = 2 Using slope intercept form, equation of line is

Q: 73. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.

73. The given eqn of limes are. 9x + 6y – 7 = 0 ______ (1) 3x + 2y + b = 0 ______ (2) Let P (x0, y0) be a point equidistant from (1) and (2) so 9x0 + 6y - 7 = ± 3 (3x0 + 2y0 + 6) When, 9x0 + 6y0 – 7 = 3 (3x0 + 2y0 + 6) ⇒ 9x0 + 6y0 - 7 = 9x0 + 6y0 + 18 ⇒ - 7 = 18 which in not true So, 9x0 + 6y0 - 7 = -3 (3x0 + 2y0 + 6) ⇒ 9x0 + 6y0 -7 = -9x0 -6y0 -18 ⇒ 18x0 + 12y0 + 11= 0. Hence, the required eqn of line through (x0, y0) & equidistant from parallel line 9x + 6y - 7 = 0 and 3x + 2y + 6 = 0 is 18x + 12y + 11 = 0.

Q: 7. Find the slope of the line, which makes an angle of 30°with the positive direction of y-axis measured anticlockwise.

7. Let l be the line making 30° with y-axis as shown in figure. Then, Angle a = ∠b + 90° (Sum of exterior angle of a triangle) ⇒∠a=30°+90° ( 30°=∠b vertically opposite angle) ⇒∠a=120° So, slope of line l = tan a = m = tan 120° = tan (180° – 60°) = –tan 60° =−√3

Updated on Jul 31, 2025 12:58 IST

By Pallavi Pathak, Assistant Manager Content

Straight Lines Class 11 continues the study of coordinate geometry, which the students have studied in the previous chapters. Coordinate geometry is a combination of algebra and geometry. In this chapter, the students will study the properties of the simplest geometric figure – the straight line. Straight Lines Class 11 Solutions explain that in a triangle ABC, if the area is zero, then the three points A, B, and C lie on the same line. The straight line is important in geometry as it is part of our daily lives in various ways.
Class 11 straight lines cover the slope of a line, the distance of a point from a line, and various forms of the equation of a line. The solutions are well-written by the experts and provide in-depth explanations of the concept. The students get the solutions to all the questions of this chapter in the NCERT book. The Straight Lines Class 11 NCERT solutions are ideal for the CBSE Board and other competitive exam preparation.
If students are looking for solved examples, important topics, and Class 11 Maths NCERT notes of all the chapters, they can get them here.

Table of content

Snapshot of NCERT Solutions for Class 11 Maths Chapter 9: Straight Lines
Class 11 Math Straight Lines: Key Topics, Weightage
Important Formulas of Straight Lines Class 11
Class 11 Maths Straight Lines Chapter – NCERT Solutions PDF
Class 11 Maths Straight Lines Exercise-wise Solution(Old NCERT)
Class 11 Maths Straight Lines Exercise 9.1 Solutions

Snapshot of NCERT Solutions for Class 11 Maths Chapter 9: Straight Lines

Here is a quick overview of the Straight Lines Class 11 NCERT:

Straight Lines Class 11 Solutions cover the concept of slope. It says that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) is given by - $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{y_{1} - y_{2}}{x_{1} - x_{2}}, x_{1} \neq x_{2}$
When a line makes an angle with the x-axis, then the slope (m) of the line is given by - $m = \tan α, α \neq 90 °^{}$
The slope of a vertical line is undefined, and that of a horizontal line is zero.
For a line L1 and L2, with slopes m1, and m2, the acute angle is given by - $\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |, m_{1} m_{2} \neq 0$
Two lines are parallel only when their slopes are equal, and they are perpendicular if and only if the product of their slopes is -1.
The condition for three points A, B, and C to be collinear is if the slope of AB = the slope of BC.
There are more such conditions and formulas covered in the Straight Lines Class 11 NCERT.

For a comprehensive NCERT solution of Class 11 and Class 12 of Physics, Chemistry & Maths, check here.

Class 11 Math Straight Lines: Key Topics, Weightage

While preparing for the Straight Lines Class 11, students must focus on understanding straight lines to master other topics in coordinate geometry and beyond. In the entrance exams, questions are twisted; hence, it is advisable to practice a variety of problems. The following are the topics covered in this chapter:

Exercise	Topics Covered
9.1	Introduction
9.2	Slope of a Line
9.3	Various Forms of the Equation of a Line
9.4	Distance of a Point From a Line

Straight Lines Class 11 Weightage in JEE Mains

Exam	Number of Questions	Weightage
JEE Mains	one to two questions	6.6%

Important Formulas of Straight Lines Class 11

Important Formulae of Straight Line for CBSE and Competitive Exams

Various Forms of Line Equations –
- Slope-intercept form: $y = mx + c$
- Point-slope form: $y - y_1 = m(x - x_1)$
- Two-point form: $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$
- Intercept form: $\frac{x}{a} + \frac{y}{b} = 1$
- Normal form: $x \cos \alpha + y \sin \alpha = p$
Slope of a Line: $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$
Angle Between Two Lines: $\tan θ = ∣ \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} ∣$
Distance of a Point $(x_1, y_1)$ from a Line $Ax + By + C = 0$ $d = \frac{∣ A x_{1} + B y_{1} + C ∣}{\sqrt{A^{2} + B^{2}}}$
Distance Between Two Parallel Lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ : $d = \frac{∣ C_{2} - C_{1} ∣}{\sqrt{A^{2} + B^{2}}}$

Class 11 Maths Straight Lines Chapter – NCERT Solutions PDF

A free Straight Lines Class 11 PDF is available here. The students must download it and prepare for their exams. The PDF will help them get the concept clarity, and hence they will be able to score high in their exams.

Chapter 9 Straight Lines Class 11 NCERT Solutions: Free PDF Download

Class 11 Maths Straight Lines Exercise-wise Solution(Old NCERT)

Class 11 Maths Straight Lines is numbered as Chapter 9 as per the latest syllabus given by CBSE. Students will find the old NCERT solutions exercise-wise below. Class 11 chapter 10 straight lines is divided on three exercises and a miscellaneous exercise which summarises all the concepts covered in the chapter. Class 11 Straight Lines Ex 10.1 focuses on basic concepts such as the equation of a line, the slope of a given straight line, and related concepts. Class 11 Ex 10.2 focuses on important concepts related to the equation of a line in different forms. Exercise 10.3 deals with more advanced concepts such as parallel and perpendicular lines and related equation forms. Students can find complete solution of all exercise below;

Class 11 Maths Straight Lines Exercise 9.1 Solutions

Class 11 Maths Straight Lines Exercise 9.1 focuses on basic concepts such as the equation of a line, the slope of a given straight line, and the slope of a line when two points and a line are given, finding parallel and perpendicular of a given line. Students must understand these topics to form a strong basic understanding of coordinate geometry. Ex 9.1 class 11 of NCERT Mathematics consists of 14 questions and students will find the solution of all the questions below.

Class 11 Straight Lines Exercise 9.1 Solutions

Q1. Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7),

(5, – 5) and (– 4, –2). Also, find its area.

A.1. Let the given points be A(–4, 5), B(0, 7), C(5, –5) and D(–4, –2).

Then quadrilated ABCD can be plotted on the graph by joining the points A, B, C and D.

