1st PUC

Students can Download Sanskrit Vyakaran वर्णविचारः, 1st PUC Sanskrit Textbook Answers, helps you to revise complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Sanskrit Textbook Answers Vyakaran वर्णविचारः

Students can Download Sanskrit Shevadhi भूमिका, 1st PUC Sanskrit Textbook Answers, helps you to revise complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Sanskrit Textbook Answers Shevadhi भूमिका

Students can Download Sanskrit Shevadhi Lesson 13 विज्ञानपथः Questions and Answers, Notes Pdf, Summary, 1st PUC Sanskrit Textbook Answers, helps you to revise complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Sanskrit Textbook Answers Shevadhi Chapter 13 विज्ञानपथः

विज्ञानपथः Questions and Answers, Notes, Summary

विज्ञानपथः Summary in Kannada

विज्ञानपथः Summary in English

Summary

We see the growth of Sanskrit since the Vedic period. Side-by-side even science developed. Our ancestors were not only great thinkers but also scholars and pundits. During the period several books were written. Let us try to know the great achievements of Indians in Astronomy, Mathematics, Biology and Environmental Science.

Astronomy and Mathematics:

Advanced thinking can be seen in Astronomy and Mathematics. The earliest references to astronomy are found in the Rig Veda. We see the reference to the ten planets of the solar system in the Vedas. The red spot in the planet Jupiter is extremely amazing and it is possible to see it in the modern photographs taken through the telescope. Information such as separation of the two satellites Mars and Earth from the moon, the revolution of planets around the Sun, the gravitational pull of the Earth are found in the Vedas. The modern scientists are under the illusion that these are their discoveries. Our ancestors knew well that planets move like a spinning top. That planets revolve in their own axis was known to them. They also knew the fact that the Sun is the source of energy on the earth and the basis of earth. We also see the reference to day, night, month, season, phases (उत्तरायण and दक्षिणायण) and years (संवत्सर). That the waxing and waning of the moon and its luminaries depends entirely on the Sun was known to them.

It is a matter of pride that the concept of numbers and the base ten system being used all over the world today have been the greatest contributions of Indians. Another special feature is blending of the numerals through padas or words. It was necessary/essential to erect different types of sacrificial altars for different sacrifices. Hence, to build such altars they knew different types of geometrical designs, their area, modifications, etc. Hence, this field was known then itself. Indians were forerunners in Algebra, explanation and analysis of equations, especially simultaneous equations. A simultaneous equation with many variables, use of geometry in Algebra and exposition of Geometry through Algebra are examples of their vast knowledge. This land became fertile on account of great masters like Aryabhatta, Brahmagupta, Mahavira, Sridhara, Bhaskaracharya and others. Amongst them, Bhaskaracharya Il excels. His speciality is that the problems are posed by describing the beauty of natural objects. This field has expanded from the days of great masters like Aryabhatta, Brahmagupta, Mahavira, Sridhara, Bhaskaracharya to the modern mathematician Srinivas Ramanujan. Because of the contribution of all it has spread widely.

Biology and Environmental Science:

Our ancestors have contributed immensely to the field of Biology. They classified plants and animals on the basis of their nature quality. Independent works such as ‘Treatise on Horses’ and ‘Treatise on Elephants’ have been written. Dalvana is known to be a specialist in the study of serpents. There are references to his expertise in 13 varieties of serpents. They considered protection of the environment as their primary duty.

They perceived God in everything – Sun, trees, water, wind, earth etc. They always considered polluting or destroying the nature/environment as sin. The ancestors did everything keeping in mind the uses and protection of plants and animals. That is why they used to say that planting five mango trees will prevent an individual from going to hell. In the construction of wells, lakes and other sources of water their contribution was really great. Not only in construction but also in their maintenance they were devoted and dedicated. They inflicted/levied fine on those who polluted the place and punished them too. They had a law for the purpose. In this way they protected the ground water level. Works like ‘Aryabhatiyam’ written by Aryabhata, ‘Panchasiddhantika’ and ‘Brihatsamhita’ by Varahamihira, ‘Kakshaputatantram’ relating to Chemistry by Nagarjuna, works on Ayurveda by Charaka and Sushruta, ‘Siddhanta Siromani’ of Bhaskaracharya bear ample testimony to the intelligence of ancient Indians. Let us all devote ourselves to the study of these works.

Students can Download Maths Chapter 10 Straight Lines Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Maths Question Bank Chapter 10 Straight Lines

Question 1.
Define a point in a plane.
Answer:
A point in a plane is an ordered pair of real numbers.
If P = (x, y) then x and y are called the coordinates of P.

Important formulae
(i) If P = (x, y) then
x = the distance of P from y-axis = abscissa
y = the distance of P from x-axis = ordinate

(ii) Distance formulae
The distance between the points P(x₁,y₁) and Q (x₂, y₂) is given by
\( P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
The distance of the point (x, y) from the origin (0, 0) is \(\sqrt{x^{2}+y^{2}}\)

(iii) Section formulae

• The coordinates of the point which divides the line joining the points
  (x₁ y₁) and (x₂, y₂) internally in the ratio m : n are
  \(\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)\)
• The coordinates of the point which divides the line joining the points externally in the ratio m : n are
  \(\left(\frac{m x_{2}-n x_{1}}{m-n}, \frac{m y_{2}-n y_{1}}{m-n}\right)\)

In particular, if m = n, the coordinates of the mid-point of the line segment joining the points (x₁ y₁) and (x₂, y₂) are
\(\left(\frac{x_{2}+x_{1}}{2}, \frac{y_{2}+y_{1}}{2}\right)\)

(iv) Centroid of a triangle whose vertices are (x₁ y₁)  (x₂, y₂) and (x₃, y₃) is
\(\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\)

(v) Area of the triangle formed by the vertices (x₁ y₁)  (x₂, y₂) and (x₃, y₃) is
\(\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{3}\right)\right|\)

If the third vertex is at the origin then area is \( \frac{1}{2}\left|x_{1} y_{2}-x_{2} y_{1}\right| \)

If the area of the triangle ABC is zero,then three points A, B and C lie in a line (i.e., they are collinear)

Question 2.
Define slope of a line.
Answer:
If a straight line makes an angle ‘θ’ with the positive direction of x-axis then ‘θ’ is called inclination of the line and ‘tan θ’ is called slope or gradient of the line. The slope of the line is denoted by ‘m’
∴ Slope = tan θ = m

Observations:
(i) 0 < θ < 180°
(ii) Slope of the line is a real number.
(iii) If θ = 90°, then the slope of the line is not defined (∵ tan 90° = ∞)
(iv) Slope of x-axis is zero
0 = 0 ⇒ tan θ = tan 0 = 0
(v)Slope of y-axis is not defined ∵ θ = 90°
⇒ tan θ = tan 90° is ∞

Question 3.
Prove that the slope of the line joining the points
\(\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right) \text { is } m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Answer:
Let P((x₁, y₁)  and Q(x₂, y₂) be two points on the line whose inclination be θ (θ may be acute or obtuse) Draw QM perpendicular to x-axis and PR perpendicular to QM.


Note: Three points A, B and C are collinear if and only if.

Question 4.
Show that two lines with slopes m₁ and m₂ are
(i) parallel iff m₁ = m₂
(ii) perpendicular iff m₁ m₂ = -1.
Answer:
(i) Let the lines l₁ and l₂ have slopes mi and m₂ respectively and their respective inclinations be α and β.

