Mahesh

Students must practice these TS Intermediate Maths 1B Solutions Chapter 4 Pair of Straight Lines Ex 4(b) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Pair of Straight Lines Solutions Exercise 4(b)

I.
Question 1.
Find the angle between the lines represented by 2x²
2x² + xy – 6y² + 7y – 2 = 0. (V.S.A.Q.)
Answer:
Comparing with the general equation
ax² + 2hxy + by² + 2gx + 2fy + c = 0 …………………….. (1)
We have a = 2, h = \(\frac{1}{2}\), b = – 6, g = 0, f = \(\frac{7}{2}\), c = – 2.
Also if θ is the angle between pair of lines (1)

Question 2.
Prove that the equation 2x² + 3xy – 2y² + 3x + y + 1 = 0 represents a pair of perpendicular lines. (V.S.A.Q.)
Answer:
We have coefficient of x² = 2 ⇒ a = 2
and coefficient of y² = – 2 ⇒ b = – 2
Since a + b = 0; the lines are perpendicular.

II.
Question 1.
Prove that the equation 3x² + 7xy + 2y² + 5x + 5y + 2 = 0 represents a pair of straight lines and find the coordinates of the point of intersection. (S.A.Q.)
Answer:
Comparing the given equation
3x² + 7xy + 2y² + 5x + 5y + 2 = 0 with the general equation ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get a = 3, h = \(\frac{7}{2}\), b = 2, g = \(\frac{5}{2}\), f = \(\frac{5}{2}\) and c = 2.
∆ = abc + 2fgh – af² – bg² – ch²

∴ The given equation represents a pair of lines and point of intersection is

Question 2.
Find the value of k, if the equation 2x² + kxy – 6y² + 3x + y + 1 = 0 represents a pair of straight lines. Find the point of intersection of the lines and the angle between the straight lines for this value of k. (E.Q.)
Answer:
Comparing
2x² + kxy – 6y² + 3x + y + 1 = 0 with the general equation, we get
a = 2, h = \(\frac{\mathrm{k}}{2}\), b = -6, g = \(\frac{3}{2}\), f = \(\frac{1}{2}\), c = 1
The given equation represents a pair of straight lines if abc + 2fgh – af² – bg² – ch² = 0 (necessary condition)
⇒ – 12 + 2\(\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\left(\frac{k}{2}\right)\) – 2\(\left(\frac{1}{4}\right)\) + 6\(\left(\frac{9}{4}\right)-\frac{k^2}{4}\) = 0
⇒ – 48 + 3k – 2 + 54 – k² = 0
⇒ – k² + 3k + 4 = 0 ⇒ k² – 3k – 4 = 0
⇒ (k – 4) (k + 1) = 0 ⇒ k = 4 or – 1

Case (i):

Case (ii):
When k = 4, then h = 2

Question 3.
Show that the equation x² – y² – x + 3y – 2 = 0 represents a pair of perpendicular lines and find their equations. (S.A.Q.)
Answer:
First we show that the given equation x² – y² – x + 3y – 2 = 0 represents a pair of lines comparing with the general equation
we get a = 1, b = -1, g = \(\frac{-1}{2}\), f = \(\frac{3}{2}\), h = 0, c = -2
∴ abc + 2fgh – af² – bg² – ch²

Hence the given equation represents a pair of lines. Coefficient of x² + coefficient of y² = 1 – 1 = 0
∴ The given equation represents a pair of perpendicular lines.
Let x² – y² – x + 3y – 2
= (x + y + c₁) (x – y + c₂)
Equating coefficients of x and y both sides
we get c₁ + c₂ = – 1 and – c₁ + c₂ = 3
Solving 2c₂ = 2 ⇒ c₂ = 1
and c₁ + c₂ = -1 ⇒ c₁ = -2
∴ Equations of lines are x + y – 2 = 0 and x – y + 1 = 0.

