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You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.3

Question 1.
Find \(\vec{a} \cdot \vec{b}\) when

Solution:

Question 2.
Find the value of λ for which the vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular, where

Solution:
When \(\vec{a}\) and \(\vec{b}\) are ⊥^r then \(\vec{a} \cdot \vec{b}\) = 0
\(\vec{a}\) ⊥^r \(\vec{b}\) ⇒ \(\vec{a} \cdot \vec{b}\) = 0
(i) (2) (1) + (λ) (-2) + (1) (3) = 0 ⇒ λ = 5/2
(ii) (2) (3) + (4) (-2) + (-1) (λ) = 0
6 – 8 – λ = 0
-λ – 2 = 0 ⇒ -λ = 2 ⇒ λ = -2

Question 3.
If \(\vec{a}\) and \(\vec{b}\) are two vectors such that |\(\vec{a}\)| = 10, |\(\vec{b}\)| = 15 and \(\vec{a} \cdot \vec{b}\) = 75\(\sqrt{2}\) , find the angle between \(\vec{a}\) and \(\vec{a}\) .
Solution:

Question 4.
Find the angle between the vectors

Solution:

Question 5.
If \(\overrightarrow{\boldsymbol{a}}, \overrightarrow{\boldsymbol{b}}, \overrightarrow{\boldsymbol{c}}\) are three vectors such that \(\vec{a}+2 \vec{b}+\vec{c}=\overrightarrow{0}\) and \(|\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=7\) find the angle between \(\vec{a}\) and \(\vec{b}\)
Solution:

Question 6.
Show that the vectors are mutually orthogonal.
Solution:

Question 7.
Show that the vectors form a right angled triangle.
Solution:

So, the given vectors form the sides of a right angled triangle

Question 8.

Solution:

Question 9.
Show that the points (2, -1, 3) (4, 3, 1) and (3, 1, 2) are collinear
Solution:
Let the given points be A, B, C

Question 10.
If \(\vec{a}, \vec{b}\) are unit vectors and θ is the angle between them, show that

Solution:

Question 11.
Let \(\vec{a}, \vec{b}, \vec{c}\) be the three vectors such that \(|\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=5\) and each one of them being perpendicular to the sum of the other two, find \(|\vec{a}+\vec{b}+\vec{c}|\)
Solution:

Question 12.
Find the projection of the vector \(\hat{i}+3 \hat{j}+7 \hat{k}\) on the vector \(2 \hat{i}+6 \hat{j}+3 \hat{k}\)
Solution:

Question 13.
Find λ, when the projection of is 5 units.
Solution:

Question 14.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.3 Additional Problems

Question 1.
Find λ so that the vectors are perpendicular to each other.
Solution:

Question 2.

Solution:

Question 3.
If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is \(\sqrt{3}\)
Solution:

Question 4.
Show that the vectors form a right angled triangle.
Solution:

⇒ The given vectors form the sides of a right of a right angled triangle.

Question 5.
Find the projection of

Solution:

Question 6.
Show that the vector \(\hat{i}+\hat{j}+\hat{k}\) is equally inclined with the coordinate axes.
Solution:

Question 7.
If \(\vec{a}, \vec{b}, \vec{c}\) are three mutually perpendicular unit vectors, then prove that \(|\vec{a}+\vec{b}+\vec{c}|=\sqrt{3}\)
Solution:

Question 8.
Show that the points whose positions vectors from a right angled triangle.
Solution:

⇒ The given points form a right angled triangle.

Question 9.

Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 1.
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid.
Solution:

Question 2.
Find \(\sqrt[3]{1001}\) approximately (two decimal places).
Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.
Write the first 6 terms of the exponential series
(i) e^5x
(ii) e^-2x
(iii) \(e^{\frac{x}{2}}\)
Solution:

Question 6.
Write the first 4 terms of the logarithmic series
(i) log(1 + 4x),
(ii) log(1 – 2x),
(iii) \(\log \left(\frac{1+3 x}{1-3 x}\right)\)
(iv) \(\log \left(\frac{1-2 x}{1+2 x}\right)\).
Find the intervals on which the expansions are valid.
Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 Additional Questions

Question 1.

Solution:

Question 2.
Write the four terms in the expansions of the following:

Solution:

Question 3.
Evaluate the following:

Solution:

Question 4.

Solution:

Question 5.

Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3

Question 1.
Find the sum of the first 20-terms of the arithmetic progression having the sum of first 10 terms as 52 and the sum of the first 15 terms as 77.
Solution:
Let ‘a’ be the first term and d be the common difference of A.P.

Question 2.

Solution:

Question 3.
Compute the sum of first n terms of the following series:
(i) 8 + 88 + 888 + 8888 + ……
(i) 6 + 66 + 666 + 6666 + …..
Solution:

Question 4.
Compute the sum of first n terms of 1 + (1 + 4) + (1 + 4 + 4²) + (1 + 4 + 4² + 4³) + ……..
Solution:

Question 5.
Find the general term and sum to n terms of the sequence \(1, \frac{4}{3}, \frac{7}{9}, \frac{10}{27}, \ldots \ldots \ldots\)
Solution:

Denominator 1, 3, 9, 27, which is a G.P. a = 1, r = 3

It is an arithmetico Geometric series. Here the n^th term is t_n = [a + (n – 1)d]r^{n – 1} where a = 1, d = 3 and r = 1/3
Now the sum to n terms is

Question 6.

Solution:

Question 7.
Show that the sum of (m + n)^th and (m – n)^th term of an A.P. is equal to twice the m^th term.
Solution:
Let the A.P. be a, a + d, a + 2d, ……..
t_{m + n} = a + (m + n – 1)d
t_{m – n} = a + (m – n – 1)d
t_m = a + (m – 1)d
2t_m = 2[a + (m – 1)d]
To prove t_{m + n} + t_{m – n} = 2t_m
LHS t_{m + n} + t_{m – n} = a + (m + n – 1)d + a + (m – n – 1)d

= 2a + d [2m – 2]
= 2[a + (m – 1)d) = 2 t_m = RHS

Question 8.
A man repays an amount of ₹ 3250 by paying ₹ 20 in the first month and then increases the payment by ₹ 15 per month. How long will it take him to clear the amount?
Solution:
a = 20, d = 15, S_n = 3250 to find n.

3n² + 5n – 1300 = 0
3n² – 60n + 65n – 1300 = 0
3n (n – 20) + 65 (n – 20) = 0
(3n + 65) (n – 20) = 0
n = – 65/3 or 20
n = – 65/3 is not possible
∴ n = 20
So he will take 20 months to clear the amount.

Question 9.
In a race, 20 balls are placed in a line at intervals of 4 meters with the first ball 24 meters away from the starting point. A contestant is required to bring the balls back to the starting place one at a time. How far would the contestant run to bring back all balls?
Solution:
t₁ = 24 × 2 = 48, t₂ =48 + 8 = 56 or (24 + 4)2, t₃ =(28 + 4)2 = 64 which is an A.P.
Here a = 48,
d = 56 – 48 = 8

The contestant has to run 2480 m to bring all the balls.