∴ tan α = m₁ and tan β = m₂
If l₁ is parallel to l₂ then α = β
∴ tan α = tan β
Conversely, if m₁= m₂, then
tan α = tan β ⇒ α = β lines are parallel

(ii) If l₁ and l₂ are perpendicular then P = 90° + α

Question 5.
Derive the formula for finding the angle between the lines in terms of their slopes
Answer:
Let L₁ and L₂ be two non-vertical lines with slopes m₁ and m₂ respectively and a₁ and a₂ be the angles made by the lines L₁ and L₂ respectively.

∴ tan α₁= m₁ and tan a₂ – m₂
Let θ be the angle between the lines L₁ and L₂ and φ, adjacent angle to θ
From the figure,α₂ = α₁ + θ


Note: The obtuse angle between the two given lines is π – θ.

Question 6.
Define intercepts made by the line with coordinate axes
Answer:
If a straight line cuts x-axis at and y-axis at B, then OA with proper sign is called x-intercept and OB with proper sign is called y-intercept made by the straight line.

x-intercept is denoted by ‘a’ and y-intercept is denoted by ‘b’

Various forms (Different forms) of a line

Question 7.
Write the equations of x-axis and y-axis
Answer:
The x-axis is the locus of a point (x, y) which moves such that its y-coordinate is always zero.

∴ Equation of x – axis is y = 0
Similarly, equation of y-axis is x = 0

Question 8.
Write the equations of horizontal line (parallel to x-axis) and vertical line (parallel toy-axis or perpendicular to x-axis)
Answer:
Let a horizontal line L is at a distance ‘a’ from the x-axis then the ordinate of every point lying on the line is either a or -a

∴ Equation of the horizontal line is y = k
where k = a or – a
Similarly, equation of the vertical line is x = h where h = b or -b

Question 9.
Derive slope-intercept form of a line (with usual notations)
Or
Prove that the equation of a line with slope m and y-intercept y = mx + c.
Answer:
Let the line L with slope ‘m’ cuts the y-axis at a distance ‘c’ from the origin. Then y-intercept is ‘o’ and the line passes through A(0, c). Let P(x, y) be any point on the line. Then

Note: This can be derived using slope-point form

Question 10.
With usual, notations derive the equations of the line in the slope-point form as
y -y₀ = m(x – x₀)
OR
Prove that equation of a straight line with slope m and passing through the point
y -y₀ = m(x – x₀)
Answer:
Suppose that A(x₀, yo) is a fixed point on a line L whose slope is m. Let P(x, y) be an arbitrary point on L.

Question 11.
Prove that the equation of a line passing through the points (x₁, y₁) and (x₂, y₂) is
\(y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)\)
Answer:
Let the line L passes through the given point A(x₁, y₁)and B(x₂, y₂)

Question 12.
Prove that the equation of a line with x-intercept ‘a’ and y-intercept ‘b’ is \(\frac{x}{a}+\frac{y}{b}=1\) (Intercept form of a line)
Answer:
Because x-intercept is ‘a’
⇒ the line passes – through (a, 0)
y-intercept is b ⇒ the line passes through (0, b)

Question 13.
Prove that the equation of a line, ‘p’ as the length of the perpendicular drawn from origin and ‘w’ as the angle made by this perpendicular with positive directive of x-axis is x cos(w) + y sin(w) = p (Equation in the normal form)
Answer:
Let the line L meets x-axis at A and y-axis at B then required equation of the line is of the form .

Question 14.
Write the general solution of a straight line and reduce it into slope-intercept form intercept and normal form. Find its slope and intercepts on the axes.
Answer:
Any equation of the form Ax + By + C = 0, where A and B are not zero simultaneously is a general solution of a line.


Remark:
If C = 0, then Ax + By = 0, which is a line passing through the origin and has zero intercepts on the axes.

iii. Normal form: Let x cos(w) + y sin(w) = p be the normal form of the line represented by the equation Ax +By+ C = 0 or Ax +By = -C
Thus both the equation are same
∴ Corresponding coefficients are in proportional

Question 15.
Define concurrent lines
Answer:
Three lines are said to be concurrent, if they pass through a common point.i.e., three lines are said to be concurrent if the point of intersection of two of them lies on the third line.

Question 16.
Derive the formula for the length of perpendicular drawn from any point (x₁, y₁) to the line Ax + By + C = 0
OR
Prove that the length of the perpendicular from any point
\(A x+B y+C=0 \text { is }\left|\frac{A x_{1}+B y_{1}+C}{\sqrt{A^{2}+B^{2}}}\right|\)
Answer:
Let L: Ax + By + C = 0 be a line, whose distance from the point P(x₁, y₁)is d.

Question 17.
Show that the distance between two parallel lines y = mx + c₁ and y = mx + c₂ is
\(d=\left|\frac{c_{1}-c_{2}}{\sqrt{1+m^{2}}}\right|\)
Answer:

∴ Distance between l₁ and l₂ is equal to the length of the perpendicular from A to l₂
∴ Distance between the lines l₁ and l₂ is

Question 18.
Show that the distance between two parallel lines ax + by + c₁= 0 and ox + by + c₂ = 0 is \(\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|\)
Answer:
Let us assume that c₁ and c₂ are positive. Then the two lines lie on the same side of the origin

Let l₁ : ax + by + c₁ = 0   …(1)
And l₂ : ax + by + c₂ = 0 … (2)
Let ON and OM be the perpendicular distances from the origin to the lines l₁ and l₂ respectively.
Distance between parallel lines = MN = |ON – OM|
Now ON = length of perpendicular from O to l₁

Question 19.
Show that equation of the line with slope ‘m’ makes x-intercept is y = m (x – d)
Answer:
Suppose a line with slope m cuts the x-axis at a distance d from the origin. Therefore point on the x-axis is (d, 0)

∴ By slope-point form, required equation of the line is of the form
y – y₁= m (x – x₁)
Here (x_1, y₁) = (d,0)
∴ (1) ⇒ y – 0 = m(x – d)
i.e., y = m(x-d)

Problems

Question 20.
Find the slope of the lines
(a) Passing through the points (3, -2) and (-1,4)
(b) Passing through the points (3, -2) and (7,-2)
(c) Passing through the points (3, -2) and (3,4)
Answer:

Question 21.
The base of an equilateral triangle with side 2a lies along y-axis such that the mid-point of the base is at the origin. Find the vertices of the triangle.
Answer:
As ‘O’ is the midpoint of the side BC of an equilateral triangle ABC, ‘A’ should lie on the x-axis

Question 22.
Find the distance between P(x₁, y₁) and Q(x₂, y₂) when
(i) PQ is parallel to y-axis
(ii) PQ is parallel to x-axis
Answer:

Question 23.
Find a point on the x-axis, which is equidistant from the points (7,6) and (3,4)
Answer:
Let A = (7, 6) and B = (3, 4) be the given points. P be a point on the x-axis The P is of the form (x, 0)
Given that P is equidistant from A and B i..e, AP = BP
Using distance formula, we have

Question 24.
Find the slope of a line, which passes through the origin and the mid-point of the line segment joining the point P(0, -4) and B (8,0)
Answer:
Given: 0(0,0), P(0, -4) and Q(8, 0)

Question 25.
Without using Pythagoras theorem. Show that the points (4, 4), (3, 5) and (-1, -1) are the vertices of a right angled triangle
Answer:
Let A = (4, 4), B = (3, 5), C – (-1, -1) be the vertices of a ΔABC

Question 26.
Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
Answer:
The line l makes an angle 30° with positive y-axis in the anticlockwise direction and it inclines 120° with the positive x-axis.