Question 4.
Show that the lines x² + 2xy – 35y² – 4x + 44y – 12 = 0 and 5x + 2y – 8 = 0 are concurrent. (S.A.Q.)
Answer:
Given x² + 2xy – 35y² – 4x + 44y – 12 = 0 and comparing this with general equation we get a = 1, h = 1, b = – 35, g = -2, f = 22, c = – 12
Point of intersection

∴ The point of intersection lies on the line 5x + 2y – 8 = 0.
Hence the given lines are concurrent.

Question 5.
Find the distance between the following pairs of parallel straight lines. (V.S.A.Q.)
(i) 9x² – 6xy + y² + 18x – 6y + 8 = 0
(ii) x²+ 2√3 xy + 3y² – 3x – 3√3 y – 4 = 0
Answer:
(i) The formula for distance between parallel lines = \(\sqrt{\frac{g^2-a c}{a(a+b)}}\)

(ii) Distance between given parallel lines

Question 6.
Show that the two pairs of lines 3x² + 8xy – 3y² = 0 and 3x² + 8xy – 3y² + 2x – 4v – 1 = 0 form a square. (S.A.Q.)
Answer:

Given equations of lines represented by OA and OB is 3x² + 8xy – 3y² = 0
⇒ (x + 3y) (3x – y) = 0
⇒ x + 3y = 0, 3x – y = 0
∴ Equation of OA is 3x – y = 0 ………………… (1)
Equation of OB is x + 3y = 0 ……………… (2)
The combined equation of CA and CB is
3x² + 8xy – 3y² + 2x – 4y – 1 = 0 …………….. (3)
Let 3x² + 8xy – 3y² + 2x – 4y – 1
= (3x – y + c₁) (x + 3y + c₂)
Equating the coefficients of x and y on both sides,
and c₁ + 3C₂ = 2
3c₁ – c₂ = – 4
Solving

⇒ c₁ = – 1, c₂ = 1
Equation of BC is 3x – y – 1 = 0 ……………………. (4)
Equation of AC is x + 3y + 1 = 0 …………………… (5)
Equations OA and BC differ by a constant.
⇒ OA is parallel to BC.
Equations OB and CA differ by a constant
⇒ OB is parallel to AC.
Also from equation (3), OACB is a rectangle and a + b = 3 – 3 = 0
OA = Length of the perpendicular from O to
AC = \(\frac{|0+0+1|}{\sqrt{1+9}}=\frac{1}{\sqrt{10}}\)
OB = Length of the perpendicular from O to
BC = \(\frac{|0+0-1|}{\sqrt{9+1}}=\frac{1}{\sqrt{10}}\)
∴ OA = OB and OACB is a rectangle
⇒ OACB is a square.

III.
Question 1.
Find the product of the length of the perpendiculars drawn from (2, 1) upon the lines (E.Q.)
12x² + 25xy + 12y² + 10x + 11y + 2 = 0
Answer:
Given equation
12x² + 25xy + 12y² + 10x + 11y + 2 = 0
represents the combined equation of AB & AC.
12x² + 25xy + 12y²
= 12x² + 16xy + 9xy + 12y²
= 4x (3x + 4y) + 3y (3x + 4y)
= (3x + 4y) (4x + 3y)
Let 12x² + 25xy + 12y² + 10x + 11y + 2 = (3x + 4y + c₁) (4x + 3y’ + C₂)
Equating the coefficients of x and y
4c₁ + 3C₂ = 10 ………………. (1)
and 3c₁ + 4c₂ = 11 ………………. (2)
Solving (1) and (2)

⇒ c₁ = 1, c₂ = 2
∴ Equation of AB is 3x + 4y + 1 = 0
and equation of AC is 4x + 3y + 2 = 0
PL = Length of the perpendicular from
P(2, 1) on AB = \(\left|\frac{6+4+1}{\sqrt{9+16}}\right|=\frac{11}{5}\)
PM = Length of the perpendicular from
P(2, 1) on AC = \(\left|\frac{8+3+2}{\sqrt{16+9}}\right|=\frac{13}{5}\)
∴ Product of lengths of perpendiculars
= PL × PM = \(\frac{11}{5} \times \frac{13}{5}=\frac{143}{25}\)