Question 10.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2^nd hour, 4^th hour and n^th hour?
Solution:
Number of bacteria at the beginning = 30
Number of bacteria after 1 hour = 30 × 2 = 60
Number of bacteria after 2 hours = 30 × 2² = 120
Number of bacteria after 4 hours = 30 × 24 = 30 × 16 = 480
∴ Number of bacteria after n^th hour = 30 × 2ⁿ

Question 11.
What will ₹ 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Solution:

Question 12.
In a certain town, a viral disease caused severe health hazards upon its people disturbing their normal life. It was found that on each day, the virus which caused the disease spread in Geometric Progression. The amount of infectious virus particle gets doubled each day, being 5 particles on the first day. Find the day when the infectious virus particles just grow over 1,50,000 units?
Solution:
a = 5, r = 2, t_n >150000, To find ‘n’

On the 15^th day it will grow over 150000 units.

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 Additional Questions

Question 1.
If the ratio of the sums of m terms and n terms of an A.P. be m² : n², prove that the ratio of its m^th and n^th terms is (2m – 1) : (2n – 1).
Solution:
Let ‘a’ be the first term and d be the common difference of A.P.

⇒ 2an + (mn – n)d = 2am + (mn – m)d
⇒ 2an – 2am = (mn – m – mn + n)d
⇒ 2a(n – m) = (n – m)d ⇒ d = 2a, [n – m ≠ 0, as n ≠ m]

Hence, the required ratio is (2m – 1) : (2n – 1)

Question 2.
Determine the number of terms of geometric progression {a_n} if a₁ = 3, a_n = 96, S_n = 189.
Solution:

Question 3.
The sum of first three terms of a G.P. is to the sum of the first six terms as 125 : 152. Find the common ratio of the G.P.
Solution:

Hence, the common ratio of the G.P. is \(\frac{3}{5}\)

Question 4.
Find the sum to n terms the series: (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + …
Solution:
(x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + … to n terms

Question 5.
Sum the series: (1 + x) + (1 + x + x²) + (1 + x + x² + x³) + … up to n terms
Solution:

Question 6.
Sum up to n terms the series: 7 + 77 + 777 + 7777 + …
Solution:
S = 1 + 77 + 777 + 7777 + … to n terms

Question 7.
If S₁, S₂, S₃ be respectively the sums of n, 2n, 3n, terms of a G.P. , then prove that S₁(S₃ – S₂) = (S₂ – S₁)₂.
Solution:
Let a be the first term and r be the common ratio of G.P.

Question 8.
If sum of the n terms of a G.P be S, their product P and the sum of their reciprocals R, the prove that \(P^{2}=\left(\frac{S}{R}\right)^{n}\)
Solution:
Let a be the first term and r be the common ratio of the G.P.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.1

Question 1.
Represent graphically the displacement of
(i) 45 cm 30 ° north of east
(ii) 80 km, 60° south of west
Solution:
(i) 45 cm 30 0 north of east

(ii) 80 km 60° south of west

Question 2.
Prove that the relation R defined on the set V of all vectors by \(\vec{a}\) R \(\vec{b}\) if \(\vec{a}=\vec{b}\) is an equivalence relation on V.
Solution:
\(\vec{a}\) R \(\vec{b}\) is given as \(\vec{a}=\vec{b}\).
(i) \(\vec{a}\) = \(\vec{a}\) ⇒ \(\vec{a}\) R \(\vec{a}\)
(i.e.,) the relation is reflexive.

(ii) \(\vec{a}=\vec{b}\) ⇒ \(\vec{b}\) = \(\vec{a}\)
(i.e.,) \(\vec{a}\) R \(\vec{b}\) – \(\vec{b}\) R \(\vec{a}\)
So, the relation is symmetric.

(iii) \(\vec{a}=\vec{b} ; \vec{b}=\vec{c} \Rightarrow \vec{a}=\vec{c}\)
(i.e„) \(\vec{a}\) R \(\vec{b}\) ; \(\vec{b}\) R \(\vec{c}\) ⇒ \(\vec{a}\) R \(\vec{c}\)
So the given relation is transitive
So, it is an equivalence relation.

Question 3.
Let \(\vec{a}\) and \(\vec{a}\) be the position vectors of the points A and B. Prove that the position vectors of the points which trisects the line segment AB are
Solution:

Question 4.
If D and E are the midpoints of the sides AB and AC of a triangle ABC, prove that

Solution:

Question 5.
Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side whose length is half of the length of the third side.
Solution:

Question 6.
Prove that the line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.
Solution:

In a quadrilateral when opposite sides are equal and parallel it is a parallelogram So, PQRS is a parallelogram, from (1) and (2).

Question 7.
If \(\vec{a}\) and \(\vec{b}\) represent a side and a diagonal of a parallelogram, find the other sides and the other diagonal.
Solution:
OABC is a parallelogram where

Question 8.
If \(\overrightarrow{\mathrm{PO}}+\overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{QO}}+\overrightarrow{\mathrm{OR}}\), prove that the points P, Q, R are collinear.
Solution:

But Q is a common point.
⇒ P, Q, R are collinear.

Question 9.
If D is the midpoint of the side BC of a triangle ABC, prove that \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{A C}}=2 \overrightarrow{\mathbf{A D}}\)
Solution:
D is the midpoint of ∆ ABC.

Question 10.
If G is the centroid of a triangle ABC, prove that \(\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}=0\)
Solution:
For any triangle ABC,
\(\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}=0\)
Now G is the centroid of ∆ABC, which divides the medians (AD, BE and CF) in the ratio 2 : 1.

Question 11.
Let A, B and C be the vertices of a triangle. Let D, E and F be the midpoints of the sides BC, CA, and AB respectively. Show that \(\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{BE}}+\overrightarrow{\mathrm{CF}}=\overrightarrow{0}\)
Solution:
In ∆ABC, D, E, F are the midpoints of BC, CA and AB respectively.

Question 12.
If ABCD is a quadrilateral and E and F are the midpoints of AC and BD respectively, then prove that \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{C B}}+\overrightarrow{\mathbf{C D}}=4 \overrightarrow{\mathrm{EF}}\)
Solution:
ABCD is a quadrilateral in which E and F are the midpoints of AC and BD respectively.

Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.1 Additional Problems

Question 1.
Shown that the points with position vectors are collinear.
Solution:
To prove the points P, Q, R are collinear we have to prove that
\(\overrightarrow{\mathrm{PQ}}\) = t \(\overrightarrow{\mathrm{PR}}\) where t is a scalar.
Let the given points be P, Q, R.

So, the points P, Q, R are collinear (i.e,) the given points are collinear.

Question 2.
If ABC and A’B’C’ are two triangles and G, G’ be their corresponding centroids, prove that \(\overrightarrow{\mathrm{AA}^{\prime}}+\overrightarrow{\mathrm{BB}^{\prime}}+\overrightarrow{\mathrm{CC}^{\prime}}=3 \overrightarrow{\mathrm{GG}}\)
Solution:
Let O be the origin.
We know when G is the centroid of ∆ ABC,

Question 3.
Prove using vectors the mid-points of two opposite sides of a quadrilateral and the mid-points of the diagonals are the vertices of a parallelogram.
Solution:
ABCD is a quadrilateral with position vectors
OA = \(\vec{a}\), OB = \(\vec{b}\), OC = \(\vec{c}\) and OD = \(\vec{d}\)
P is the midpoint of BC and R is the midpoint of AD.
Q is the midpoint of AC and S is the midpoint of BD.
To prove PQRS is a parallelogram. We have to prove that \(\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{SR}}\)

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5

Choose the correct or the most suitable answer from the given four alternatives.