Question 27.
Find the value of x for which the points (x, -1), (2,1) and (4,5) are collinear
Answer:
Given points are collinear
⇒ Area of triangle formed by them is zero

Question 28.
Without using distance formula, show that points (-2, -1), (4, 0) (3, 3) and (-3, 2) are the vertices of a parallelogram.
Answer:
Given points are A(-2, -1) B(4, 0), C(3, 3) and D(-3, 2)

Clearly
Slope of AB = Slope of CD and Slope of BC = Slope of DA
∴ opposite sides of the quadrilateral ABCD are parallel
⇒ ABCD is a parallelogram

Question 29.
Find the angle between the x-axis and the line joining the points (3, -1) and (4, -2)
Answer:
The given points are A(3, -1) and B = (4, -2)
\(\text { Slope of } A B=\frac{-2-(-1)}{4-3}=\frac{-1}{1}=-1\)
Let θ be the angle made by the line AB with the positive x-axis then slope of
AB = tan θ … (2)
∴ From (1) and (2), we have
tan θ = -1 = – tan 45°
= tan (180°-45°)
= tan (135°)
= θ =135°
AB makes an angle of 135° with positive direction of x-axis

Question 30.
The slope of a line is double of the slope of the another line. If tangent of the angle between them is \(\frac{1}{3}\),find the slopes of the lines
Answer:
Let the slopes of the two lines be m and 2m and 0 be the angle between the lines. Then

Question 31.
If the angle between two lines is \(\frac{\pi}{4}\) and slope of one the lines \(\frac{1}{2}\)is find the slope of other line
Answer:
We have, the acute angle between two lines with slopes m₁ and m₂ is

Question 32.
Line through the points (-2, 6) and (4, S) is perpendicular to the line through the points (8,12) and (x, 24). Find the value of x
Answer:
Slope of the line passing through the points (-2, 6) and (4, 8) is

Question 33.
A line passes through (x₁, y₁) and (h, k). If slope of the line is m, show that k-y₁ =m(h-x₁)
Answer:
Slope of the line passing through (x₁ > y₁) and (h, k) is
\(=\frac{k-y_{1}}{h-x_{1}}=m(\text { given })\)
k-y₁=m(h-x_l)

Question 34.
If the points (h, 0), (a, b) and (0, k) lie on a line, show that
\(\frac{a}{h}+\frac{b}{k}=1\)
Answer:
Let A(h, 0), B(a, b) and C(0, k) be the points lie on a line

Question 35.
Consider the following population and year graph find the slope of the line AB and using it, find what will be the population in the year 2010?
Answer:

Problems on various forms of the equation of a line.

Question 36.
Find the equation of the line which satisfy the given conditions
(i) Write the equation for x and y axes
(ii) Passing through the point (-4, 3) with slope \(\frac{1}{2}\)
(iii) Passing through (0, 0) with slope m
(iv) Passing through \((2, \sqrt{3})\) and inclined with the x-axis at an angle of 15°.
(v) Intersecting the x-axis at a distance of 3 units to the left of origin with slope = -2
(vi) Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of x-axis
(vii) Passing through the points (-1, 1) and (2,-4)
(viii) Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with positive x-axis is 30°.
Answer:
(i) Equation of x-axis is y = 0
And equation of y-axis is x =0

(ii) Required equation is of the form



(viii) Given: Perpendicular from the orgin = p = 5 Angle made by the perpendicular and positive x – axis = w = 30º

Question 37.
Find the equation of two line perpendicular distance from the origin is 4 units and the angle the normal makes with positive direction of x-axis is 15°
Answer:

Question 38.
Find the equation of the line, which makes -3 and 2 on the x and y-axes respectively.
Answer:
Given: x-intercept = a = -3
y – intercept = b = 2
∴ Using intercept form, required line is

Question 39.
Write the equation of the lines for which \(\tan \theta=\frac{1}{2}, \theta\) is the inclination of the line and 2
(i) y-intercept is \(-\frac{3}{2}\)
(ii) x-intercept is 4.
Answer:
Using intercept form, required equations are of the form
y = mx+c and y = m(x-d)
\(\text { i.e., } y=\frac{1}{2} x-\frac{3}{2} \text { and } y=\frac{1}{2}(x-4)\)

Question 40.
Find the equation of the line passing through (-2,3) with slope -4.
Answer:
Given: (x₁, y₁) = (-2, 3), m = 4
Required equation is of the form
y – y₁₍ = m(x-x₁) (slope-point form)
i.e., y-3 = – 4(x + 2)
i.e., y-3 =  – 4x – 8
i.e., 4x+y+5=0

Question 41.
Write the equation of the line through the points (1, -1) and (3,5)
Answer:
Given: (x₁, y₁)= (1, -1); (x₂, y₂) = (3, 5)
∴ Required equation is of the form

Question 42.
The vertices of ΔPQR are P(2,1) Q(-2, 3) and R(4, 5). Find the equation of the median triangle the vertex R
Answer:
Given P(2, 1), Q(-2, 3) and R(4, 5)

Equation of the median through R is the equation of the line through m(0, 2) and r(4, 5)
∴ By two-point form, equation of the line is

Question 43.
Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2,5) and (-3,6)
Answer:
Given points are A(-3, 5), 5(2, 5) and C(-3, 6)
Now, Slope of BC =
\(B C=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{6-5}{-3-2}=-\frac{1}{5}\)
∴ Slope of any line perpendicular to BC = 5
(y condition for perpendicularity is  m₁ x m₂ = -1)
The required equation of the line through A(-3, 5) with m- 5 is
y – 5 = 5(x + 3) (by slope-point form)
⇒ 5x-y + 20 = 0

Question 44.
A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divide it in the ratio 1: n. Find the equation of the line
Answer:
Let A(1, 0) and B (2, 3) be the given point
The required line divides the segment AB in the ratio 1 : n at P

Question 45.
Find the equation of the line that cuts off equal intercepts on the coordinate axes and passes through the point (2,3).
Answer:
Given : x-intercept = y-intercept
i.e., a = b, then the equation of the line is
\(\frac{x}{a}+\frac{y}{a}=1\)
⇒ x + y = a  … (1)
Since (1) passes through the point (2, 3), then we have,
2 + 3 = a ⇒ a = 5
∴ From (1). required line is x + y = 5.

Question 46.
Find the equation of the line passing through (2, 2) and cutting off intercepts on the axes whose sum is 9.
Answer:
Let the intercepts on the axes be a and b
Given : a + b = 9 ⇒ b = 9-a
Intercepts of the line on the axes a, 9 – a
By intercept form, required line

Question 47.
Find the equation of the line through the point (0, 2) making an angle \(\frac{2 \pi}{3} \)with 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.
Answer:
Slope of the line

Question 48.
The perpendicular from the origin to a line meets it at the point (-2, 9), find the equation of the line.
Answer:
Let OM be the perpendicular drawn from 0 to 1

Question 49.
The length L cms of copper rod is a linear function of its Celsius temperature C, In an experiment, it L =942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.
Answer:
Since L is a linear function of C then
L = mC + b  ……………… (1)
(slope-intercept form) when C = 20, L = 124.942, then (1)
⇒ 124.924 = 20 m + b ……………… (2)
when C = 110, L = 125.134,then (1)
⇒ 125.134 = 110C + b ……………… (3)
From (2), 124,924 = 20 x 0.00213 + b
⇒ b = 124.942 – 0.00426 = 124.899
∴ (1) ⇒ L = 0.00213C + 124.899 is the required relation between L and C.