We connect diagonal AC such that

area (ABCD) = (ΔABC) + (ΔADC)

Now,

$(ΔABC) = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= \frac{1}{2} | - 4 [7 - (- 5)] + 0 [- 5 - 5] + 5 [5 - 7] | unit^{2}$

$= \frac{1}{2} | - 4 (7 + 5) + 0 + 5 (5 - 7) | unit^{2}$

$= \frac{1}{2} | (- 4) \times 12 + 5 \times (- 2) |unit^{2}$

$= \frac{1}{2} | - 48 - 10 |unit^{2}$

$= \frac{1}{2} | - 58 | unit^{2}$

$= \frac{58}{2} unit^{2} = 29 unit^{2}$

Similarly, $(ΔACD) = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= \frac{1}{2} | - 4 [- 5 - (2)] + 5 [- 2 - 5] + (- 4) [5 - (- 5)] | unit^{2}$

$= \frac{1}{2} | - 4 (- 5 + 2) + 5 (- 2 - 5) - 4 (5 + 5) |unit^{2}$

$= \frac{1}{2} | (- 4) \times (- 3) + 5 \times (- 7) - 4 \times (10) |unit^{2}$

$= \frac{1}{2} | 12 - 35 - 40 |unit^{2}$

$= \frac{1}{2} | - 63 | unit^{2}$

$= \frac{63}{2} unit^{2}$

Hence, area (ABCD) = $29 + \frac{63}{2} unit^{2}$

$= \frac{58 + 63}{2} unit^{2} = \frac{121}{2} unit^{2}$

Q2. 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.

A.2. Let ABC be the equilateral triangle of side 2a and 0 be the origin. Then

AB = BC = AC = 2a

O is the mid-point of AB we have AO = a

BO = a

We know that A and B lies on y-axis so they have co-ordinate of the form (0, y).

Hence, co-ordinate of A is (0, a) and that of B is (0, –a)

Since OC, bisects AB at right angle, by Pythagoras theorem,

AC²= OA² + OC²

$\Rightarrow {(2 a)}^{2} = {(a)}^{2} +OC^{2}$

$\RightarrowOC^{2} = 4 a^{2} - a^{2}$

And as C we on x-axis it has co-ordinate of the form (x, 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.

Q4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

A.4. Let A(x, 0) be the point on x-axis when is equidistant from P(7, 6) and Q(3, 4)

Then, PA = QA

Squaring both sides, we get,

$\Rightarrow x^{2} - 14 x + 49 + 36 = x^{2} - 6 x + 9 + 16$

$\Rightarrow 49 + 36 - 9 - 16 = 14 x - 6 x$

$\Rightarrow 60 = 8 x$

$\Rightarrow x = \frac{60}{8} = \frac{15}{2}$

$∴$ The required point on x-axis is $(\frac{15}{2}, 0) .$

Commonly asked questions

20. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30^o with positive direction of the x-axis.

20.

The slope of the line is m = tan 30° = $\frac{1}{√3}$

And x-intercept, c = 2

Using slope intercept form, equation of line is

73. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.

73. The given eqⁿ of limes are.

9x + 6y – 7 = 0 ______ (1)

3x + 2y + b = 0 ______ (2)

Let P (x₀, y₀) be a point equidistant from (1) and (2) so

9x₀ + 6y - 7 = ± 3 (3x₀ + 2y₀ + 6)

When, 9x₀ + 6y₀ – 7 = 3 (3x₀ + 2y₀ + 6)

⇒ 9x₀ + 6y₀ - 7 = 9x₀ + 6y₀ + 18

⇒ - 7 = 18 which in not true

So, 9x₀ + 6y₀ - 7 = -3 (3x₀ + 2y₀ + 6)

⇒ 9x₀ + 6y₀ -7 = -9x₀ -6y₀ -18

⇒ 18x₀ + 12y₀ + 11= 0.

Hence, the required eqn of line through (x₀, y₀) & equidistant from parallel line 9x + 6y - 7 = 0

and 3x + 2y + 6 = 0 is 18x + 12y + 11 = 0.

74. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

Let A have the co-ordinate (x, o)

By laws of reflection

∠PAB = ∠ QAB = θ

And ∠ CAQ + θ = 90°

As normal is ⊥ to surface (x-axis)

⇒ ∠CAQ = 90° - θ

and ∠CAP = 90° + θ

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

Let l be the line making 30° with y-axis as shown in figure. Then,

Angle a = $∠ b$ + 90° (Sum of exterior angle of a triangle)

$\Rightarrow ∠ a = 30 ° + 90 °$ ( $30 ° = ∠ b$ vertically opposite angle)

$\Rightarrow ∠ a = 120 °$

So, slope of line l = tan a

= m = tan 120°

= tan (180° – 60°)

= –tan 60°

$= -\sqrt3$

75. Prove that the product of the lengths of the perpendiculars drawn from the points

Kindly go through the solution

76. A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

76.

If point P be the junction between the lines

2x – 3y + 4 = 0 ______ (1)

3x + 4y – 5 = 0 ______ (2)

Solving (1) and (2) using 3 × (1) – 2 × (2) we get,

6x – 9y + 12 – (6x + 8y – 10) = 0

–17y + 22 = 0

y = $\frac{22}{17}$

And 2x = 3y– 4

=> 2x = 3 × $\frac{22}{17}$ – 4

x = $\frac{33}{17}$ – 2 = $\frac{33 - 34}{17}$ = $- \frac{1}{17}$

Hence, the co-ordinate of the junction is P $(- \frac{1}{17}, \frac{22}{17})$

The eqⁿ of the path to be reach is

6x – 7y + 8 = 0 _____ (3)

Then, least distance will be perpendicular path.

So, slope of ⊥ path = $- 1$
$slope of line (3)$

$= \frac{- 1}{(6 / 7)} = \frac{- 7}{6}$

Hence eqⁿ of shortest/least distance path from P $(- \frac{1}{17}, \frac{22}{17})$ is

$y - \frac{22}{17} = \frac{- 7}{6} (x + \frac{1}{17}) .$

$\Rightarrow 6 y - \frac{132}{17} = - 7 x - \frac{7}{17} .$

$\Rightarrow 7 x + 6 y - \frac{132}{17} + \frac{7}{17} = 0$

$\Rightarrow 7 x + 6 y - \frac{125}{17} = 0 .$

119x + 102y – 125 = 0

5. 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).

Let 0 (0, 0) be the origin and A be the mid-point of line joining P (0, –4) and B (8, 0)

Then, co-ordinate of A = $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) = (\frac{0 + 8}{2}, \frac{- 4 + 0}{2}) = (4, - 2)$

$∴$ Slope of OA, m = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{- 2 - 0}{4 - 0} = - \frac{2}{4} = \frac{- 1}{2}$

45. Find angles between the lines √3x + y = 1and x + √3y = 1.

Kindly go through the solution

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

. Let the given point be A (4, 4), B (3, 5) and C (–1, –1)

Then, slope of AB, m₁ = $\frac{5 - 4}{3 - 4} = \frac{1}{- 1} = - 1$

Slope of AC, m₂ = $\frac{- 1 - 4}{- 1 - 4} = \frac{- 5}{- 5} = 1$

And slope of BC, m₃ = $= \frac{- 1 - 5}{- 1 - 3} = \frac{- 6}{- 4} = \frac{3}{2}$

As m₁ – m₂ = –1 × 1 = –1

$\Rightarrow m_{1} = \frac{- 1}{m_{2}}$

We conclude that AB and AC are perpendicular to each other.

Hence, ABC is a right-angle triangle right-angled at A

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

8. Let P (x, –1), Q (2, 1) and R (4, 5) be the collinear points. Then,

Slope of PQ = Slope of QR

$\Rightarrow \frac{1 - (- 1)}{2 - x} = \frac{5 - 1}{4 - 2}$

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

$\Rightarrow 2 \times 2 = 4 (2 - x)$

$\Rightarrow 4 = 8 - 4 x$

$\Rightarrow x = \frac{4}{4}$

$\Rightarrow x = 1$

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

Let A (–2, –1), B (4, 0), C (3, 3) and D (–3, 2) be the given points.