Question 1.
If a_ij = \(\frac{1}{2}\) (3i – 2j) and A = [a_ij]_2×2 is

Solution:

Question 2.
What must be the matrix X, if 2X + \(\left[\begin{array}{ll}{1} & {2} \\ {3} & {4}\end{array}\right]=\left[\begin{array}{ll}{3} & {8} \\ {7} & {2}\end{array}\right]\) ?

Solution:

Question 3.
Which one of the following is not true about the matrix \(\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {0} & {0} \\ {0} & {0} & {5}\end{array}\right]\)?
(a) a scalar matrix
(b) a diagonal matrix
(c) an upper triangular matrix
(d) A lower triangular matrix
Solution:
(b) a diagonal matrix

Question 4.
If A and B are two matrices such that A + B and AB are both defined, then …………
(a) A and B are two matrices not necessarily of same order.
(b) A and B are square matrices of same order.
(c) Number of columns of a is equal to the number of rows of B.
(d) A = B.
Solution:
(b) A and B are square matrices of same order.

Question 5.
If A = \(\left[\begin{array}{rr}{\lambda} & {1} \\ {-1} & {-\lambda}\end{array}\right]\), then for what value of λ, A² = 0?
(a) 0
(b) ±1
(c) -1
(d) 1
Solution:

Question 6.
If and (A + B)² = A² + B², then the values of a and b are ……………….
(a) a = 4, b = 1
(b) a = 1, b = 4
(c) a = 0, b = 4
(d) a = 2, b = 4
Solution:

Question 7.
If is a matrix satisfying the equation AA^T = 9I, where I is 3 × 3 identity matrix, then the ordered pair (a, b) is equal to ………….
(a) (2, -1)
(b) (-2, 1)
(c) (2, 1)
(d) (-2, -1)
Solution:

Question 8.
If A is a square matrix, then which of the following is not symmetric?
(a) A + A^T
(b) AA^T
(c) A^TA
(d)A – A^T
Solution:
(b)

Question 9.
If A and B are symmetric matrices of order n, where (A ≠ B), then …………….
(a) A + B is skew-symmetric
(b) A + B is symmetric
(c) A + B is a diagonal matrix
(d) A + B is a zero matrix
Solution:
(b)

Question 10.
If and if xy = 1, then det (AA^T) is equal to …………..
(a) (a – 1)²
(b) (a² + 1)²
(c) a² – 1
(d) (a² – 1)²
Solution:

Question 11.
The value of x, for which the matrix is singular is ………….
(a) 9
(b) 8
(c) 7
(d) 6
Solution:
(b) Hint: Given A is a singular matrix ⇒ |A| = 0

⇒ e^x-2.e^2x+3 – e^2+x.e^7+x = 0
⇒ e^3x+1 – e^9+2x = 0 ⇒ e^3x+1 = e^9+2x
⇒ 3x + 1 = 9 + 2x
3x – 2x = 9 – 1 ⇒ x = 8

Question 12.
If the points (x, -2), (5, 2), (8, 8) are collinear, then x is equal to …………
(a) -3
(b) \(\frac{1}{3}\)
(c) 1
(d) 3
Solution:
(d) Hint: Given that the points are collinear
So, area of the triangle formed by the points = 0

Question 13.

Solution:

Question 14.
If the square of the matrix is the unit matrix of order 2, then α, β and γ should satisfy the relation.
(a) 1 + α² + βγ = 0
(b) 1 – α² – βγ = 0
(c) 1 – α² + βγ = 0
(d) 1 + α² – βγ = 0
Solution:

Question 15.

(a) Δ
(b) kΔ
(c) 3kΔ
(d) k³Δ
Solution:

Question 16.
A root of the equation is …………….
(a) 6
(b) 3
(c) 0
(d) -6
Solution:

Question 17.
The value of the determinant of is ……………
(a) -2abc
(b) abc
(c) 0
(d) a² + b² + c²
Solution:

Question 18.
If x₁, x₂, x₃ as well as y₁, y₂, y₃ are in geometric progression with the same common ratio, then the points (x₁, y₁), (x₂, y₂), (x₃, y₃) are
(a) vertices of an equilateral triangle
(b) vertices of a right angled triangle
(c) vertices of a right angled isosceles triangle
(d) collinear
Solution:
(d)

Question 19.
If \(\lfloor.\rfloor\) denotes the greatest integer less than or equal to the real number under consideration and -1 ≤ x < 0, 0 ≤ y < 1, 1 ≤ z ≤ 2, then the value of the determinant is …………..
(a) \(\lfloor z\rfloor\)
(b) \(\lfloor y\rfloor\)
(c) \(\lfloor x\rfloor\)
(d) \(\lfloor x\rfloor+ 1\)
Solution:
(a) Hint: From the given values
>

Question 20.
If a ≠ b, b, c satisfy then abc = ……………..
(a) a + b + c
(b) 0
(c) b³
(d) ab + bc
Solution:
(c) Hint: Expanding along R₁,
a(b² – ac) – 2b (3b – 4c) + 2c (3a – 4b) = 0
(b² – ac) (a – b) = 0
b² = ac (or) a = b
⇒ abc = b(b²) = b³

Question 21.
If then B is given by ………………..
(a) B = 4A
(b) B = -4A
(c) B = -A
(d) B = 6A
Solution:

Question 22.
IfA is skew-symmetric of order n and C ¡s a column matrix of order n × 1, then C^T AC is ……………..
(a) an identity matrix of order n
(b) an identity matrix of order 1
(e) a zero matrix of order I
(d) an Identity matrix of order 2
Solution:
(c) Hint : Given A is of order n × n
C is of order n × 1
so, CT is of order 1 × n

Let it be equal to (x) say
Taking transpose on either sides
(C^T, AC)^T (x)^T .
(i.e.) C^T(A^T)(C) = x
C^T(-A)(C) = x
⇒ C^TAC = -x
⇒ x = -x ⇒ 2x = 0 ⇒ x = 0

Question 23.
The matrix A satisfying the equation is ……………

Solution:

Question 24.
If A + I = , then (A + I) (A – I) is equal to …………….

Solution:

Question 25.
Let A and B be two symmetric matrices of same order. Then which one of the following statement is not true?
(a) A + B ¡s a symmetric matrix
(b) AB ¡s a symmetric matrix
(c) AB = (BA)^T
(d) A^TB = AB^T
Solution:
(b)

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.3

Question 1.
Show that the lines are 3x + 2y + 9 = 0 and 12x + 8y – 15 = 0 are parallel lines.
Solution:

Here m₁ = m₂ ⇒ the two lines are parallel.