Question 50.
The Fahrenheit temperature F and absolute temperature K satisfy a linear equation. Give that K = 273 when F = 32 and that K = 373 when F = 212. Express K in terms of F and find the value of F, when K = 0.
Answer:
Take the equation in the form

Question 51.
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?
Answer:
Let the owner sell L litre of milk at Rs. p/ litre, then linear relationship between L and p be

OR
Let A=(14,980),B = (16,12200 then slope of

Let owner can sell L litres of milk at Rs. 17 per litre. Then the point P(17, L) will lie on the AB.
⇒ A, B and P are collinear
⇒ Sloe of AB = slope of BP
⇒ \(120=\frac{L-1220}{17-16}=L-1220\)
⇒ L = 120 + 1220 = 1340 litres
∴ The milk owner can sell 1340 litres of milk at Rs. 17 per litre.
Remark: This can be solve, using two-point form also.

Question 52.
P(a,b) is the midpoint of a line segment between axes. Show that equation of the line is
\(\frac{x}{a}+\frac{y}{b}=2\)
Answer:
Let the line meets the x-axis at A(h, 0) and y-axis at B(0, k), then the equation of the line segment of the line segment AB in intercept form is

Question 53.
Point R(h, k) divides a line segment between axes in the ratio 1 : 2. Find equation of the line.
Answer:
Let the line meets the x-axis at A and y-axis at B. Then
A = (a, 0) and B = (0, b)
Given : R divides AB in the ratio 1 : 2 then by section formula,

Question 54.
By using the concept of equation of a line, prove that the three points (3,0), (- 2, – 2) and (8,2) are collinear.
Answer:
Let A(3, 0), B(- 2, 2) and C(8, 2) be the given points
Using two-point form, we can find the equation of AB as

If C(8, 2) lies on the line (1) then we can conclude A, B and C are collinear.
Now putting x = 8 and y =2 in (1), we get
5(2) = 2(8) – 6
10 = 16 – 6= 10
10 – 10=0
= Clieson(1)
= A, B and C are collinear

Question 55.
Reduce the equations into slope-intercept form and And their sloes and y-intercept.
(i) x + 7y = 0
(ii) 6x – 3y – 5 = 0
(iii) y=0
Answer:

(iii) y = 0
Rewrite as y = (0)x + (0)
Compare with y = mx + c, we get
m = 0 = slope, c = 0 = y – intercept

Question 56.
Reduce the equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y -12 = 0
(ii) 3x – 4y +10 = 0
(iii) 4x – 3y = 6
(iv) 3y+ 2 = 0
Answer:

Question 57.
Reduce the equations into normal form, find their perpendicular distances from the origin and angle between perpendicular and the possible x-axis.
(i)\(x-\sqrt{3} y+8=0\)
(ii) y – 2 = 0
(iii) x – y = 4
(iv)\(\sqrt{3} x+y-8=0\)
Answer:

(ii) Given : y – 2 = O
y=2
⇒ o.x +1.y = 2
Compare with x cos(ω) + y sin(ω) = p
we get, cos(ω) = 0, sin(ω) = 1, p = 2
we know that,
cos(90°) = 0 and sin(90°) = 1
ω = 90° and p = 2
∴ Required normal form is,
x cos(90°) + y sin(90°) = 2

 

Question 58.
Find the distance of the point (- 1,1) from the line 12(x + 6) = 5 (y – 2).
Answer:
Given equation is,
12(x + 6) = 5(y – 2)
⇒ 12x + 72 = 5m-10
⇒ 12x – 5y + 82 = 0
Compare with Ax + By + C = 0,
we get A = 12, B = -5, C = 82
Given point is (—1,1) = (x1, y1)
∴ Required distance

Question 59.
Find the points on the x-axis whose from the line \(\frac{x}{3}+\frac{y}{4}=1\) are 4 units.
Answer:
Rewrite the given equation as 4x ± 3y = 12 or 4x ± 3y – 12 = 0
Any point on the x-axis is of the form (x, 0)
Given that the distance of (x, 0) from the line 4x + 3y – 12 = 0 is 4

Question 60.
Find the distance between the parallel lines
(i) 15x ± 8y – 34 = 0 and 15x ± 8y ± 31 = 0
(ii) 3x – 4y ± 7 = 0 and 3x – 4y + 5 = 0
(iii) l(x + y) + p = 0 and l(x + y)-r = 0
Answer:
(i) Given equations are
15x ± 8y – 34 = 0 and 15x ± 8y + 31 = 0
Here, A = 15, B = 8, C₁ = -34, C₂ =31
∴ Required distance is

Question 61.
Find the equation of the line parallel to the line 3x – 4y ± 2 = 0 and passing through the point (- 2,3).
Answer:

Question 62.
Find the equation of the line perpendicular to the line x – 7y + 5 = 0 and having x-intercept 3.
Answer:
Given line is x-7y + 5 = 0
\(\text { Slope }=m=-\frac{a}{b}=-\frac{1}{-7}=\frac{1}{7}\)
Slope of the line perpendicular to the given line = – 7 (∵ m₁ x m₂=-1)
Since x-intercept is 3 then (3, 0) is the point on the required line.
∴ Required equation the line passing through (3, 0) having slope
– 7 is, y – 0 = -7(x-3)
i.e., 7x ± y – 21 = 0

Question 63.
Find angles between the lines \(\sqrt{3} x+y=1\) and \(x+\sqrt{3} y=1\)
Answer:

Question 64.
Find the angle between the lines
\(y-\sqrt{3} x-5=0 \text { and } \sqrt{3} y-x+6=0\)
Answer:

Question 65.
The line through the points (h, 3) and (4,1) intersects the line 7x-9y = 19 at right angle. Find the value of h.
Answer:

Since the given line and the line passing through the given points are perpendicular to each other, then

Question 66.
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
Answer:
Slope of the given line Ax + By + C = 0 is  \(m=-\frac{A}{B}\)
∴ Slope of the line through  \(\left(x_{1}, y_{1}\right)=-\frac{A}{B}\)
(Because, the line through (x, y) and given the line are parallel)
Thus, equation of line through (x₁,y₁) with

Question 67.
Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find the equation of the other line.
Answer:
We know that the acute angle ‘0’ between two lines with slopes m₁ and m₂ is given by

Question 68.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1,2).
Answer:
Let A = (3, 4), B = (-1, 2) be the given points.