Slope of DC = $\frac{3 - 2}{3 - (- 3)} = \frac{1}{3 + 3} = \frac{1}{6}$

As slope of AB = slope of DC

We conclude that AB | | DC.

Similarly, slope of BC = $\frac{3 - 0}{3 - 4} = \frac{3}{- 1} = - 3$

Slope of AD = $\frac{2 - (- 1)}{- 3 - (- 2)} = \frac{2 + 1}{- 3 + 2} = \frac{3}{- 1} = - 3$

As slope of BC = slope of AD we conclude that BC | | AD.

Hence, as the pair of opposite sides of ABCD are parallel we can conclude that the given points are the vertices of a parallelogram.

11. 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.

11.

Let the two slopes be m and 2m. And if θ is the angle between the two lines.

$t a nθ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$

$\Rightarrow \frac{1}{3} = | \frac{2 m - m}{1 + 2 m . . m} | = | \frac{m}{1 + 2 m^{2}} |$

$\Rightarrow \frac{m}{1 + 2 m^{2}} = \pm \frac{1}{3} ($ $| x | = \pm x)$

Case I. When $\frac{m}{1 + 2 m} = \frac{1}{3}$

$\Rightarrow 3 m = 2 m^{2} + 1$

$\Rightarrow 2 m^{2} - 3 m + 1 = 0$

$\Rightarrow 2 m^{2} - 2 m - m + 1 = 0$

$\Rightarrow 2 m (m - 1) - (m - 1) = 0$

$\Rightarrow (m - 1) (2 m - 1) - 0$

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

Case II. When $\frac{m}{1 + 2 m^{2}} = - \frac{1}{3}$

$\Rightarrow 3 m = - (1 + 2 m^{2})$

$\Rightarrow 3 m = - 1 - 2 m^{2}$

$\Rightarrow 2 m^{2} + 3 m + 1 = 0$

$\Rightarrow 2 m^{2} + 2 m + m + 1 = 0$

$\Rightarrow 2 m (m + 1) + (m + 1) = 0$

$\Rightarrow (m + 1) (2 m + 1) = 0$

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

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

$∴$ The possible slopes of the two lines (m, 2m) are (1, 2), $(\frac{1}{2}, 1), (- 1, - 2)$ and $(- \frac{1}{2}, - 1)$

63. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.

63. The given eqⁿ of the lines are.

3x + y - 2 = 0 _____ (1)

Px + 2y - 3 = 0 ______ (2)

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

Point of intersection of (1) and (3) is given by,

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

=> 5x - 5 = 0

=> x = $\frac{5}{5}$

=> x = 1

So, y = 2 - 3x = 2 -3 (1) = 2 - 3 = 1.

i e, (x, y) = (1, -1).

As the three lines interests at a single point, (1, -1) should line on line (2)

i e, P * 1 + 2 * (-1)- 3 = 0

P - 2 - 3 = 0

P = 5

66. Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.

66. The given eqn of the line is.

4x + 7y – 3 = 0 _____ (1)

2x – 3y + 1 = 0 _______ (2)

Solving (1) and (2) using eqn (1) 2 x eqn (2) we get,

(4x + 7y – 3) 2 [ (2x – 3y + 1)] = 0

4x + 7y – 3 – 4x + 6y – 2 = 0

13y = 5

y = $\frac{5}{13}$

And 2x – 3 $(\frac{5}{13})$ + 1 = 0

2x = $\frac{15}{13}$ – 1 = $\frac{2}{13}$

$\Rightarrow x = \frac{1}{13}$

Point of intersection of (1) and (2) is $(\frac{1}{13}, \frac{5}{13})$

Since, the line passing through $(\frac{1}{13}, \frac{5}{13})$ has equal intercept say c then it is of the form

$\frac{x}{c} + \frac{y}{c} = 1 .$

x + y = c

$\Rightarrow \frac{1}{13} + \frac{5}{13} = c$

c = $\frac{6}{13}$

the read eqn of line is x + y = $\frac{6}{13}$

13x + 13y – 6 = 0

70. Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.

70. The given equation of the line is

l₁: x + y = 4

Let P (x₀, y₀) be the point of intersect of l₁ and the line to be drawn.

Then, x₀+ y₀ = 4 ⇒ y₀ = 4? x

Given, distance between P (x₀, y₀) and Q (? 1, 2) is 3

ie,

⇒ (x₀ + 1)^{2 +} (y? 2)²⁼9

⇒x²₀+1+ 2x₀ + (4? x? 2)² = 9

⇒ x²₀+ 2x₀+ 1 + (2? x₀)² = 9

⇒x²₀+ 2x₀+ 1 + 4 + x²₀? 4x₀?9 = 0

⇒ 2 x²₀ ?2x₀ ? 4 = 0

x²₀ ? x₀? 2 = 0

x²₀ + x₀ ? 2x₀? 2 = 0

x₀ (x +1)? 2 (x₀ +1) = 0

(x₀ +1) (x₀ ? 2) = 0

x₀ = 2 and x₀ =? 1

When, x₀ = 2, y₀= 4 ?2 = 2.

and when x₀ =? 1, y₀ = y? (?1) =5.

The points of interaction of line l₁which are at distance 3 unit from Q (? 1, 2) are P₁ (2,2) and P₂ (? 1, 5)

71. Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror.

71.

62. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0.

62.

The given eqⁿ of the lines are

y - x = 0 _____ (1)

x + y = 0 ______ (2)

x - k = 0 ______ (3)

The point of intersection of (1) and (2) is given by

(y - x) - (x + y) = 0

⇒ y - x -x -y = 0

y = 0 and x = 0

ie, (0, 0)

The point of intersection of (2) and (3) is given by

(x + y) – (x – k) = 0

y + k = 0

y = –k and x = k

i.e, (k, –k)

The point of intersection of (3) and (1) is given by

x = k

and y = k

ie, (k, k).

Hence area of triangle whose vertex are (0, 0), (k, –k)

and (k, k) is

59. Find perpendicular distance from the origin to the line joining the points (cosθ, sin θ) and (cos ϕ, sin ϕ).

Kindly go through the solution

24. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

24.

Slope of line passing through points (2, 5) and (–3, 6) is

$m_{1} = \frac{- 1}{m_{1}} = h$

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

So, slope of line perpendicular to line through (2, 5) and (–3, 6) is

$m_{2} = - \frac{1}{(\frac{- 1}{5})} = 5$

$∴$ Equation of line with slope 5 and passing through (–3, 5) is

$\Rightarrow 5 = \frac{y - 5}{x - (- 3)} = \frac{y - 5}{x + 3}$

$\Rightarrow 5 (x + 3) = y - 5$

$\Rightarrow 5 x + 15 - y + 5 = 0$

$\Rightarrow 5 x - y + 20 = 0$

41. Find the points on the x-axis, whose distances from the line $\frac{x}{3} + \frac{y}{4} = 1$ are 4 units.

Kindly go through the solution

25. 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.

25.