Question 2.
Find the equation of the straight line parallel to 5x – 4y + 3 = 0 and having x – intercept 3.
Solution:
Equation of a line parallel to ax + by + c = 0 will be of the form ax + by + k = 0
So equation of a line parallel to 5x – 4y + 3 = 0 will be of the form 5x – 4y = k

Question 3.
Find the distance between the line 4x + 3y + 4 = 0 and a point
(i) (-2, 4)
(ii) (7, -3)
Solution:
The distance between the line ax + by + c = 0 and the point(x₁, y₁) is given by

(i) Now the distance between the line 4x + 3y + 4 = 0 and (-2, 4) is

(ii) The distance between the line 4x + 3y + 4 = 0 and (7, -3) is

Question 4.
Write the equation of the lines through the point (1, -1)
(i) Parallel to x + 3y – 4 = 0
(ii) Perpendicular to 3x + 4y = 6
Solution:
(i) Any line parallel to x + 3y – 4 = 0 will be of the form x + 3y + k = 0.
It passes through (1,-1) ⇒ 1 – 3 + k = 0 ⇒k = 2
So the required line is x + 3y + 2 = 0

(ii) Any line perpendicular to 3x + 4y – 6 = 0 will be of the form 4x – 3y + k = 0.
It passes through (1, -1) ⇒ 4 + 3 + k = 0 ⇒ k = -7.
So the required line is 4x – 3y – 7 = 0

Question 5.
If (- 4, 7) is one vertex of a rhombus and if the equation of one diagonal is 5x – y + 7 = 0, then find the equation of another diagonal.
Solution:
In a rhombus, the diagonal cut at right angles.
The given diagonal is 5x – y + 7 = 0 and (- 4, 7) is not a point on the diagonal.
So it will be a point on the other diagonal which is perpendicular to 5x – y + 7 = 0.
The equation of a line perpendicular to 5x – y + 7 = 0 will be of the form x + 5y + k = 0. It passes through (-4, 7) ⇒ -4 + 5(7) + k = 0 ⇒ k = -31
So the equation of the other diagonal is x + 5y – 31 = 0

Question 6.
Find the equation of the lines passing through the point of intersection lines 4x – y + 3 = 0 and 5x + 2y +7 = 0, and
(i) Through the point (-1, 2)
(ii) Parallel to x – y + 5 = 0
(iii) Perpendicular to x – 2y + 1 = 0.
Solution:
To find the point of intersection of the lines we have to solve them

Substituting x = -1 in equation (2) we get
-5 + 2y = -7
⇒ 2y = – 7 + 5 = -2
⇒ y = -1
So the point of intersection is (-1, -1)

(ii) Equation of a line parallel to x – y + 5 = 0 will be of the form x – y + k= 0.
It passes through (-1, -1) ⇒ -1 + 1 + k = 0 ⇒ k = 0.
So the required line is x – y = 0 ⇒ x = y.

(iii) Equation of a line perpendicular to x – 2y+ 1 =0 will be of the form 2x + y + k = 0. It passes through (-1, -1) ⇒ -2 – 1 + k = 0 ⇒ k = 3.
So the required line is 2x + y + 3 = 0

Question 7.
Find the equations of two straight lines which are parallel to the line 12x + 5y + 2 = 0 and at a unit distance from the point (1, -1).
Solution:
Equation of a line parallel to 12x + 5y + 2 = 0 will be of the form 12x + 5y + k = 0.
We are given that the perpendicular distance form (1, -1) to the line 12x + 5y + k = 0 is 1 unit.

So the required line will be 12x + 5y + 6 = 0 or 12x + 5y – 20 = 0

Question 8.
Find the equations of straight lines which are perpendicular to the line 3x + 4y – 6 = 0 and are at a distance of 4 units from (2, 1).
Solution:
Given equation of line is 3x + 4y – 6 = 0.
Any line perpendicular to 3x + 4y – 6 = 0 will be of the form 4x – 3y + k = 0
Given perpendicular distance is 4 units from (2, 1) to line (1)

∴ 20 = + (5 + k) or 20 = – (5 + k)
⇒ k = 20 – 5 or k = -(20 + 5)
k = 15 or k : = -25
∴ Required equation of the lines are 4x – 3y + 15 = 0 and 4x – 3y – 25 = 0

Question 9.
Find the equation of a straight line parallel to 2x + 3y = 10 and which is such that the sum of its intercepts on the axes is 15.
Solution:
The equation of the line parallel to 2x + 3y = 10 will be of the form 2x + 3y = k .

Question 10.
Find the length of the perpendicular and the co-ordinates of the foot of the perpendicular from (-10, -2) to the line x + y – 2 = 0.
Solution:
Length of the perpendicular from (-10, -2) to x + y – 2 = 0 is

Question 11.
If p₁ and p₂ are the lengths of the perpendiculars from the origin to the straight lines. sec θ +y cosec θ = 2a and x cos θ – y sin θ = a cos 2θ, then prove that _{p_{1}^{2}+p_{2}^{2}=a^{2}}
Solution:
p₁ = length of perpendicular from (0, 0) to x sec θ + y cosec θ = 2a

Question 12.
Find the distance between the parallel lines
(i) 12x + 5y = 7 and 12x + 5y + 7 = 0
(ii) 3x – 4y + 5 = 0 and 6x – 8y – 15 = 0.
Solution:

Question 13.
Find the family of straight lines
(i) Perpendicular
(ii) Parallel to 3x + 4y – 12 = 0.
Solution:
Equation of lines perpendicular to 3x + 4y – 12 = 0 will be of the form 4x – 3y + k = 0, k ∈ R Equation of lines parallel to 3x + 4y – 12 = 0 will be of the form 3x + 4y + k = 0, k ∈ R

Question 14.
If the line joining two points A (2, 0) and B (3, 1) is rotated about A in anti-clockwise direction through an angle of 15°, then find the equation of the line in new position.
Solution:

This line is rotated about 15° in anti clockwise direction
⇒ New slope = tan (45° + 15°) = tan 60° = \(\sqrt{3}\) (i.e) m = \(\sqrt{3}\).
Point A = (2, 0)

Question 15.
A ray of light coming from the point (1, 2) is reflected at a point A on the x-axis and it passes through the point (5,3). Find the co-ordinates of the point A.
Solution:
The image of the point P (1, 2) will be P’ (1, -2).
Since ∠OAP = ∠XAQ (angle of inches = angle of reflection) So ∠OAP’ = ∠XAQ = a (Vertically opposite angles)
⇒ P’, A, Q lie on a same line.

Now equation of the line P’, Q is [where P’ = (1, -2), Q = (5, 3)]

Since we find point of intersection with x axis we put y = 0.

Question 16.
A line is drawn perpendicular to 5x = y + 7. Find the equation of the line if the area of the triangle formed by this line with co-ordinate axes is 10 sq. units.
Solution:
Equation of the given lines 5x = y+ 7 ⇒ 5x – y = 7.
So its perpendicular will be of the form x + 5y = 7

Question 17.
Find the image of the point (-2, 3) about the line x + 2y – 9 = 0.
Solution:
The coordinates of image of the point (x₁, y₁) with respect to the line ax + by + c = 0 can be

Question 18.
A photocopy store charges Rs. 1.50 per copy for the first 10 copies and Rs. 1.00 per copy after the 10th copy. Let x be the number of copies, and let y be the total cost of photocopying.
(i) Draw graph of the cost as x goes from 0 to 50 copies.
(ii) Find the cost of making 40 copies
Solution:

Question 19.
Find atleast two equations of the straight lines in the family of the lines y = 5x + b, for which b and the x-coordinate of the point of intersection of the lines with 3x – 4y = 6 are integers.
Solution:
y = 5x + b …….. (1)
3x-4y = 6 …….. (2)
Solving (1) and (2)
Substituting y value from (1) in (2) we get


Since, x coordinate and 6 are integers 6 + 46 must be multiple of 17

Question 20.
Find all the equations of the straight lines in the family of the lines y = mx – 3, for which m and the x-coordinate of the point of intersection of the lines with x – y = 6 are integers.
Solution:
Equation of the given lines are
y= mx – 3 …….. (1)
and x – y = 6 ……. (2)
Solving (1) and (2)
x – (mx – 3) = 6

Since m and x coordinates are integers
1 – m is the divisor of 3 (i.e) ± 1, ± 3

So equation of lines are (y = mx – 3) ,y = mx – 3
(i) When m = 0, y = -3
(ii) When m = 2, y = 2x – 3
(iii) When m = -2, y = -2x – 3 or 2x + y + 3 = 0
(iv) When m = 4, y = 4x – 3

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.3 Additional Questions

Question 1.
Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, -1).
Solution:
Slope of the line joining the points (2, 3) and (3, -1) is

Slope of the required line which is perpendicular to it

Equation of the line passing through the point (5, 2) is

Hence, the required equation is x – 4y + 3 = 0.