Question 69.
Find the co-ordinates of the foot of perpendicular from the point (- 1, 3) to the line
3x – 4y -16 = 0.
Answer:
Let M be the foot of the perpendicular drawn from p(-1, 3) to the given line P (-1.3)

3x – 4y – 16 = 0 ……… (1)
Let the coordinates of M be (h, k)
Since M hes on the line (1), then
3h – 4A –  16 = 0 ………….(2)
Also,
Slope of PM x Slope of (1) = – 1

Question 70.
The perpendicular drawn from the origin to the line y = mx + c meets it at the point
(- 1, 2). Find the values of m and c.
Answer:
Given line is y = mx + c ……………. (1)
∵ the point P(-1, 2) lies on (1), then 2 = m(-1) + c
⇒ 2 = – m + c ……………. (2)
Slope of (1) = m and

Question 71.
If p and q are the lengths of perpendiculars from the origin to the lines
x cos θ – y sin θ =  k cos 2θ and
x sec θ +y cosec θ = k, respectively, prove that p² + 4q² = k².
Answer:
Given lines are
x cos θ –  y sin θ = k cos 2θ …………(1)
x sec θ+y cosec θ = k ……….. (2)

Question 72.
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’
Answer:
A(2,3), B{4,-1) and C(1, 2) be the given points. Let AD be the altitude from A.
∴ AD is perpendicular to BC

Question 73.
If ‘p’ is the length of perpendicular from the origin to the line whose intercepts on the axes area ‘a’ and ‘b’ then show that \(\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\)
Answer:
By intercept form, equation of the line is
\(\frac{x}{a}+\frac{y}{b}=1\)
i.e., bx + ay = ab
i.e., bx + ay-ab- 0
Given :
p = length of perpendicular from (0, 0) to the line

Miscellaneous examples:

Question 1.
If the lines 2x + y – 3 = 0, 5x + ky – 3 = 0 and 3x – y – 2 = 0 are concurrent, find the value of k.
Answer:
Given lines are

2x + y – 3 = 0 ………….. (1)
5x + ky – 3 = 0 ………….. (2)
3x – y – 2 = 0 …………..(3)
Solve (1) and (3), we get
5x-5 = 0 ⇒ x = 1
and from (1), y = 3-2(1) = 3-2 = 1
∴  x = 1,y = 1
Point of intersection of lines (1) and (3) is (1,1)
Since the given lines are concurrent, the point (1,1) will satisfy (2), so that,
5(1) + k(1)-3 = 0 ⇒ k = -2

Question 2.
If three lines whose equations are y = m₁x + c_l, y=m₂x + c₂ and y=m₃ x + c₃ are concurrent, then show that m₁(c₂ – c₃)+m₂(c₃ – c₁)+m3(c₁ – c₂)=O
Answer:

Question 3.
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
Answer:
Give line is,
(k-3)x-(4-k²)y + k² – 7k+ 6 = 0 …(1)
(a) line (1) is parallel to x-axis if coefficient of x is zero
If coefficient of x is zero
(∵ any line parallel to x-axis is of the form y = b)
k – 3 = 0
⇒ k = 3

(b) Line (1) is parallel to y-axis if coefficient of y is zero
Any line parallel to y-axis is of the form x = a)
∴ 4- k² = 0
⇒ k = ± 2

(c) Line (1) passes through the origin (0, 0) then
0 – 0 + k²-7k + 6 = 0
i.e.,  k² – 7k + 6 = 0
i.e., (k – 6)(k- 1) = 0
⇒ k = 1,6

Question 4.
Find the value of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line \( \sqrt{3} x+y+2=0 \)
Answer:
Given Lines

(-cos 30°) + (-sin 30°) y = 1
∴ cos 210° x+sin 210 y=1
Comparing with x cosθ + y sinθ = p
we get
θ = 120°and p = 1

Question 5.
Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and – 6, respectively.
Answer:
Let a and b be the intercepts of the line on the axes. Then

Question 6.
What are the points on the y-axis whose distance from the line \(\frac{x}{3}+\frac{y}{4}=1\)
Answer:

Question 7.
Find the perpendicular distance from the origin of the line joining the points
(cos θ, sin θ) and (cosφ, sinφ).
Answer:
Let A = (cos θ, sin θ) and B = (cosφ, sinφ) be the given points. Then equation of AB is of the form,

Question 8.
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.
Answer:
Given equations are,
x – 7y + 5 = 0 ……………. (1)
3x + y = 0 ……………. (2)
From (2), y = -3x …………….(3)
Using (3) in (1), we

Question 9.
Find the equation of a line drawn perpendicular to the line \(\frac{x}{4}+\frac{y}{6}=1 \) through the point, where it meets the y-axis.
Answer:

Question 10.
Find the area of the triangle formed by the lines y-x = 0, x + y = 0 and x – k = 0.
Answer: Given lines are
y-x = 0 ………………. (1)
x+ y = 0  …………………….. (2)
First find the points of intersection of the given lines
Solve (1) and (2), we get x = y = 0
Solve (1) and (3), we get x= y =k
Solve (2) and (3), we get x-k, y = -k
∴ Vertices of the triangle are
(0, 0), (k, k) and (k, -k)
∴ Required area

Question 11.
Find the value of p so that the three lines
3x + y – 2 = 0, px + 2y – 3 = 0 and
2x – y – 3 = 0 may intersect at one point.
Answer:
Given lines are,
3x + y – 2 = 0  ………………. (1)
px + 2y – 3 = 0 …………………… (2)
2x – y – 3 = 0  ……………….. (3)
Now, solve (1) and (3): we get
5x – 5 = 0
⇒ x = 1
Using x = 1 on (1), we get y = -1
Point of intersection of (1) and (3) is U,y) = (1,-1)
The three lines are concurrent if the point (1,-1) lies on (2).
i.e., p( 1) + 2(-1)- 3 = 0
⇒ p = 5

Question 12.
Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x – 2y = 3.
Answer:

Question 13.
Find the equation of the line passing through the point of intersection of the lines
4 x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Answer:
Given lines are
4x + 1y-3 = 0  …………….(1)
2x – 3y + 1 = 0 ………………. (2)
First, find the point of intersection of (1) and (2),
∴ 3 x (1) + 7 x (2) gives
12 x + 14x – 9 + 7 = 0

Since the required line has equal intercepts on the axes then its equation is of the form,

Question 14.
In what ratio, the line joining (- 1, 1) and (5,7) is divided by the line jc + y = 4?
Answer:
Let A = (-1,1), B(5, 7) be the given points.
Let the given line x+y = 4 divides AB at C in the ratio k : 1

Question 15.
Find the distance of the line 4x + 1y + 5 = 0 from the point (1,2) along the line 2x-y = 0.
Answer:
Given lines are,
4x + 7y +5 = 0 ………..(1)
2x – y = 0 ……………….(2)

Question 16.
Find the direction in which a straight line must be drawn through the point (- 1, 2) so that is point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.
Answer:
Let the required direction make an angle 0 with the positive direction of x-axis, then the equation of the line through (- 1, 2) and said direction

Required line from (1) is
y – 2 = 0 or + 1 = 0
i.e., Either parallel to x-axis or parallel to y-axis.

Aliter:
Required equation of the line through (x₁, y₁), making an angle θ with the positive direction x-axis and at a distance r from (x, y) is of the form

Question 17.
The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (- 4, 1). Find the equation of the perpendicular sides (legs) of the triangle.
Answer:
Let P = (1, 3) and Q = (-4,1) be the ends of the hypotenuse.