Question 2.
Show that the straight lines y² – 4y + 3 = 0 and x² + 4xy + 4y² + 5x + 10y + 4 = 0 form a parallelogram and find the lengths of its sides. (E.Q.)
Answer:

The equation of first pair of lines is y² – 4y + 3 = 0
⇒ (y – 1) (y – 3) = 0 ⇒ y = 1 ………………….. (1)
and y = 3 ………………….. (2)
∴ AB, CD are parallel.
The equation of second pair of lines is
x² + 4xy + 4y² + 5x + 10y + 4 = 0
⇒ (x + 2y)² + 5 (x + 2y) + 4 = 0
⇒ (x + 2y)² + 4 (x + 2y) + (x + 2y) + 4 = 0
⇒ (x + 2y) [x + 2y + 4] + 1 [(x + 2y) + 4] = 0
⇒ (x + 2y + 1) (x + 2y + 4) = 0
⇒ x + 2y + 1 = 0, x + 2y + 4 = 0
Equation of AD is x + 2y + 1 = 0 ……………………. (3)
Equation of BC is x + 2y + 4 = 0 ……………………. (4)
∴ AD and BC are parallel.
Solving (1) and (3), x + 2 + 1 = 0
⇒ x = – 3
∴ Co-ordinates of A = (-3, 1)
Solving (2) and (3), x + 6 + 1 = 0 ⇒ x = -7
and coordinates of D = (-7, 3)
Solving (2) and (4) x + 6 + 4 = 0 ⇒ x = -10
∴ Coordinates of C = (-10, 3)
Solving (1) and (4), we get x + 2 + 4 = 0
⇒ x = -6
∴ Coordinates of B = (- 6, 1)
Hence A = (-3, 1), B = (-6, 1), C = (-10, 3), D = (-7, 3)

∴ AC ≠ BD
Hence a parallelogram is formed with the
lines y² – 4y + 3 = 0 and x² + 4xy + 4y² + 5x + 10y + 4 = 0.

Question 3.
Show that the product of the perpendicular distances from the origin to the pair of straight lines represented by ax² + 2hxy +
by² + 2gx + 2fy + c = 0 is \(\frac{|c|}{\sqrt{(a-b)^2+4 h^2}}\) (E.Q.)
Answer:
Let ax² + 2hxy + by² + 2gx + 2fy + c = 0 represent the lines
l₁x + m₁y + n₁ = 0 ……………. (1)
and l₂x + m₂y + n₂ = 0 ……………. (2)
∴ ax² + 2hxy + by² + 2gx + 2fy + c
= (l₁x + m₁y + n₁) (l₂x + m₂y + n₂)
∴ l₁l₂ = a, m₁m₂ = b, l₁m₂ + l₂m₁ = 2h,
l₁n₂ +l₂n₁ = 2g, m₁n₂ + m₂n₁ = 2f, n₁n₂ = c
∴ Perpendicular distance from origin to (1) is
= \(\frac{\left|\mathrm{n}_1\right|}{\sqrt{l_1^2+\mathrm{m}_1^2}}\)
Perpendicular distance from origin to (2) is
= \(\frac{\mathrm{n}_2}{\sqrt{l_2^2+\mathrm{m}_2^2}}\)
Product of perpendiculars

Question 4.
If the equation ax² + 2hxy + by2 + 2gx + 2fy + c = 0 represent a pair of intersecting lines, then show that the square of the distance of their point of intersection from the origin is \(\). Also show that the square of this distance is \(\frac{f^2+g^2}{h^2+b^2}\) if the given lines are perpendicular. (E.Q.)
Answer:
Let the equation
ax² + 2hxy + by² + 2gx + 2fy + c = 0 represent the lines
l₁x + m₁y + n₁ = 0 ………………. (1)
and l₂x + m₂y + n₂ = 0 ……………….. (2)
∴ (l₁x + m₁y + n₁) (l₂x + m₂y + n₂)
= ax² + 2hxy + by² + 2gx + 2fy + c
Comparing coefficients
l₁l₂ = a, m₁ m₂ = b, n₁n₂ = c
l₁m₂ + l₂m₁ = 2h, l₁n₂ + l₂n₁ = 2g and
m₁n₂ + m₂n₁ = 2f
From (1) and (2) on solving

Here students can locate TS Inter 1st Year Accountancy Notes Chapter 6 Bank Reconciliation Statement to prepare for their exam.