Question 2.
Find the points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10. Solution:

So, the required point is (3, 1)
Now taking(-) sign, we have

So, the required point is (- 7, 11)
Hence, the required points on the given line are (3, 1) and (-7, 11).

Question 3.
Find the equation of the line passing through the point of intersection 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x +4y = 7.
Solution:

On putting the value of λ in equation (iv) we get
(2x + y – 5) + 1 (x + 3y + 8) = 0
⇒ 2x + y – 5 + x + 3y + 8 = 0
⇒ 3x + 4y + 3 = 0
Hence, the required equation is 3x + 4y + 3 = 0

Question 4.
A line passing through the points (a, 2a) and (-2, 3) is perpendicular to the line 4 x + 3y + 5 = 0, find the value of a.
Solution:

Question 5.
Find the equation of the straight line which passes through the intersection of the straight lines 2x + y = 8 and 3x – 2y + 7 = 0 and is parallel to the straight line 4x + y – 11 = 0.
Solution:
Equation of line through the intersection of straight lines
2x + y = 8 and 3x – 2y + 7 = 0 is
2x + y – 8 + k (3x – 2y + 7) = 0
x(2 + 3k) + y (1 – 2k) +(-8 + 7k) = 0

⇒ 28x + 7y – 74 = 0

Question 6.
Find the equation of the straight line passing through intersection of the straight lines 5x – 6y = 1 and 3x + 2y + 5 = 0 and perpendicular to the straight line 3x – 5y + 11 = 0.
Solution:
Equation of line through the intersection of straight lines 5x – 6y = 1 and 3x + 2y + 5 = 0 is
5x – 6y – 1 + k (3x + 2y + 5) = 0
x (5 + 3k) + y (-6 + 2k) + (-1 + 5k) = 0
This is perpendicular to 3x – 5y + 11 = 0
That is, the product of their slopes is -1

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.2

Question 1.
Find the equation of the lines passing through the point (1, 1)
(i) With y-intercept (- 4)
(ii) With slope 3
(iii) And (-2, 3)
(iv) And the perpendicular from the origin makes an angle 60° with x-axis.
Solution:
(i) Given y intercept = – 4,
Let x intercept be a

(ii) Slope m = 3, passing through (x₁, y₁) = (1, 1)
Equation of the line is y – y₁ = m(x – x₁)
(i.e) y- 1 = 3(x – 1) ⇒ y – 1 = 3x – 3 ⇒ 3x – y = 2

(iii) Passing through (1, 1) and (-2, 3)

Question 2.
If P (r, c) is mid point of a line segment between the axes, then show that \(\frac{x}{r}+\frac{y}{c}=\) 2.
Solution:
P (r, c) is the mid point of AB.
⇒ A = (2r, 0) and B = (0, 2c)
(i.e) x intercept = 2r and
y intercept = 2c .

Question 3.
Find the equation of the line passing through the point (1, 5) and also divides the co-ordinate axes in the ratio 3 : 10.
Solution:
Let x intercept be 3a and y intercept be 10a

Question 4.
If p is length of perpendicular from origin to the line whose intercepts on the axes are a and b, then show that \(\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\)
Solution:

Question 5.
The normal boiling point of water is 100°C or 212°F and the freezing point of water is 0°C or 32°F.
(i) Find the linear relationship between C and F
(ii) Find the value of C for 98.6° F and
(iii) The value of F for 38°C .
Solution:
Given when C = 100, F = 212 and when C = 0, F = 32

Question 6.
An object was launched from a place P in constant speed to hit a target. At the 15th second it was 1400m away from the target and at the 18th second 800m away. Find
(i) The distance between the place and the target
(ii) The distance covered by it in 15 seconds,
(iii) Time taken to hit the target.
Solution:
Taking time = x and distance = y
We are given at x = 15, y = 1400 and at x = 18, y = 800

Question 7.
Population of a city in the years 2005 and 2010 are 1,35,000 and 1,45,000 respectively. Find the approximate population in the year 2015. (assuming that the growth of population is constant).
Solution:
Taking year as x and population as y
We are given when x = 2005,
y = 1,35,000 and
when x = 2010,
y = 1,45,000

y – 135000 = 2000 (x – 2005)
y = 2000(x – 2005) + 135000
At x = 2015, y = 2000 (2015 – 2005) + 135000
(i.e) y = 2000 (10) + 135000 = 20000 + 135000 = 1,55,000
The approximate population in the year 2015 is 1,55,000

Question 8.
Find the equation of the line, if the perpendicular drawn from the origin makes an angle 30° with x – axis and its length is 12.
Solution:
The equation of the line is x cos α + y sin α = p

Question 9.
Find the equation of the straight lines passing through (8, 3) and having intercepts whose sum is 1.
Solution:
Given sum of the intercepts = 1 ⇒ when x intercept = a then y intercept = 1 – a

8 (1 – a) + 3a = a (1 – a)
8 – 8a + 3a = a – a²

Question 10.

Solution:

⇒ The points A, B, C lie on a line
⇒ The points A, B, C are collinear

Question 11.
A straight line is passing through the point A (1, 2) with slope \(\frac{5}{12}\). Find points on the line which are 13 units away from A.
Solution:

Question 12.
A 150m long train is moving with constant velocity of 12.5 m/s. Find
(i) The equation of the motion of the train,
(ii) Time taken to cross a pole,
(iii) The time taken to cross the bridge of length 850 m is?
Solution:
(i) Now m = \(\frac{y}{x}\) = 12.5m / second,
The equation of the line is y = mx + c ….(1)
Put c = -150, m = 12.5 m,
The equation of motion of the train is y = 12.5x – 150

(ii) To find the time taken to cross a pole we take y = 0 in (1)
⇒ 0 = 12.5x – 150 ⇒ 12.5x = 150

(iii) When y = 850 in (1)
850 = 12.5 x – 150 ⇒ 12.5x = 850 + 150 = 1000

Question 13.
A spring was hung from a hook in the ceiling. A number of different weights were attached to the spring to make it stretch, and the total length of the spring was measured each time shown in the following table.

(i) Draw a graph showing the results.
(ii) Find the equation relating the length of the spring to the weight on it.
(iii) What is the actual length of the spring.
(iv) If the spring has to stretch to 9 cm long, how much weight should be added?
(v) How long will the spring be when 6 kilograms of weight on it?
Solution:
Taking weight (kg) as x values and length (cm) as y values we get (x₁, y₁) = (2, 3), (x₂, y₂) = (4, 4)

The equation of the line passing through the above two points is

(iii) When x = 0, 2y = 4 ⇒ y = 2 cm

(iv) When y = 9 cm, x – 18 = – 4
x = -4 + 18 = 14 kg

(v) When x = 6 (kg), 6 – 2y = – 4, -2y = -4 – 6 = -10
⇒ 2y = 10 ⇒ y = 10/2 = 5 cm.