From the figure, there are two possible right angled triangles PRQ and PSQ.
Clearly R = (1, 1) and 3 = (-4, 3)

Case (i) : In the right angled ΔPQR, required equations are:
Equations to PR and QR
Equation of PR is x = 1 and
Equation of QR is y = 1

Case (ii) : In the right angled ΔPSQ, required equations are:
Equation to PS and QS.
Equation of QS is x = -4 and
Equation of PS is y = 3.

Question 18.
Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming that line to be a plane mirror.
Answer:
Let Q = (h, k) be the image of the point P(3, 8) in the line x + 3y-7 = 0 ………… (1)

Question 19.
If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.
Answer:
Given lines are
l₁ : 3x-y +1 = 0 … (1)
_l2: x – 2y + 3 = 0 … (2)
l₃: mx – y + u = 0  … (3)

Question 20.
The 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.
Answer:
Given lines are
x + y – 5 – 0 ………. (1)
3x – 2y + 7 = 0  …………. (2)
Also,
Distance of (1) from P + distance of (2) from R = 10

Question 21.
Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and
3x + 2y+ 6 = 0.
Answer:
Given lines are
\(3 x+2 y-\frac{7}{3}=0 \quad \ldots(1) \quad(9 x+6 y-7=0)\)
2y + 6 = 0 ……………(2)
Since the required line is equidistant from the parallel lines (1) and (2) then it is of the form,
3x + 2y+ k = 0 …(3)
Given that distance between (1) and (3) equal to distance between (2) and (3).

Question 22.
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 co-ordinates of A
Answer:
We know that the incident ray and the reflected ray equally inclined to the normal at A. Let A = (h, 0)

If AC makes an angle (-) with the normal at A, then it makes an angled (90° – θ) with positive x-axis and AB makes an angle (90° + θ) with positive x-axis.
Now, slope of AB = tan (90° + θ) = – cot θ
Slope of AC = tan (90° – θ) = cot θ
∴ Slope of AB + slope of AC = θ

Question 23.
Prove that the product of the lengths of the perpendiculars drawn from the points
\((\sqrt{a^{2}-b^{2}}, 0) \text { and }(-\sqrt{a^{2}-b^{2}}, 0)\) to the line
\(\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1 \text { is } b^{2}\)
Answer:

Question 24.
A person standing at the junction (crossing) of the two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the paths whose equation is 6x-7y + 8 = 0 in the least time. Find the equation of the path that he should follow.
Answer:
Given lines are

Students can Download Sanskrit Shevadhi Lesson 12 कन्नडकण्वः Questions and Answers, Notes Pdf, Summary, 1st PUC Sanskrit Textbook Answers, helps you to revise complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Sanskrit Textbook Answers Shevadhi Chapter 12 कन्नडकण्वः

कन्नडकण्वः Questions and Answers, Notes, Summary

कन्नडकण्वः Summary in Kannada

कन्नडकण्वः Summary in English

Introduction

The land of Karnataka is a sacred one. It is the land of sandalwood trees. The people of Karnataka are noble and virtuous. It is the duty of each and every Kannadiga to protect the land, water, language as well as the fame of Karnataka. Before independence, the British ruled this land. In those days some people believed that it was a shame to speak in Kannada. B.M. Srikantaiah was born at this time. If Kannada holds its head high in the court of Indian literature today, it is in no small measure the result of his pioneering efforts. He is an ideal for the youth of today. This is the main reason why a concise history of his life is given here.

 

Summary

Popularly known as ‘B.M.Shri’, B.M. Srikantaiah was born on 3rd January 1884 as the son of Mylarayya and Bhagirathamma. His ancestors lived in a place called Belur in Nagamangala taluk of Mandya district. His father was a lawyer. The childhood of Sri was spent at Srirangapattana.

After completing his high school education he passed the first examination in Arts (F.A.). He then went to Bangalore and joined Central College. He took his B.A. degree from the college in 1903. The Central College offered no post-graduate courses in those days. Srikantaiah had therefore to go to Madras for further studies. He took his M.A. in 1907. After obtaining the L.L.B. and L.L.M. degrees from Madras, he returned to Mysore.

When he was in Madras he started thinking deeply about Kannada literature. He always thought about the development of Kannada language – “How can the wealth lying with English language come to Kannada? What should be the roadmap for developing Kannada literature? How can Kannada be developed with the help of English? This is what was always in his thoughts.

Shri first studied Physics and then the English language. Still his love for Kannada language did not diminish. He was not only interested in serving Kannada language but also put in continuous efforts in elevating it. Even the experts in English were surprised to see his expertise in English literature. Prof. Macintosh, an expert in English language, said thus – “Shri was probably the head of the English department in the University of Oxford in his previous birth”. This was his depth of knowledge.

Shri started his career as the head of English department on January 15, 1909. Protecting the sanctity of his career he was always engrossed in learning and remained a student till the end.

All the wellknown poets and writers of Kannada language were his students. The love Shri had for his students remained unparalleled. He always encouraged bright and talented students by praising them. Prof.A.N. Murthyrao, the popular Kannada writer, was a student par excellence of Shri.

When Murthyrao was his student he would show all the articles written by him in English to his teacher. Seeing the articles Shri was surprised. Having read the articles, in the classroom he praised the student wholeheartedly: “This kind of excellent composition is not possible even from me”.

Kuvempu, the winner of the Jnanpith Award and known as the Rashtrakavi, was a student of Shri. In 1937, Shri was the president of the Kannada Sahitya Sammelana held at Kalburgi. At the sammelana the drama ‘Gadhayuddha’ was enacted as also Kuvempu’s ‘Yamana Solu’ – defeat of Yama. The organizers came up on the stage to garland Shri. But, Shri himself called Kuvempu on the stage and garlanded him. Such was the kind of encouragement he gave his students. This is worthy of emulation by all.

Shri occupied the chair of Assistant Professor, Professor and later the Vice Chancellor of Mysore University. He started the P.G. Department in Kannada in the University of Mysore.

Shri’s great desire was that the youth must engage themselves in the cause of Kannada. He arranged seminars/conferences/congregations for the sake of students. Under his editorship, the ‘Mysore Viswavidyanilaya Granthamala’ was started to enable common people lay their hands on stories, different forms of poetry, dramas and works on criticism. He brought out books relating to knowledge and science that helped the people immensely.

His ‘English Gitagalu’ a collection of poems, is a translation of popular poets like Shakespeare and others into Kannada.’Honganasu’ is his independent work. Amongst the tragedies, his drama ‘Gadhayuddha’ is prominent in the world of Kannada drama. Having translated ‘Ashwathama’, ‘Parasikaru’ and such other tragedies into Kannada, he created a legacy of tragedies even in Kannada language. In the year 1943 he retired from service but not from his hobby. The dream of Bharat Ratna Sir M.Visveshwariah came true in 1915 with the inauguration of Kannada Sahitya Parishat. Shri served as the president of Kannada Sahitya Parishat and devoted his entire life in protecting, preserving and enriching Kannada literature. He worked constantly in this direction.

Another topic must be dealt with here – the present modern Karnataka was divided into various parts by the Britishers before independence. The Britishers used English as the administrative language. Everywhere English was used. Just as sage Kanva brought Sakuntala, who was abandoned by her parents and who was looked after by birds and beasts in the forest, to the hermitage and brought her up, similarly Shri nourished the Kannada language which was in a miserable condition, enriched it and reinstated it to its original position. It is for this reason the winner of Jnanpith Prashasti Sri Masti Venkatesh Iyengar praised him as ‘Kannada Kanva’. The then Maharaja of Mysore honoured him with the title ‘Rajasevasakta’.