Let P(1, 0) and Q(2, 3) be the given points. Then, slope of line joining PQ,

$m_{1} = \frac{3 - 0}{2 - 1} = \frac{3}{1} = 3$

If A divides PQ with ratio 1:n, co-ordinate of A is

$(\frac{m x_{2} + n x_{1}}{m + n}, \frac{m y_{2} + n y_{1}}{m + n})$

$= (\frac{1 \times 2 + n \times 1}{1 + n}, \frac{1 \times 3 + n \times 0}{1 + n})$

$= (\frac{2 + n}{1 + n}, \frac{3}{1 + n})$

So, slope of line perpendicular to line joining points P and Q

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

Hence, equation of line passing through point A and (perpendicular to line) joining points P and Q is given by

$m_{2} = \frac{y - y_{0}}{x - x_{0}}$

$\Rightarrow - \frac{1}{3} = \frac{y - (\frac{3}{1 + n})}{x - (\frac{2 + n}{1 + n})}$

$\Rightarrow - \frac{1}{3} = \frac{y (n + 1) - 3}{x (n + 1) - (2 + n)}$

$\Rightarrow - [x (n + 1) - (2 + n)] = 3 [y (n + 1) - 3]$

$\Rightarrow - x (n + 1) + (2 + n) = 3 y (n + 1) - 3 \times 3$

$\Rightarrow x (n + 1) + 3 (n + 1) y - (2 + n) - 9 = 0$

$\Rightarrow x (n + 1) + 3 (n + 1) y - n - 2 - 9 = 0$

$\Rightarrow x (n + 1) + 3 (n + 1) y - n - 11 = 0$

61. Find the equation of a line drawn perpendicular to the line $\frac{x}{3} + \frac{y}{6} = 1$ through the point, where it meets the y-axis.

61. The given Eqⁿ of the line is $\frac{x}{4} + \frac{y}{6}$ = 1 ______ (1)

so, Slope of line = - $\frac{6}{4} = - \frac{3}{2} .$

The line ⊥ to line (1) say l₂ has

Slope of l₂ = $\frac{- 1}{(- 3 / 2)} = \frac{2}{3} .$

Let P (0, y) be the point of on y-axis where it is cut by the line (1)

Then, $\frac{0}{4} + \frac{y}{6} = 1$

y = 6

i.e, the point P has co-ordinate (0, 6)

Eqⁿ of line ⊥ to $\frac{x}{4} + \frac{y}{6} = 1$ and cuts y-axis at P (0,6) is

y – 6 = $\frac{2}{3}$ (x – 0)

3y – 18 = 2x

2x – 3y + 18 = 0

64. If three lines whose equations are y = m₁x + c₁, y = m₂x + c₂ and y = m₃x + c₃ are concurrent, then show that m₁(c₂ – c₃) + m₂ (c₃ – c₁) + m₃ (c₁ – c₂) = 0.

64. The given eqⁿ of the three lines are

y = m₁ x + c₁ ______ (1)

y = m₂ x + c₂ ______ (2)

y = m₃ x + c₃ ______ (3)

The point of intersection of (2) and (3) is given by.

y - y = (m₂x + c₂) - (m₃ x + c₃)

(m₂ - m₃) x = c₃ - c₂

$\Rightarrow x = \frac{c_{3} - c_{2}}{m_{2} - m_{3}} .$

Hence, y = $\frac{m_{2} (c_{3} - c_{2})}{(m_{2} - m_{3})} + c_{2}$

$= \frac{m_{2} (c_{3} - c_{2}) + c_{2} (m_{2} - m_{3})}{m_{2} - m_{3} .}$

$= \frac{m_{2} c_{3} - m_{3} c_{2}}{m_{2} - m_{3}} .$

$i e, (\frac{c_{3} - c_{2}}{m_{2} - m_{3}}, \frac{m_{2} c_{3} - m_{3} c_{2}}{m_{2} - m_{3}})$

As the three lines are concurrent, the point of intersection of (2) and (3) lies on line (1) also

i e, $\frac{m_{2} c_{3} - m_{3} c_{2}}{m_{2} - m_{3}} = m_{1} (\frac{c_{3} - c_{2}}{m_{2} - m_{3}}) + c_{1}$

$\Rightarrow \frac{- m_{1} (c_{2} - c_{3}) + c_{1} (m_{2} - m_{3})}{m_{2} - m_{3}} = \frac{- m_{2} c_{3} - m_{3} c_{2}}{m_{2} - m_{3}} .$

m₁ (c₂ - c₃) - c₁ (m₂ - m₃) + m₂ c₃ - m₃ c₂ = 0

m₁ (c₂ - c₃) - m₂ c₁ + m₃ c₁ + m₂ c₃ - m₃ c₂ = 0

m₁ (c₂ - c₃) + m₂ (c₃ - c₁) + m₃ (c₁ - c₂) = 0

43. Find equation of the line parallel to the line 3x − 4y + 2 = 0 and passing throughthe point (–2, 3).

43.

The slope of the line 3x - 4y + 2 = 0 is,

$m =- \frac{}{} \frac{3}{4}$
$= - \frac{3}{- 4} = \frac{3}{4}$

So, slope of line parallel to 3x - 4y + 2 = 0 is also m= $\frac{3}{4}$ .

Hence, this parallel line passes through ( -2, 3) we can write,

$m = \frac{y - y_{0}}{x - x_{0}}$

$\frac{3}{4} = \frac{y - 3}{x - (- 2)} = \frac{y - 3}{x + 2}$

3 (x + 2) = 4 (y - 3)

3x + 6 = 4y - 12

3x - 4y + 6 + 12 = 0.

3x - 4y + 18 = 0. Which is the required equation of line.

51. The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c.

51.

Let 0 (o, o) be the origin and P (-1, 2) be the given point on the line y = mx + c.

Then, slope of OP, = $= \frac{2 - 0}{- 1 - 0}$

Slope of OP = -2

As the line y = mx + c is ⊥ to OP we can write

60. Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0

60. The given eqn of lines are

x - 7y + 5 = 0 ______ (1) ⇒ x = 7y - 5

and 3x + y = 0 _________ (2)

Solution (1) and (2) we get,

3 [7y – 5] + y = 0 .

⇒ 21y - 15 + y = 0

⇒ 22y = 15

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

12.

Let the line l passes through points (x₁, y₁) and (h, x)

Then slope of l,

$m = \frac{k - y_{1}}{h - x_{1}}$

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

Hence, proved

26. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

26.

Let the line cuts x and y axis at intercept c. Then a = b = c

Then by intercept form the equation of line is

$\frac{x}{a} + \frac{y}{b} = 1$

$\Rightarrow \frac{x}{c} + \frac{y}{c} = 1$

$\Rightarrow x + y = c (1)$

Since equation (1) passes through point (2, 3).

Hence 2 + 3 = c

$\Rightarrow c = 5$

So, substituting value of c in equation (1) we get

$\Rightarrow x + y = 5$

$\Rightarrow x + y - 5 = 0$

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

Let a and b be the intercepts on x and y-axis

Then a + b = 9

$\Rightarrow b = 9 - a (1)$

Using intercept form equation of line with a and b intercepts are

$\frac{x}{a} + \frac{y}{b} = 1$

$\Rightarrow \frac{x}{a} + \frac{y}{9 - a} = 1 (2)$

As point (2, 2) passes the line having equation of form equation (2) we can write as

$\frac{2}{a} + \frac{2}{9 - a} = 1$

$\Rightarrow \frac{2 (9 - a) + 2 a}{a (9 - a)} = 1$

$\Rightarrow 18 - 2 a + 2 a = (9 - a) a$

$\Rightarrow 18 = 9 a - a^{2}$

$\Rightarrow a^{2} - 9 a + 18 = 0$

$\Rightarrow a^{2} - 3 a - 6 a + 18 = 0$

$\Rightarrow a (a - 3) - 6 (a - 3) = 0$

$\Rightarrow (a - 3) (a - 6) = 0$

So, a = 3, 6

Case I

When a = 3, b = 9 – 3 = 6. Then the equation of line is

$\frac{x}{3} + \frac{y}{c} = 1$

$\Rightarrow \frac{2 x + y}{6} = 1$

$\Rightarrow 2 x + y = 6$

$\Rightarrow 2 x + y - 6 = 0$

Case II

When a = 6, b = 9 – 6 = 3. Then equation of line is

$\frac{x}{6} + \frac{y}{3} = 1$

$\Rightarrow \frac{x + 2 y}{6} = 1$

$\Rightarrow x + 2 y = 6$

$\Rightarrow x + 2 y - 6 = 0$

Hence, equation of line is 2x + y – 6 or x + 2y – 6 = 0

36. By using the concept of equation of a line, prove that the three points (3, 0), (– 2, – 2) and (8, 2) are collinear.

36.