Question 14.
A family is using Liquefied petroleum gas (LPG) of weight 14.2 kg for consumption. (Full weight 29.5 kg includes the empty cylinders tare weight of 15.3 kg.). If it is use with constant rate then it lasts for 24 days. Then the new cylinder is replaced
(i) Find the equation relating the quantity of gas in the cylinder to the days.
(ii) Draw the graph for first 96 days.
Solution:

Question 15.
In a shopping mall there is a hall of cuboid shape with dimension 800 × 800 × 720 units, which needs to be added the facility of an escalator in the path as shown by the dotted line in the figure. Find
(i) The minimum total length of the escalator,
(ii) The heights at which the escalator changes its direction,
(iii) The slopes of the escalator at the turning points.

Solution:
(i) the minimum total length of the escalator.
Shape of the hall in the shopping mall is cuboid. When you open out the cuboid, the not of the cuboid will be as shown in the following diagram.

The path of the escalator is from OA to AB to BC to CD

The minimum length = 3280 units

(ii) The height at which the escalator changes its direction.

(iii) Slope of the escalator at the turning points

Since ∆OAE = ∆ABB’ = ∆BCC’ = ∆CAD
Slope at the points B, C will be \(\frac{9}{40}\)

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.2 Additional Questions Solved

Question 1.
Find the equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of x – axis.
Solution:

Question 2.
Find the equation of the line which passes through the point (- 4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point.
Solution:
Let AB be a line passing through a point (-4, 3) and meets x-axis at A (a, 0) and y-axis at B (0, b).

Question 3.
If the intercept of a line between the coordinate axes is divided by the point (-5, 4) in the ratio 1 : 2, then find the equation of the line.
Solution:
Let a and b be the intercepts on the given line.


Hence, the required equation is 8x – 5y + 60 = 0

Question 4.
Find the equation of the straight line which passes through the point (1, -2) and cuts off equal intercepts from axes.
Solution:
Intercept form of a straight line is \(\frac{x}{a}+\frac{y}{b}\) = 1, where a and b are the intercepts on the axis

If equation (1) passes through the point (1, -2) we get

So, equation of the straight line is x v

Hence, the required equation x + y + 1 = 0

Question 5.
Find the distance of the line 4x – y = 0 from the point P(4, 1) measured along the line making an angle 135° with the positive x-axis
Solution:
The equation in distance form of the line passing through P(4, 1) and making an angle of 135° with the positive x-axis

Question 6.
The line 2x – y = 5 turns about the point on it, whose ordinate and abscissa are equal, through an angle of 45° in the anti-clockwise direction, find the equation of the line in the new position.
Solution:
If the line 2x – y = 5 makes an angle θ with x – axis. Then, tan θ = 2. Let P (α, α) be a point on the line 2x – y = 5.
Then, 2 α – α = 5 ⇒ α = 5

So, the coordinates of P are (5, 5). If the line 2x – y – 5 = 0 is rotated about point
P through 45° in anti clockwise direction, then the line in its new position makes angle θ + 45° with x – axis. Let m’ be the slope of the line in its new position. Then,

Thus, the line in its new pdsition passes through P (5, 5) and has slope m’ = -3
So, its equation y – 5 = m’ (x – 5) or, y – 5 = -3 (x – 5) or, 3x + y – 20 = 0

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.1

Question 1.
Find the locus of P, if for all values of a, the co-ordinates of a moving point P is
(i) (9 cos α, 9 sin α)
(ii) (9 cos α, 6 sin α)
Solution:
(i) Let P(h, k) be the moving point.
We are given h = 9 cos α and k = 9 sin α and

∴ locus of the point is x² + y² = 81

(ii) Let P(h , k) be a moving point.
We are given h = 9 cos α and k = 6 sin α

Question 2.
Find the locus of a point P that moves at a constant distant of
(i) Two units from the x-axis
(ii) Three units from the y-axis.
Solution:
(i) Let the point (x, y) be the moving point. Equation of a line at a distance of 2 units from x-axis is k = 2
So the locus is y = 2
(i.e.) y – 2 = 0
435

(ii) Equation of a line at a distance of 3 units from y-axis is h = 3
So the locus is x = 3 (i.e.) x – 3 = 0

Question 3.
If θ is a parameter, find the equation of the locus of a moving point, whose coordinates are x – a cos³ θ, y = a sin³ θ
Solution:

Question 4.
Find the value of k and b, if the points P (-3, 1) and Q (2, b) lie on the locus of x² – 5x + ky = 0.
Solution:
Given P (-3, 1) lie on x² – 5x + ky = 0
⇒ (-3)² – 5(-3) + k(1) = 0
9 + 15 + k = 0 ⇒ k = -24
Q (2, b) lie on x² – 5x + ky = 0
(2)² – 5(2) + k(b) = 0 ⇒ 4 – 5(2) – 24b = 0

Question 5.
A straight rod of length 8 units slides with its ends A and B always on the x and y-axis respectively. Find the locus of the mid point of the line segment AB.
Solution:
Let P (h, k) be the moving point A (a, 0) and B (0, b) P is the mid point of AB.

Question 6.
Find the equation of the locus of a point such that the sum of the squares of the distance from the points (3, 5), (1, -1) is equal to 20.
Solution:
Let P (h, k) be the moving point
Let the given point be A (3, 5) and B (1, -1)
We are given PA² + PB² = 20
⇒ (h – 3)² + (k – 5)² + (h – 1)² + (k + 1)² = 20
⇒ h² – 6h + 9 + k² – 10k + 25 + h² – 2h + 1 + k² + 2k + 1 = 20
(i.e.) 2h² + 2k² – 8h – 8k + 36 – 20 = 0
2h² + 2k² – 8h – 8k + 16 = 0
(÷ by 2 ) h² + k² – 4h – 4k + 8 = 0
So the locus of P is x² + y² – 4x – 4y + 8 = 0

Question 7.
Find the equation of the locus of the point P such that the line segment AB, joining the points A (1, -6) and B (4, -2), subtends a right angle at P.
Solution:
Let P (h, k) be the moving point

Given A (1, – 6) and B (4, – 2),
Since ∆APB = 90°, PA² + PB² = AB²
(i.e.) (h – 1)² + (k + 6)² + (h – 4)² + (k + 2)² = (4 – 1)² + (-2 + 6)²
(i.e) h² + 1 – 2h + k² + 36 + 12k + h² + 16 – 8h + k² + 4 + 4k = 3² + 4² = 25
2h² + 2k² -10h + 16k + 57 – 25 = 0
2h² + 2k² – 10h + 16k + 32 = 0
(÷ by 2)h² + k² – 5h + 8k + 16 = 0
So the locus of P is x² + y² – 5x + 8y + 16 = 0

Question 8.
If O is origin and R is a variable point on y² = 4x, then find the equation of the locus of the mid-point of the line segment OR.
Solution:
Let P(h, k) be the moving point
We are given O (0, 0). Let R = (a, b)

Substituting a, b values is y² = 4x
we get (2k)² = 4 (2h)
(i.e) 4k² = 8h
(÷ by 4) k² = 2h
So the locus of P is y² = 2x

Question 9.