Let us follow in the footsteps of Shri and bear love for Kannada language in mind, spread the language every-where and fulfil his statement “Sirigannadam Gelge” in the State of Karnataka. Let us make our lives useful, purposeful and fruitful.

Students can Download Sanskrit Shevadhi Lesson 11 वचनामृतम् Questions and Answers, Notes Pdf, Summary, 1st PUC Sanskrit Textbook Answers, helps you to revise complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Sanskrit Textbook Answers Shevadhi Chapter 11 वचनामृतम्

वचनामृतम् Questions and Answers, Notes, Summary

वचनामृतम् Summary in Kannada and English

Students can Download Sanskrit Shevadhi Lesson 10 सन्मित्रम् Questions and Answers, Notes Pdf, Summary, 1st PUC Sanskrit Textbook Answers, helps you to revise complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Sanskrit Textbook Answers Shevadhi Chapter 10 सन्मित्रम्

सन्मित्रम् Questions and Answers, Notes, Summary

सन्मित्रम् Summary in Kannada

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Introduction

The present extract is taken from Samskrita Samanya Jnanam’, a collection of stories. ‘A learned man is respected everywhere for his scholarship’ is the main theme of this story. The story helps us in learning the importance of a true friend.

There arose an argument between the king and his minister in the assembly. The minister said – “A learned man by virtue of his scholarship is respected all over but a king is not.” However, King Senajit did not approve of the statement and ordered the minister Devavrita to prove it. When the minister maintained silence he ordered the minister to leave the country. The minister went to the house of his friends but two of them did not entertain him for fear of incurring the wrath of the king. Finally, he went to the house of one of his best friends, Chandradhara. Chandradhara, ignoring the order of the king, gave resort to him. Later, the king realized his mistake and he personally went to the minister’s house and begged his pardon and took him to the palace. ‘आपत्सु मित्रं जानीयात्’ which means one comes to know who is a true friend during the time of calamity (difficulty). This is the moral of this story.

 

Summary

In the East there was a city called Vaishali. There, a king by name Senajit ruled the kingdom. This king was strong and powerful. There was a well read, truthful and intelligent minister called Devavrata in his court. Devavrata was so intelligent that he did not do anything in haste. Senajit was extremely friendly with him. He always attained success in all his undertakings with the help of the smart and intelligent minister. Hence the king spent his time happily.

Who knows the ways of destiny? (Who can predict destiny?) Once, in the court of the king, there was a conference of the pundits, poets and scholars well versed in the Sastras/various branches of learning. There arose arguments between them. The minister said -(A scholar is not equal to a king) It is not possible for a king to be on par with a scholar. A learned man, by his scholarship, is respected not only in his country but in other countries as well but a king is not. A king is respected only in his country. Without accepting the words of the minister, Senajit said – (It is the king who deserves worship and not the scholar) Without a king a scholar does not deserve respect. Show the proof of your argument.

‘Yes’, said the minister, and maintained silence. But, the king without thinking, ordered the minister to leave the country immediately. The minister, Devavrata, had three friends Vachikamitra, Somadeva and Chandradhara. Devavrata went to the house of his friend Vachikamitra. Devavrata was very close to him. The king would not know about Devavrata living in this bunglow till his last breath (the end of his life). But, after a few days Devavrata realised his friend’s fear of being a traitor to the king and remembering his help left the house and went to the house of Somadeva. Somadeva welcomed Devavrata by offering a seat. Somadeva too (spent time happily) enjoyed the company of Devavrata having food, drinks and other things. ‘Indeed, despite opposition from the King I will extend help. Am I not capable of doing this much?’ Somadeva said. Considering the generosity of his friend when such good words were uttered, Devavrata was very happy. But, very quickly Somadeva said, ‘Oh friend, I will not forget your help. In case the king comes to know about your stay in my house, he will definitely put me to the gallows. I know that it is an offence to offer shelter to a person who has transgressed the command of the king. Hence, quickly leave my house and go elsewhere’.

Hearing the words of Somadeva, Devavrata, unable to bear the sorrow and anger, left his house and stated ‘Ah, fie on me and my two selfish friends’. He decided, “By anger alone such relationships have to be faced. I will look for well wishers somewhere else’. When he was thinking thus he saw Chandradhara, his childhood friend. He had never been honoured. He became extremely happy seeing Devavrata. Being disappointed with his friends, Devavrata came to the house of Chandradhara.

Chandradhara offered him a seat and water for washing his hands and feet. The happiness experienced by Chandradhara was inexplicable.

Devavrata recalled certain incidents in his life and said – ‘Men who have erred in life cheat other people blind in friendship adopting different methods. Some people act according to the situation. They do not desire the welfare of other people’. Definitely this world is full of meaningless traditions. Thus thinking Devavrata expressed his helplessness. He also explained the behaviour of his two friends and the order/command of the king.

Having heard his story Chandradhara consoled him. Ignoring the order of the king he said, Oh friend, if for saving the life of my learned friend my life is at stake, then also I will not swerve from the path of my duty (responsibility). My relatives consider it as my luck if life is given up for the sake of a friend. In reality a friend alone is very important on this earth (in this world). Devavrata, along with his family, carried on his life harmoniously. People honoured and respected Devavrata. The king also felt happy learning the attitude of the friend who was ready to save somebody’s life by offering his own life. The king Senajit, after knowing all this, was unable (became helpless) to carry out the kingly duties/administration without Devavrata. The king came to the house of Chandradhara and pleaded with Devavrata to forgive him and at the same time thought in his mind – Ah! What an attitude of mine is this? With this he not only respected Devavrata but also comforted him. Devavrata, remembering the help of Chandradhara again and again, invited him to his house and went back to the palace with the king.

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Students can Download Sanskrit Shevadhi Lesson 8 सान्तःपुरः शरणागतोऽस्मि Questions and Answers, Notes Pdf, Summary, 1st PUC Sanskrit Textbook Answers, helps you to revise complete Karnataka State Board Syllabus and score more marks in your examinations.

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सान्तःपुरः शरणागतोऽस्मि Summary in Kannada

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Introduction:

Sanskrit ‘Kavya’ is classified into two types namely, ‘श्रव्य’ and ‘दॄश्य’. That which is capable of being read and experienced through the eyes is known as ‘ दॄश्यकाव्यl’. In Sanskrit ‘ दॄश्यकाव्य’ is generally famous as नाटक

The present lesson is an extract from the fourth Act of the play ‘Balacharitam’ which is written by the famous playwright Bhasa.

Bhasa has written thirteen plays in all. The learned T. Ganapathi Shastri printed and published these plays for the first time. He opines that Bhasa belongs to the 5th century B.C.

A wicked serpent Kaliya by name takes resort along with his family in the Yamuna river. The water of the river Yamuna becomes poisoned due to the venom of Kaliya. The residents of Gokula face problems due to this. Hearing this, Damodara comes there and destroys the arrogance of Kaliya.