Let the given points be A (3, 0), B (–2, –2) and C (8, 2). Then by two point form we can write equation of line passing point A (3, 0) and B (–2, –2) as

$y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$

$\Rightarrow y - 0 = \frac{- 2 - 0}{- 2 - 3} (x - 3)$

$\Rightarrow y = \frac{- 2}{- 5} (x - 3)$

$\Rightarrow 5 y = 2 (x - 3)$

$\Rightarrow 5 y = 2 x - 6$

$\Rightarrow 2 x - 5 y - 6 = 0 (1)$

If the three points A, B and C are co-linear, C will also lieonm the line formed by AB or satisfies equation (1).

Hence, putting x = 8 and y = 2 we have

L.H.S. = 2 × 8 – 5 × 2 – 6

= 16 – 10 – 6

= 0 = R.H.S.

The given three points are collinear.

44. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and havingx intercept 3.

44.

The slope of line x- 7y + 5 = 0 is

So, equation of line with slope m₂ and having x-intercept 3 is

y = m (x-d), d = x- intercept

⇒y = - 7 (x- 3)

⇒y = - 7x + 21

⇒ 7x + y- 21 = 0.

47. Prove that the line through the point (x₁, y₁) and parallel to the line Ax + By + C = 0 is A (x –x₁) + B (y – y₁) = 0.

47.

The slope of line 4x + by + c = 0 is

m = -A/B

As the required line is parallel to the line Ax + by + c = 0

They have the same slope ie, m = -A/B

So, equation of line with slope m and passing through (x₁, y₁)

is given by point-slope from as,

⇒ - A (x-x₁) B (y -y₁)

⇒ A (x-x₁) + B (y-y₁)= 0.

Hence proved.

48. Two lines passing through the point (2, 3) intersects each other at an angle of 60^o. If slope of one line is 2, find equation of the other line.

48.

Given, slope of line 1, m₁ = 2.

Let m₂ be the slope of line 2.

If θ is the angle between the two lines then we can write,

55. Find the values of k for which the line (k–3) x – (4 – k2) y + k2 –7k + 6 = 0 is

(a) Parallel to the x-axis,

(b) Parallel to the y-axis,

(c) Passing through the origin.

55. We have (k - 3) x - (4 - k²) y + k² - 7 y + 6 = 0.

(i) When the line is parall to x-axis, all x coefficient = 0. then,

(k - 3)x - (4 -k²)y + k²- 7y + 6 = 0 x.x - a x y where a = constant

Equating the co-efficient,

K – 3 = 0

=> k = 3

(ii) When the line is parallel to y-axis all y co-efficient = 0 then

- (4 -k)² = 0

=> – 4 + x² = 0

k² = 4

k = ± 2.

(iii) When the line pares through origin, (0, 0) need satisfy the given eqn then,

k² - 7k + 6 = 0

k² - k – 6k + 6 = 0

k (k- 1) - 6 (k - 1) = 0

(k = 1) (k - 6) = 0

k = 1 and k = 6

65. Find the equation of the lines through the point (3, 2) which make an angle of 45^o with the line x – 2y = 3.

65. x – 2y = 3

y = $\frac{x}{2}$ - $\frac{3}{2}$ ______ (1)

Slope of line (1) is $\frac{1}{2} \cdot$

Let the line through P (3, 2) have slope m

Then, angle between the line = $| \frac{m - \frac{1}{2}}{1 + m \frac{1}{2}} |$

$\Rightarrow t a n 4 5^{°} = | \frac{2 m - 1}{2 + m} |$

$\Rightarrow 1 = | \frac{2 m - 1}{2 + m} |$

$\Rightarrow \frac{2 m - 1}{2 + m} = \pm 1 .$

When, $\frac{2 m - 1}{2 + m} = 1$ =>2m – 1 = 2 + m=> m = 3.

The eqn of line through (3, 2) is

y – 2 = 3 (x – 3) 3x – y – 7 = 0.

When $\frac{2 m - 1}{2 + m}$ = – 1=> 2m – 1 = – 2 – m =>3m = – 1 m = $- \frac{1}{3}$

The equation of line through (3,2) is,

y – 2 = $- \frac{1}{3}$ (x – 3) => 3y – 6 = – X + 3

x + 3y – 9 = 0

67. Show that the equation of the line passing through the origin and making an angle ø with the line $y = m x + c i s \frac{y}{x} = \frac{m \pm t a n,}{m ? t a n,} .$

67. The given eqⁿ of line is.

l₁ : y = mx + c.

Slope of l₁ = m

Let m? be the slope of line passing through origin (0, 0) and making angle θ with l₁

Thus, (y 0) = m? (x 0)

y = m? x

m? = $y$
$x$ ______ (1)

And tanθ = $| \frac{m^{'} - m}{1 + m^{'} \cdot m} |$ = $\frac{m^{'} - m}{1 + m^{'} m}$

When, tanθ = $\frac{m^{'} - m}{1 + m^{'} m} .$

tanθ + m? m tanθ = m’ - m

m + tanθ = m? - m?m tanθ

m' = $\frac{m + t a n θ}{1 - m t a n θ} .$

When tan θ = $- (\frac{m^{'} - m}{1 + m^{'} m})$

tan θ + m? m tanθ = -m? + m

m' = $\frac{m - t a n θ}{1 + m t a n θ} .$

Hence combining the two we get,

$m^{'} = \frac{m \pm t a n θ}{1 ? m t a n θ} .$

$\Rightarrow \frac{y}{x} = \frac{m \pm t a n θ}{1 ? m t a n θ} .$ {-: eqⁿ (1) }

18. Passing through (2 , 2√3) and inclined with the x-axis at an angle of 75^o.

18.

The slope of the line is, m = tan 75° – tan (45° + 30°)

68. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4?

68. The given eqⁿ of line is

l1: x + y = 4

Let R divides the line joining two points P (?1,1) and Q (5,7) in ratio k:1. Then,

Co-ordinate of R = ( $(\frac{5 k - 1}{k + 1}, \frac{7 k + 1}{k + 1})$ )

As l₁ divides line joining PQ, then R lies on l₁

i e, $(\frac{5 k - 1}{k + 1}, \frac{7 k + 1}{k + 1})$ =4

5k ?1 + 7k + 1= 4 (k + 1)

12k = 4k + 4

8k = 4

k = $\frac{1}{2}$

The ratio in which x + y = 4 divides line joining (?1,1) ad (5,7) is $\frac{1}{2}$ :1 i.e., 1: 2.

15. Write the equations for the x-and y-axes.

Exercise 9.2

15.

For any points on x-axis the y-co-ordinate or the ordinate is always 0. Hence, the x-axis have the equation y = 0.

Similarly for xy points on y-axis the x-co-ordinate or the abscissa is always 0. Hence, the y-axis have the equation x = 0.

19. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.

19.

Given m = –2

x-intercept, d = –3

Using slope intercept form, equation of line is

y = m (x – d)

$\Rightarrow y = (- 2) [x - (- 3)]$

$\Rightarrow y = (- 2) (x + 3)$

$\Rightarrow y = - 2 x - 6$

$\Rightarrow 2 x + y + 6 = 0$

21. Passing through the points (–1, 1) and (2, – 4).

.Here (x₁, y₁) = (–1, 1) and (x₂, y₂) = (2, –4)

Using two point form, equation of line is

$y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$

$\Rightarrow y - 1 = \frac{- 4 - 1}{2 - (- 1)} (x - (- 1))$

$\Rightarrow y - 1 = \frac{- 5}{2 + 1} (x + 1)$

$\Rightarrow 3 (y - 1) = - 5 (x + 1)$

$\Rightarrow 3 y - 3 = - 5 x - 5$

$\Rightarrow 5 x + 3 y - 3 + 5 = 0$

$\Rightarrow 5 x + 3 y + 2 = 0$

35. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.

35. Equation of line with intercept form is

$\frac{x}{a} + \frac{y}{b} = 1 (1)$

As R (h, x) divides line segment joining point A (a, 0) and B (0, b) in the ratio 1 : 2 we can write,

$(h, k) = (\frac{1 \times 0 + 2 \times a}{1 + 2}, \frac{1 \times b + 2 \times 0}{1 + 2})$

So, $h = \frac{0 + 2 a}{3} k = \frac{b + 0}{3}$

$\Rightarrow 3h = 2 a b = 3 k$

$\Rightarrow a = \frac{3 h}{2} b = 3 k$

Hence, putting value of a and b in equation (1) we get,

$\frac{x}{3 h / 2} + \frac{y}{3k} = 1$

$\Rightarrow \frac{x}{3h} + \frac{y}{3k} = 1$

16. Passing through the point (– 4, 3) with slope $\frac{1}{2}$ .

Given,

$m = \frac{1}{2}, (x_{0}, y_{0}) = (- 4, 3)$

By point slope form equation of line is

$m = \frac{y - y_{0}}{x - x_{0}}$

$\Rightarrow \frac{1}{2} = \frac{(y - 3)}{x - (- 4)}$

$\Rightarrow x + 4 = 2 (y - 3)$

$\Rightarrow x + 4 = 2 y - 6$

$\Rightarrow x - 2 y + 4 + 6 = 0$

$\Rightarrow x - 2 y + 10 = 0$

42. Find the distance between parallel lines

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0 (ii) l (x + y) + p = 0 and l (x + y) – r = 0.

42.

(i) Given, equation of lines are

15x + 8y- 34 = 0

15x + 8y + 31 = 0

So, c₁ = 34 and c₂ = 31, A = 15 and B = 8

53. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

53. Let P be the point on the BC dropped from vertex A.

Slope of BC $= 2 -$
$\frac{(-1)}{1 - 4}$

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

$= \frac{3}{- 3}$

= $-$ 1.

As A P $⊥$ BC,

Slope of AP= $\frac{- 1}{slope of B C} = \frac{- 1}{- 1} = 1 .$

Using slope-point form the equation of AP is,

$1 = \frac{y - 3}{x - 2}$

$\Rightarrow$ x $-$ 2 = y $-$ 3

$\Rightarrow$ x – y – 2 + 3 = 0 $\Rightarrow$ x – y + 1 = 0

The equation of line segment through B(4, -1) and C(1, 2) is.

$y - (- 1) = \frac{2 - (- 1)}{1 - 4} (x - 4) .$

$\Rightarrow y + 1 = \frac{2 + 1}{- 3} (x - 4)$

$\Rightarrow (y + 1) = \frac{3}{- 3} (x - 4) .$

$\Rightarrow y + 1 = - (x - 4)$

$\Rightarrow x y + 1 = - x + 4$

$\Rightarrow - x + y + 1 - 4 = 0$

$\Rightarrow$ $x + y - 3 = 0$

So, A=1, B=1 and C= $-$ 3.

Hence, length of AP=length of $⊥$ distance of A(2,3) from BC.

56. Find the values of q and p, if the equation x cosØ + y sinØ = p is the normal form of the line √3 $x + y + 2 = 0 .$

56. The given equation of the line is 3x + y + 2 = 0

57. Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and – 6, respectively.

57. Let a and b be the x & y intercept. Then,

$\frac{x}{a} + \frac{y}{b} = 1 .$ _____ (1)

Given, a + b = 1. ______ (2) b = 1 -a _____ (3)

and ab = -6 _____ (4)

Putting eqn (B) in (iii) we get a

a (1- a) = - 6

a - a² = - 6

a² - a - 6 = 0

a² + 2a - 3a - 6 = 0

a (a + 2) - 3 (a + 2) = 0

(a + 2) (a -3) = 0

(a + 2) (a -3) = 0.

a = 3 or a = -2.

When a = 3, b = 1- a = 1 - 3 = - 2

When a = - 2, b = 1 - (-2) = 1 + 2 = 3

So, (a, b) = (3, -2) and (-2, 3)

Hence, eqⁿ (1) becomes,

$\frac{x}{3} + \frac{y}{- 2} = 1$ and $\frac{x}{- 2} + \frac{y}{3} = 1 .$

2x – 3y = 6 and 2y - 3x = 6

Gives the read eqⁿ of lines

69. Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.

69.

The given eqⁿ of the lines are.

4x + 7y + 5 = 0______ (1)

2x - y = 0 ______ (2)

Solving (1) and (2) we get,

4 x + 7 (2 x)+5 = 0

4x +14 x + 5= 0

x = $\frac{- 5}{18}$

and y = 2x = $2 (\frac{- 5}{18}) = \frac{- 5}{9}$

72. If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y +7 = 0 is always 10. Show that P must move on a line.

72. The given eqⁿ of the lines are.

x + y ? 5 = 0 _______ (1)

3x ? 2y + 7 = 0 ______ (2)

Given, sum of perpendicular distance of P (x, y) from the two lines is always 10 .

The above eqⁿ can be expressed as a linear combination Ax + By + C = 0 where A, B & C are constants representing a straight line

P (x, y) mover on a line.

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

Exercise 9.1

1. Let the given points be A(–4, 5), B(0, 7), C(5, –5) and D(–4, –2).

Then quadrilated ABCD can be plotted on the graph by joining the points A, B, C and D.

We connect diagonal AC such that

area (ABCD) = (ΔABC) + (ΔADC)

Now,

$(ΔABC) = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= \frac{1}{2} | - 4 [7 - (- 5)] + 0 [- 5 - 5] + 5 [5 - 7] | unit^{2}$

$= \frac{1}{2} | - 4 (7 + 5) + 0 + 5 (5 - 7) | unit^{2}$

$= \frac{1}{2} | (- 4) \times 12 + 5 \times (- 2) |unit^{2}$

$= \frac{1}{2} | - 48 - 10 |unit^{2}$

$= \frac{1}{2} | - 58 | unit^{2}$

$= \frac{58}{2} unit^{2} = 29 unit^{2}$

Similarly, $(ΔACD) = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= \frac{1}{2} | - 4 [- 5 - (2)] + 5 [- 2 - 5] + (- 4) [5 - (- 5)] | unit^{2}$

$= \frac{1}{2} | - 4 (- 5 + 2) + 5 (- 2 - 5) - 4 (5 + 5) |unit^{2}$

$= \frac{1}{2} | (- 4) \times (- 3) + 5 \times (- 7) - 4 \times (10) |unit^{2}$

$= \frac{1}{2} | 12 - 35 - 40 |unit^{2}$

$= \frac{1}{2} | - 63 | unit^{2}$

$= \frac{63}{2} unit^{2}$

Hence, area (ABCD) = $29 + \frac{63}{2} unit^{2}$

$= \frac{58 + 63}{2} unit^{2} = \frac{121}{2} unit^{2}$

2. 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.

2. Let ABC be the equilateral triangle of side 2a and 0 be the origin. Then

AB = BC = AC = 2a

O is the mid-point of AB we have AO = a

BO = a

We know that A and B lies on y-axis so they have co-ordinate of the form (0, y).

Hence, co-ordinate of A is (0, a) and that of B is (0, –a)

Since OC, bisects AB at right angle, by Pythagoras theorem,

AC²= OA² + OC²

$\Rightarrow {(2 a)}^{2} = {(a)}^{2} +OC^{2}$

$\RightarrowOC^{2} = 4 a^{2} - a^{2}$

And as C we on x-axis it has co-ordinate of the form (x, 0)

3. 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.

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

Let A (x, 0) be the point on x-axis when is equidistant from P (7, 6) and Q (3, 4)

Then, PA = QA

Squaring both sides, we get,

$\Rightarrow x^{2} - 14 x + 49 + 36 = x^{2} - 6 x + 9 + 16$

$\Rightarrow 49 + 36 - 9 - 16 = 14 x - 6 x$

$\Rightarrow 60 = 8 x$

$\Rightarrow x = \frac{60}{8} = \frac{15}{2}$

$∴$ The required point on x-axis is $(\frac{15}{2}, 0) .$

14. Consider the following population and year graph (Fig 9.10), find the slope of the line AB and using it, find what will be the population in the year 2010 ?

14.

From figure,

Slope of AB = $\frac{97 - 92}{1995 - 1985}$

$= \frac{5}{10}$

$= \frac{1}{2}$

As, A, B and C lie on same line

Slope of AB = Slope of BC

$\Rightarrow \frac{1}{2} = \frac{a - 97}{2010 - 1995} = \frac{a - 97}{15}$

$\Rightarrow 15 = 2 (a - 97)$

$\Rightarrow 15 = 2 a - 194$

$\Rightarrow 194 + 15 = 2 a$

$\Rightarrow 209 = 2 a$

$\Rightarrow a = \frac{209}{2} = 104.5$

Hence, population in 2010 will be 104.5 crores.

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

31.

Assuming the price per litre say P in x-axis and the corresponding demand say D in y-axis, we have two point (14, 980) and (16, 1220) in xy plane. Then the points (P, D) will satisfy the equation.

$\RightarrowD - 980 = \frac{1220 - 980}{16 - 14} (P - 14)$

$\RightarrowD - 980 = \frac{240}{2} (P - 14)$

$\RightarrowD - 120 (P - 14) + 980$

$\RightarrowD - 12 0P - 1680 + 980$

$\RightarrowD = 12 0P - 700$

Which is the required relation

Where P = 17, we have

D = 120 × 17 – 700

D = 1340

Hence, the owner can sell 1340 litres of milk weekly at 17/litre

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

13.

Since the three points say P (h, o), Q (a, b) and R (o, k) lie on a line.

Slope of PQ = slope of QR

$\Rightarrow \frac{b - o}{a - h} = \frac{k - b}{o - a}$

$\Rightarrow \frac{b}{a - h} = \frac{k - b}{- a}$

$\Rightarrow - a b = (k - b) (a - h)$

$\Rightarrow - a b = k a - k h - a b + b h$

$\Rightarrow k a + b h = k h$

Dividing both sides by kh we get,

$\Rightarrow \frac{k a}{k h} + \frac{b h}{k h} = \frac{k h}{k h}$

$\Rightarrow \frac{a}{h} + \frac{b}{k} = 1$

17. Passing through (0, 0) with slope m.

17.

Given slope = m and $(x_{0}, y_{0}) = (0, 0)$

By point slope from, equation of line is

$m = \frac{y - y_{0}}{x - x_{0}}$

$\Rightarrow m = \frac{y - 0}{x - 0}$

$\Rightarrow m = \frac{y}{x}$

$\Rightarrow m x = y$

$\Rightarrow$ $y = m x$

22. Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30⁰.

23.

Given P = 5 and W = 30°

Using normal form, equation of line is

x cos W + y sin W = P

23. The vertices of ΔPQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.

23.

Let RM be the median draw from vertex R so that M is the mid-point of line segment PQ. Then,

$= (\frac{2 - 2}{2}, \frac{1 + 3}{2}) = (\frac{0}{2}, \frac{4}{2})$

= (0, 2)

So equation of line passing through R (4, 5) and M (0, 2) is

$y - 5 = \frac{2 - 5}{0 - 4} (x - 4)$

$\Rightarrow y - 5 = \frac{- 3}{- 4} (x - 4)$

$\Rightarrow 4 (y - 5) = 3 (x - 4)$

$\Rightarrow 4 y - 20 = 3 x - 12$

$\Rightarrow 3 x - 4 y + 20 - 12 = 0$

$\Rightarrow 3 x - 4 y + 8 = 0$

29. The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.

29.

The slope of line passing through (0, 0) and (–2, 9) is

$m_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{9 - 0}{- 2 - 0} = \frac{9}{- 2}$

The line perpendicular to the line having slope m1 will have the slope

$m_{2} = \frac{- 1}{m_{1}} = \frac{- 1}{- 9 / 2} = \frac{2}{9}$

So, the equation of line with slope m2 and passing $(x_{0}, y_{0}) = (- 2, 9)$ is

$m_{2} = \frac{y - y_{0}}{x - x_{0}}$

$\Rightarrow \frac{2}{9} = \frac{y - 9}{x - (- 2)} = \frac{y - 9}{x + 2}$

$\Rightarrow 2 (x + 2) = 9 (y - 9)$

$\Rightarrow 2 x + 4 = 9 y - 81$

$\Rightarrow 2 x - 9 y + 4 + 81 = 0$

$\Rightarrow 2 x - 9 y + 85 = 0$

30. 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.

30.

Assuming L along y-axis and C along x-axis, we have two points (124.942, 20) and (125.134, 110) in xy-plane. By two-point form, the point L and C satisfies the equation.

y-124.942= $\frac{(125.134 - 124.942)}{(110 - 20)}$ (x-20)

$\Rightarrow$ y-124.942= $\frac{0.192}{90}$ (x - 20)

$\Rightarrow$ y-124.942= $\frac{0.032}{15}$ (x - 20)

$\Rightarrow$ 15y – 1874.13=0.032x -0.64

$\Rightarrow$ 15y= 0.032x +1873.49

$\Rightarrow$ y = 0.0021x+124.8993

$\Rightarrow$ L = 0.0021C + 124.8993

32. P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is $\frac{x}{a} + \frac{y}{b} = 2$ .

32.

Since P (a, b) is the mid-point of the line segment say AB with points A (0, y) and B (x, 0) we can write,

$(a, b) = (\frac{x + 0}{2}, \frac{y + 0}{2})$

$\Rightarrow a = \frac{x}{2}$

$b = \frac{y}{2}$

$\Rightarrow x = 2 a$
$y = 2 b$

So, the equation of line with x and y intercept 2a and 2b using intercept form is

$\frac{x}{2a} + \frac{y}{2b} = 1$

$\Rightarrow \frac{x}{a} + \frac{y}{b} = 2$

Hence, proved

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

33. Assuming the price per litre say P in x-axis and the corresponding demand say D in y-axis, we have two point (14, 980) and (16, 1220) in xy plane. Then the points (P, D) will satisfy the equation.

$\RightarrowD - 980 = \frac{1220 - 980}{16 - 14} (P - 14)$

$\RightarrowD - 980 = \frac{240}{2} (P - 14)$

$\RightarrowD - 120 (P - 14) + 980$

$\RightarrowD - 12 0P - 1680 + 980$

$\RightarrowD = 12 0P - 700$

Which is the required relation

Where P = 17, we have

D = 120 × 17 – 700

D = 1340

Hence, the owner can sell 1340 litres of milk weekly at? 17/litre

34. P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is $\frac{x}{a} + \frac{y}{b} = 2$ .

34.

Since P (a, b) is the mid-point of the line segment say AB with points A (0, y) and B (x, 0) we can write,

$(a, b) = (\frac{x + 0}{2}, \frac{y + 0}{2})$

$\Rightarrow a = \frac{x}{2} ? b = \frac{y}{2}$

$\Rightarrow x = 2 a ? y = 2 b$

So, the equation of line with x and y intercept 2a and 2b using intercept form is

$\frac{x}{2a} + \frac{y}{2b} = 1$

$\Rightarrow \frac{x}{a} + \frac{y}{b} = 2$

Hence, proved

37. Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts.

(i) x + 7y = 0, (ii) 6x + 3y – 5 = 0, (iii) y = 0.

Exercise 9.3

37. (i) Given, x + 7y = 0.

7y = -x

y = $\frac{-1}{7}$ x + 0.

Comparing the above equation with y = mx + c we get, slope, m = - $\frac{1}{7}$ and c = 0, y-intercept

(ii) Given, 6x + 3y - 5 = 0

3y = -6x + 5

y = - $\frac{6}{3}$ x + $\frac{5}{3}$ = -2x + $\frac{5}{3}$

Comparing the above equation with y = mx + c we get, slope, m = -2 and $c = \frac{5}{3}$ , y-intercept

(iii) Given, y = 0

y = 0xx + 0

Comparing the above equation with y = mx + c we get, Slope, m = 0 and c = 0, y-intercept.

38. Reduce the following equations into intercept form and find their intercepts onthe axes.

(i) 3x + 2y – 12 = 0, (ii) 4x – 3y = 6, (iii) 3y + 2 = 0.

38. (i) Given, 3x + 2y 12 = 0.

3x + 2y = 12

Dividing both sides by 12 we get,

$\Rightarrow \frac{3 x}{12} + \frac{2 y}{12} = \frac{12}{12}$

$\Rightarrow \frac{x}{4} + \frac{y}{6} = 1 .$

Comparing the above equation with $\frac{x}{a} + \frac{y}{b} = 1$ = we get, x-intercept, a = 4 and y-intercept b = 6.

(ii) Given, 4x - 3y = 6

Dividing the both sides by 6.

$\frac{4 x}{6} - \frac{3 y}{6} = \frac{6}{6}$

$\Rightarrow \frac{2 x}{3} - \frac{y}{2} = 1$

$\Rightarrow \frac{x}{3 / 2} + \frac{y}{(- 2)} = 1 .$

Comparing above equation by $\frac{x}{a} + \frac{y}{b} = 1$ we get, x-intercept a = $\frac{3}{2}$ and y-intercept, b = -2

(iii) Given, 3y + 2 = 0.

3y = -2

$y = - \frac{2}{3}$

As the equation of line is of form y = constant, it is parallel to x-axis and has no x-intercept.

y-intercept = - $\frac{2}{3}$

View other answers

39. Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

(i) x – √3y + 8 = 0, (ii) y – 2 = 0, (iii) x – y = 4.

As w lies in IV^th quadrant

Cos w = cos 45° and sin w = - sin 45°

= cos (360°- 45°) = sin (360°- 45°)

= cos 315° = sin 315°

So, w = 315°

40. Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

40.

The given equation of the line is.

12 (x + 6) = 5 (y- 2)

⇒ 12x + 72 = 5y- 9

⇒ 12x- 5y + 72 + 9 = 0

⇒ 12x- 5y + 82 = 0

The perpendicular distance of point (-1, 1) from the line is given by

46. The line through the points (h, 3) and (4, 1) intersects the line 7x − 9y −19 = 0 at right angle. Find the value of h.

46. Slope of line 1 (passing) through (h. 3) and (4, 1) is

49. Find the equation of the right bisector of the line segment joining the points (3, 4)and (–1, 2).

49.

Let P(3, 9) and Q (-1, 2) be the point. Let M bisects PQ at M so,

Co-ordinate of M = $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

$= (\frac{3 + (- 1)}{2}, \frac{4 + 2}{2})$

$= (\frac{3 - 1}{2}, \frac{6}{2}) = (\frac{2}{2}, \frac{6}{2}) = (1, 3)$

Now, slope of AB, m = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{2 - 4}{- 1 - 3} = \frac{- 2}{- 4} = \frac{1}{2} .$

As the bisects AB perpendicular it has slope $- \frac{1}{m}$ and it passes through M(1, 3) it has the equation of the form,

$\frac{- 1}{m} = \frac{y - y_{σ}}{x - x_{0}}$

$\frac{- 1}{\frac{1}{2}} = \frac{y - 3}{x - 1}$

$\Rightarrow - 2 = \frac{y - 3}{x - 1}$

$\Rightarrow - 2 (x - 1) = y - 3$

$\Rightarrow - 2 x + 2 = y - 3$

$\Rightarrow 2 x + y - 3 - 2 = 0$

2x + y - 5 = 0

50. Find the coordinates of the foot of perpendicular from the point (–1, 3) to theline 3x – 4y – 16 = 0.

50. Let P(-1, 3) be the given point and Q(x, y,) be the Co-ordinate of the foot of perpendicular

So, slope of line 3x - 4y - 16 = 0 is

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

And slope of line segment joining P(-1, 3) and Q(x, y,) is

$m_{2} = \frac{y_{1} - 3}{x_{1} - (- 1)} = \frac{y_{1} - 3}{x_{1} + 1}$

As they are perpendicular we can write as,

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

$\frac{3}{4} = - \frac{1}{(y_{1} - {3 / x}_{} + 1)}$

$\Rightarrow \frac{3}{4} = - \frac{(x_{1} + 1)}{(y_{1} - 3)}$

(y₁-3)3 = - 4(x₁ +1)

3y₁- 9 = - 4x₁- 4.

4x₁ + 3y₁-9 + 4 = 0

4x₁ + 3y₁-5 = 0 ___ (1)

As point Q(x₁, y₁) lies on the line 3x- 4y - 16 = 0 it must satisfy the equation hence,

3x₁- 4y₁- 16 = 0 ____ (2)

Now, multiplying equation (1) by 4 and equation (2) by 3 and adding then,

4× (4x₁ + 3y₁- 5) + 3(3x₁- 4y₁- 16) = 0.

16x₁ + 12y₁- 20 + 9x₁- 12y₁- 48 = 0

25x₁ = 48 + 20

$\Rightarrow x_{1} = \frac{68}{25}$ .

Putting value of x₁ in equation (1) we get,

$4 \times \frac{68}{25} + 3 y_{1} - 5 = 0$

4× 68 + 25× 3y₁ = 25 ×5.

25× 3y₁ = 125 - 272

$y_{1} = \frac{- 147}{25 \times 3}$

$\Rightarrow y_{1} = - \frac{49}{25}$

52. If p and q are the lengths of perpendiculars from the origin to the lines x cos q − ysinq = k cos2 and x sec q + y cosec q = k, respectively, provethat p² + 4q² = k².

52. The given equation lines are.

line 1: xcosθ-y sin θcos 2θ

⇒ xcosθ-y sin θ - kcos 2θ = 0

The perpendicular distance from origin (0,0) to line 1 is

54. If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that $\frac{1}{p^{2}} = \frac{1}{b^{2}} + \frac{1}{a^{2}} .$

54. The equation of line whose intercept on axes are a and b is given by,

$\frac{x}{a} + \frac{y}{b} = 1 .$

Multiplying both sides by ab we get,

$\frac{a b x}{a} + \frac{a b y}{b} = a b$

$\Rightarrow b x + a y - a b = 0 .$

58. What are the points on the y-axis whose distance from the line $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units.

58. Let (0, y) be the point on y-axis which is at a distance 4 unit from the line $\frac{x}{3} + \frac{y}{4} = 1$

Then, the line $\frac{x}{3} + \frac{y}{4} = 1$

4x + 3y = 12.

4x + 3y - 12 = 0

28. 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.

28. Kindly consider the following