Solution:

Question 10.
If P (2, -7) is a given point and Q is a point on 2x² + 9y² = 18, then find the equations of the locus of the mid-point of PQ.
Solution:
P = (2, -7); Let (h, k) be the moving point Q = (a, b)

⇒ a = 2h – 2,
b = 2k + l
Q is a point on 2x² + 9y² = 18 (i.e) (a, b) is on 2x² + 9y² = 18
⇒ 2(2h – 2)² + 9 (2k + 7)² = 18
(i.e) 2 [4h² + 4 – 8h] + 9 [4k² + 49 + 28k] – 18 = 0
(i.e) 8h² + 8 – 16h + 36k² + 441 + 252k – 18 = 0
8h² + 36k² – 16h + 252k + 431 = 0
The locus is 8x² + 36y² – 16x + 252y + 431 = 0

Question 11.
If R is any point on the x-axis and Q is any point on the y-axis and Pis a variable point on RQ with RP = b, PQ = a. then find the equation of locus of P.
Solution:
P = (x, 0), Q = (0, y), R (h, k) be a point on RQ such that PR : RQ = b : a

From the right angled triangle OQR, OR² + OQ² = QR²

Question 12.
If the points P (6, 2) and Q (-2, 1) and R are the vertices of a ∆PQR and R is the point on the locus y = x² – 3x + 4, then find the equation of the locus of centroid of ∆PQR.
Solution:
P (6, 2), Q (-2, 1). Let R = (a, b) be a point on y = x² – 3x + 4.

But (a, b) is a point on y = x² – 3x + 4
b = a² – 3a + 4
(i.e) 3k – 3 = (3h – 4)² – 3(3h – 4) + 4
(i.e) 3k – 3 = 9h² + 16 – 24h – 9h + 12 + 4
⇒ 9h² – 24h – 9h + 32 – 3k + 3 = 0
(i.e) 9h² – 33h – 3k + 35 = 0,
Locus of (h, k) is 9x² – 33x – 3y + 35 = 0

Question 13.
If Q is a point on the locus of x² + y² + 4x – 3y + 7 = 0 then find the equation of locus of P which divides segment OQ externally in the ratio 3 : 4, where O is origin.
Solution:
Let (h, k) be the moving point O = (0, 0);
Let PQ = (a, b) on x² + y² + 4x – 3y + 7 = 0

Question 14.
Find the points on the locus of points that are 3 units from x – axis and 5 units from the point (5, 1).
Solution:
A line parallel to x-axis is of the form y = k. Here k = 3 ⇒y = 3
A point on this line is taken as P (a, 3). The distance of P (a, 3) from (5, 1) is given as 5 units
⇒ (a – 5)² + (3 – 1)² = 5²
a² + 25 – 10a + 9 + 1 – 6 = 25
a² – 10a + 25 + 4 – 25 = 0
a² – 10a + 4 = 0

Question 15.
The sum of the distance of a moving point from the points (4, 0) and (-4, 0) is always 10 units. Find the equation of the locus of the moving point.
Solution:
Let P (h, k) be a moving point
Here A = (4, 0) and B = (-4, 0)
Given PA + PB = 10

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.1 Additional Questions

Question 1.
If the sum of the distance of a moving point in a plane from the axis is 1, then find the locus of the point.
Solution:
Let coordinates of a moving point P be (x, y).
Given that the sum of the distances from the axis to the point is always 1.

∴ |x| + |y| = 1 ⇒ x + y = 1
⇒ -x – y = 1 ⇒ x – y = 1
Hence, these equations give us the locus of the point P which is a square.

Question 2.
A point moves so that square of its distance from the point (3, -2) is numerically equal to its distance from the line 5x -12y = 3. The equation of its locus is ……..
Solution:
The given equation of line is 5x – 12y = 3 and the given point is (3, -2).
Let (a, b) be any moving point.


⇒ 13a² + 13b² – 78a + 52b + 169 = 5a – 12b – 3
⇒ 13a² + 13b² – 83a + 64b + 172 = 0
So, the locus of the point is 13x² + 13y² – 83x + 64y + 172 = 0

Question 3.
Find the Locus of the mid points of the portion of the line x cos θ + y sin θ = p intercepted between the axis.
Solution:
Given equation of the line is x cos θ + y sin θ = p … (i)
Let C (h, k) be the mid point of the given line AB where it meets the two axis at A (a, 0) and B (0, b).
Since (a, 0) lies on eq (i) then “a cos θ + θ = p”


B (0, b) also lies on the eq (i) then 0 + b sin θ = p

Since C (h, k) is the mid point of AB

Putting the values of a and b is eq (ii) and (iii) we get P

Squaring and adding eq (iv) and (v) we get

Question 4.

Solution:

Here, α is a variable. To find the locus of P (h, k), we have to eliminate α.
From (i), we obtain

Question 5.

Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4

Question 1.
Find the area of the triangle whose vertices are (0, 0), (1, 2) and (4, 3)
Solution:
Area of triangle with vertices

∴ Area of A with vertices (0, 0), (1, 2) and (4, 3) is

(as the area cannot be negative).

Question 2.
If (k, 2), (2, 4) and (3, 2) are vertices of the triangle of area 4 square units then determine the value of k.
Solution:

Question 3.
Identify the singular and non-singular matrices:

Solution:
(i) For a given square matrix A,
1. If |A| = 0 then it is a singular matrix.
2. If |A| ≠ 0 then it is a non singular matrix.

⇒ A is a singular matrix.

Which is a skew symmetric matrix
∴ |A| = 0 ⇒ A is a singular matrix.

Question 4.
Determine the value of a and b so that the following matrices are singular:

Solution:

expanding along R₁
b(4 + 4) + 7 (-6 – 1) = 0 (given)
8b + 7 (-7) = 0
(i.e.,) 8b – 49 = 0 ⇒ 8b = 49 ⇒ b = 49/8

Question 5.
If cos 2θ = 0, determine \(\left[\begin{array}{ccc}{\theta} & {\cos \theta} & {\sin \theta} \\ {\cos \theta} & {\sin \theta} & {0} \\ {\sin \theta} & {0} & {\cos \theta}\end{array}\right]^{2}\)
Solution:

Question 6.
Find the value of the product;
Sol:

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 Additional Problems

Question 1.
Identify the singular and non-singular matrix.

Solution:
If the determinant value of a square matrix is zero it is called a singular matrix. Otherwise it is non-singular.

Question 2.
Show that
Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.5

Question 1.
The sum of the digits at the 10th place of all numbers formed with the help of 2, 4, 5, 7 taken all at a time is ……..
(a) 432
(b) 108
(c) 36
(d) 18
Solution:
(b) 108
Hint. Number of digits given = 2, 4, 5, 7
Number of 4 digit numbers formed = 4! =24
So each digit occur \(\frac{24}{4}\) = 6 times
Sum of the digits = 2 + 4 + 5 + 7 = 18
So sum of the digits in each place = 18 × 6 = 108

Question 2.
In an examination there are three multiple choice questions and each question has 5 choices. Number of ways in which a student can fail to get all answer correct is ……..
(a) 125
(b) 124
(c) 64
(d) 63
Solution:
(b) 124
Hint. Each question has 5 options in which 1 is correct
So the number of ways of getting correct answer for all the three questions is 5³ = 125
So the number of ways in which a student can fail to get all answer correct is < 125 (i.e.) 125 – 1 = 124