Characters:

Sankarshan – Balarama
Damodara – Sri Krishna
Kaliya – A huge black snake
Gopikas – Cowherd girls
Vriddha Gopalaka – Old herdsman

 

Summary:

(Then enter the cowherd girls)

Cowherd Girls – Oh my master, Don’t! Don’t go into that river! Indeed it is the home of a great and wicked snake.
Damodara – Girls! Don’t be anxious.
Cowherd Girls – Lord Sankarshana! Please prevent Lord Damodara.

(Entering)

Sankarshana – No need for worry or anxiety. Show him your love.
Damodara – For the welfare of the people I shall subdue this snake as quickly as possible.

(Enters the pool)

Cowherd Girls – Oh, look at the rising smoke! Damodara – Oh! The pool is so deep.
The river Yamuna, boiling with fire of poison within, grey with Kaliya’s smoky breath. I will change this river with ripples of liquid sapphires and the luster of dark blue silk in folds (1)

(Exit)
(Then enters the old herdsman)

Old Herdsman – Oh, Lord! He entered into the river though the girls tried to stop him. Don’t be rash in going on. Here, the tigers, wild boars and elephants that drink the water of this river die on the spot. I can’t see him. What shall I do now? Well, I shall climb this Kumbhapalasha tree and get a view.
(Climbing and seeing) Alas! there is smoke rising up.
Sankarshana – Ladies! Take a look.
Having seized the snake, stirring up the water from the bottom, standing on the hood of the dark blue snake, Damodara looks like Indra who is standing on a cloud. (2)

(Enter Damodara holding Kaliya)

Damodara – Here he is.
Ignoring the anger of Kaliya – which is full of poison, I will place my feet at the edge of its head, and perform the herdsmen’s dance, hallisaka. (3)
Cowherd Girls – Wonderful, master indeed wonderful! He is performing hallisaka dance standing on Kaliya’s five hoods. Damodara – Now, I too will pluck some flowers.
Kaliya – Ah! Just as the Lokaloka Mountain occupied the entire earth at the time of churning of the milky ocean, like the bowstring of Shiva, Adisesha circumambulated the Mandara mountain, today, I will, with my hoods as strong as the trunk of Iravata, circumambulate you and dispatch you to the abode of God of death – Yama. (4)
Old Herdsman – (coming down) Well done my lord well done! I will also come and assist you.
Oh! I am scared my lord! Indeed I am scared! I will go and convey all these things to Nandagopa.

(Exit)

Damodara – Having destroyed the fish and crocodiles from the interior of the river Yamuna, I will, by force, throw this wicked serpent breathing heavily as it is at the peak of arrogance and moving its large circular hood, at once on the ground. (5)
Kaliya – Here I am.

By whose rage the human body is set on fire and the whole earth consumed, I will burn you up with a circle of flames; let the group of Maruts protect you and your world. (6)
Damodara – Kaliya! If you are strong, then burn this single arm of mine.
Kaliya – Ha, Ha, Ha! I could burn the entire earth with its seven great mountains extended to the four oceans. Why can’t I burn your arm? Wait for a while, I will reduce you to ashes.

(Emits venomous flames)

Damodara – Alas! You have shown your valour!
Kaliya – Have mercy! My lord Narayana, have mercy on me.
Damodara – You were arrogant because of your strength.
Kaliya – Have mercy O my lord, I was ignorant. With all my wives I seek your protection.
Damodara – Kaliya, why did you enter this Yamuna river?
Kaliya – I was afraid of Garuda who is your vehicle. Hence I entered this river. I seek your help. Please free me from the fear of Garuda.
Damodara – Let it be so.
Hey great snake! When he looks at my footprint on your head, Garuda himself will give you security. From today onwards you must not disturb people, cows and brahmins.
Kaliya – O my Lord! This water is poisoned with my venom. I will suck out the poison and leave this river Yamuna.
Damodara – You may go now.
Kaliya – As my Lord commands.

(Exits with his family)

Students can Download Sanskrit Shevadhi Lesson 7 ज्यौतिषिकस्य दिनम् Questions and Answers, Notes Pdf, Summary, 1st PUC Sanskrit Textbook Answers, helps you to revise complete Karnataka State Board Syllabus and score more marks in your examinations.

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Introduction

‘An Astrologer’s Day’ is a short story written by the famous English writer Padmshree Dr. R.K. Narayan. Avinashvarna has translated the story from English to Sanskrit. The story was published in the October 2010 issue of the monthly magazine ‘संभाषण संदेश’.

According to the saying ‘प्रत्यक्षम् ज्यौतिषं शास्त्रम्’ Indian astrology is very scientific. Not knowing anything about astrology fake astrologers cheat people. This fact has been given here in the form of an attractive story.

 

Summary

The astrologer arrived at his roadside office at midday. He worked under a banyan tree near Market Road. It was lighted by the sputtering flare of the groundnuts vendor nearby. His office equipment included cowrie shells, a piece of cloth painted with obscure mystic charts, a notebook, and a bundle of palmyra writing. He wore a saffron-colored turban around his head; his forehead was painted with sacred ash and vermillion. Customers came to him due to the sparkle in his eyes which they mistook for a prophetic light.

The astrologer carried on his business based on guesswork of what people wanted to hear. He never wanted to be an astrologer, but fate forced the profession on him.

The astrologer was born in a small village, and was destined to live there with his family. However, one day he left the village, without a word to anyone, and travelled hundreds of miles to settle in the city and started working under the banyan tree.

He based his profession of astrology on his knowledge of mankind’s troubles. These troubles included problems arising from marriage, money, and the tangles of human ties. His method was simple. He did not utter a word until his customer spoke to him for at least ten minutes. Then, based on this speech, he formulated answers, questions and predictions. His normal predictions were:”In many ways you are not getting the fullest results for your efforts,” or “Most of your troubles are due to your nature… You have an impetuous nature and a rough exterior.”

When the groundnuts-vendor blew out the fire that lit the astrologer’s workplace, he knew it was time to go home. As he was packing his cowrie shells, he looked up and saw a man before him. He tried to attract the man in with a promise of telling his fortune. The astrologer said that the stranger should give him twenty rupees if he liked his answers. The stranger agreed but said that the astrologer would have to give him twenty rupees if he was not satisfied with the answers. The stranger lighted up a cheroot, and by the light of the match, the astrologer caught a glimpse of his client’s face. For a moment, the world seemed to come to a standstill for the astrologer after looking at his client’s face.

The stranger told the astrologer that he wanted to hear the answer to just one question, whether or not he would succeed in his present search. The astrologer hesitated for a few moments. Then he asked the stranger if he was once pushed into a well and left for dead. The stranger’s interest picked up. He asked if a knife was used in the fight; the stranger answered in the affirmative. The stranger asked when would he find the man who tried to kill him, and the astrologer replied that he would see him only in the next world, because the person had died in an accident. The stranger was frustrated at hearing that he would not be able to avenge his attempted murder. But, he was surprised when the astrologer addressed him by name and advised him to return home to his village, as he foresaw great danger to his life if he stayed in the city.

The man left, giving the astrologer a handful of notes, and the astrologer went home. After dinner, the astrologer and his wife discussed the day’s events. He told her that a burden had been lifted from him that day. He further said that he thought he had killed a man many years ago, and that was why he had left his village. He told her that he had now seen the man alive, and thus, he is no longer guilty of murder. He told her that the incident had taken place when he was a youngster, when a drunken quarrel over a gambling debt had gone too far. She was horrified, but the astrologer went to sleep with a clear conscience.