TS Inter 1st Year Accountancy Notes Chapter 6 Bank Reconciliation Statement

→ Bank Reconciliation Statement is a statement prepared to reconcile the difference between the balance as per the bank column of the cash book and pass book on any given date.

→ There are certain reasons for the difference in the pass book balance and the cash book balance.

→ Favourable balance means debit balance as per cash book and credit balance as per credit balance.

→ Unfavourable balance/overdraft balance means credit balance as per cash book and debit balance as per pass book.

TS Inter 1st Year Accountancy Notes Chapter 6 బ్యాంక్ నిల్వల సమన్వయ పట్టిక

→ నిర్ణీత తేదీన నగదు చిట్టి బ్యాంకు వరసల నిల్వ, పాస్బుక్ నిల్వలకు గల తేడాలను సమన్వయపరుస్తూ తయారుచేసే పట్టికను బ్యాంకు నిల్వల సమన్వయ పట్టిక అంటారు.

→ నగదు చిట్టాలోని నిల్వకు, పాస్బుక్లో లోని నిల్వకు గల తేడా చూపడానికి కొన్ని కారణాలున్నవి.

→ నగదు పుస్తకము డెబిట్ నిల్వను, పాస్బుక్ క్రెడిట్ నిల్వను చూపితే దానిని అనుకూల నిల్వ అంటారు.

→ నగదు పుస్తకము క్రెడిట్ నిల్వను, పాస్బుక్ డెబిట్ నిల్వను చూపితే దానిని ప్రతికూల నిల్వ అంటారు.

Here students can locate TS Inter 1st Year Accountancy Notes Chapter 5 Cash Book to prepare for their exam.

TS Inter 1st Year Accountancy Notes Chapter 5 Cash Book

→ Cash book is a very important subsidiary book. The object of the cash book is to keep a daily record of transactions relating to cash receipts and cash payments. Cash book acts as both journal and a ledger.

→ There are different kinds of cash books :

1. Simple cash book.
2. Two-column cash book with cash and discount columns.
3. Two-column cash books with Bank and discount columns.
4. Three-column cash book.
5. Petty cash book.

→ The entry which appears on both sides of the three-column cash book is known as a contra entry. It is required for transactions relating to cash or cheques deposited into the bank and cash withdrawn for office use.

→ All small payments are recorded in a separate cash book known as the Analytical Petty cash book.

TS Inter 1st Year Accountancy Notes Chapter 5 నగదు పుస్తకము

→ నగదు పుస్తకము చాలా ముఖ్యమైన సహాయక చిట్టా. రోజువారీ నగదు వసూళ్ళు చెల్లింపు వ్యవహారములు నమోదు చేయడమే నగదు పుస్తకము ముఖ్య ఉద్దేశ్యము.

→ నగదు పుస్తకములో దిగువ రకాలు ఉన్నవి;

1. సాధారణ నగదు చిట్టా,
2. నగదు, డిస్కౌంటు వరుసలు గల నగదు చిట్టి,
3. బాంకు, డిస్కౌంటు వరుసలు గల నగదు చిట్టా,
4. మూడు వరుసలు గల నగదు చిట్టా,
5. చిల్లర నగదు చిట్టా.

→ ఒక చిట్టాపద్దును మూడు వరుసలు గల నగదు చిట్టాలో రెండు వైపులా నమోదు చేస్తే దానిని ఎదురు వద్దు అంటారు. ఎదురు పద్దును దిగువ సందర్భాలలో రాయాలి.

• నగదు లేదా చెక్కులను బాంకులో జమ చేసినపుడు,
• ఆఫీసు అవసరాలకై బాంకు నుంచి నగదు తీసినపుడు.

→ వివిధ రకాల చిల్లర ఖర్చులను నమోదు చేయడానికి తయారుచేసే ప్రత్యేక నగదు పుస్తకాన్ని చిల్లర నగదు చిట్టా అంటారు.