Question 3.
The number of ways in which the following prize be given to a class of 30 boys first and second in mathematics, first and second in physics, first in chemistry and first in English is ……….
(a) 30⁴ × 29²
(b) 30³ × 29³
(c) 30² × 29⁴
(d) 30 × 29⁵
Solution:
(a) 30⁴ × 29²
Hint.
I and II in maths can be given can be given in 30 × 29 ways.
I and II in physics can be given in 30 × 29 ways.
I and chemistry can be given in 30 ways.
I in English can be given in 30 ways.
So total number of ways = 30 × 29 × 30 × 29 × 30 × 30 = 30⁴ × 29²

Question 4.
The number of 5 digit numbers all digits of which are odd is ………
(a) 25
(b) 5⁵
(c) 5⁶
(d) 625
Solution:
(b) 5⁵
Hint. The odd number are 1, 3, 5, 7, 9
Number of odd numbers = 5
We need a five digit number So the number of five digit number = 5⁵

Question 5.
In 3 fingers, the number of ways four rings can be worn is …… ways.
(a) 4³ – 1
(b) 3⁴
(c) 68
(d) 64
Solution:
(b) 3⁴
Hint. Each letter can be ported in 3 ways
∴ 4 letter is 3⁴ ways

Question 6.

(a) 7 and 11
(b) 6 and 7
(c) 2 and 11
(d) 2 and 6
Solution:
(b) 6 and 7

Question 7.
The product of r consecutive positive integers is divisible by ………
(a) r!
(b) (r – 1)!
(c) (r + 1)!
(d) r!
Solution:
(a) r!
Hint.
1(2) (3) ….. (r) = r! which is ÷ by r!

Question 8.
The number of 5 digit telephone numbers which have none of their digits repeated is
(a) 90000
(b) 10000
(c) 30240
(d) 69760
Solution:
(d) 69760
Hint.
The number of 5 digit telephone numbers which have none of their digits repeated is ¹⁰P₅ = 30240
Thus the required number of telephone number is 10⁵ – 30240 = 69760

Question 9.
If a² – ^aC₂ = a² – ^aC₄ then the value of ‘a’ is ….
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
(b) 3
Hint.
a₂ – a = 2 + 4 = 6
a₂ – a – 6 = 0
(a – 3) (a + 2) = 0 ⇒ a = 3

Question 10.
There are 10 points in a plane and 4 of them are collinear. The number of straight lines joining any two points is ……..
(a) 45
(b) 40
(c) 39
(d) 38
Solution:
(b) 40
Hint.

Question 11.
The number of ways in which a host lady invite 8 people for a party of 8 out of 12 people of whom two do not want to attend the party together is
(a) 2 × ¹¹C₇ + ¹⁰C₈
(b) ¹¹C₇ + ¹⁰C₈
(c) ¹²C₈ – ¹⁰C₆
(d) ¹⁰C₆ + 2!
Solution:
(c) ¹²C₈ – ¹⁰C₆
Hint.
Number of way of selecting 8 people from 12 in ¹²C₈
∴ out of remaining people 8 can attend in ¹⁰C₈
The number of ways in which two of them do not attend together = ¹²C₈ – ¹⁰C₆

Question 12.
The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines …….
(a) 6
(b) 9
(c) 12
(d) 18
Solution:
(d) 18
Hint.
Number of parallelograms = ⁴C₂ × ³C₂
= 6 × 3 = 18

Question 13.
Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is …….
(a) 11
(b) 12
(c) 10
(d) 6
Solution:
(b) 12
Hint.

Question 14.
Number of sides of a polygon having 44 diagonals is ……….
(a) 4
(b) 4!
(c) 11
(d) 22
Solution:
(c) 11
Hint:

Question 15.
If 10 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, then the total number of points of intersection are ………
(a) 45
(b) 40
(c) 10!
(d) 2¹⁰
Solution:
(a) 45
Hint:

Question 16.
In a plane there are 10 points are there out of which 4 points are collinear, then the number of triangles formed is …….
(a) 110
(b) ¹⁰C₃
(c) 120
(d) 116
Solution:
(d) 116
Hint:

Question 17.
In ²ⁿC₃ : ⁿC₃ = 11 : 1 then n is ………
(a) 5
(b) 6
(c) 11
(d) 1
Solution:
(b) 6
Hint.

Question 18.
^{(n – 1)}C_r + ^{(n – 1)}C_{(r – 1)} is ………
(a) ^{(n + 1)}C_r
(b) ^{(n – 1)}C_r
(c) ⁿC_r
(d) ⁿC_{r – 1}
Solution:
(c) ⁿC_r

Question 19.
The number of ways of choosing 5 cards out of a deck of 52 cards which include at least one king is …….
(a) ⁵²C₅
(b) ⁴⁸C₅
(c) ⁵²C₅ + ⁴⁸C₅
(d) ⁵²C₅ – ⁴⁸C₅
Solution:
(d) ⁵²C₅ – ⁴⁸C₅
Hint.
Selecting 5 from 52 cards = ⁵²C₅
selecting 5 from the (non-king cards 48) = ⁴⁸C₅
∴ Number of ways is ⁵²C₅ – ⁴⁸C₅

Question 20.
The number of rectangles that a chessboard has ……
(a) 81
(b) 99
(c) 1296
(d) 6561
Solution:
(c) 1296
Hint. Number of horizontal times = 9
Number of vertical times = 9
Selecting 2 from 9 horizontal lines = ⁹C₂
Selecting 2 from 9 vertical lines = ⁹C₂

Question 21.
The number of 10 digit number that can be written by using the digits 2 and 3 is ……..
(a) ¹⁰C₂ + ⁹C₂
(b) 2¹⁰
(c) 2¹⁰ – 2
(d) 10!
Solution:
(b) 2¹⁰
Hint.
Selecting the number from (2 and 3)
For till the first digit can be done in 2 ways
For till the second digit can be done in 2 ways ….
For till the tenth digit can be done in 2 ways
So, total number of ways in 10 digit number = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2¹⁰

Question 22.
If P_r stands for ^rP_r then the sum of the series 1 + P₁ + 2P₂ + 3P₃ + … + nP_n is ……..
(a) P_{n + 1}
(b) P_{n + 1} – 1
(c) P_{n – 1} + 1
(d) ^{(n + 1)}P_{(n – 1)}
Solution:
(b) P_{n + 1} – 1
Hint:
1 + 1! + 2! + 3! + … + n!
Now 1 + 1 (1!) = 2 = (1 + 1)!
1 + 1 (1!) + 2(2!) = 1 + 1 + 4 = 6 = 3!
1 + 1(1!) + 2(2!)+ 3(3!) = 1 + 1 + 4 + 18 = 24 = 4!
1 + 1(1!) + 2(2!) + 3(3!) ….+ n(n!) = (n + 1) ! – 1
= ^{n + 1}P_{n + 1} – 1 = P_{n + 1} – 1

Question 23.
The product of first n odd natural numbers equals …….

Solution:

Hint:

Question 24.
If ⁿC₄, ⁿC₅, ⁿC₆ are in AP the value of n can be ………..
(a) 14
(b) 11
(c) 9
(d) 5
Solution:
(a) 14
Hint:

30 + n² – 9n + 20 – 12n + 48 = 0
n² – 21 n + 98 = 0
(n – 7) (n – 14) = 0
n = 7 (or) 14

Question 25.
1 + 3 + 5 + 7 + + 17 is equal to ………
(a) 101
(b) 81
(c) 71
(d) 61
Solution:
(b) 81
Hint: