Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 6 Two Dimensional Analytical Geometry Ex 6.3 Text Book Back Questions and Answers, Notes.

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

Question 1.

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

Answer:

The equations of the given lines are

3x + 2y + 9 = 0 ——- (1)

12x + 8y – 15 = 0 ——- (2)

m_{1} = m_{2}

∴ The given lines are parallel.

Question 2.

Find the equation of the straight line parallel to 5x – 4y + 3 = 0 and having x – intercept 3.

Answer:

The equation of any line parallel to 5x – 4y + 3 = 0 is

5x – 4y + k = 0 ……….. (1)

The x – intercept of line (I) is obtained by putting

y = 0 in the equation.

(1) ⇒ 5x – 4(0) + k = 0

5x = – k ⇒ x = \(-\frac{k}{5}\)

Given that the x – intercept in 3

∴ \(-\frac{k}{5}\) = 3 ⇒ k = – 15

∴ The equation of the required line is

5x – 4y – 15 = 0

Question 3.

Find the distance between the line 4x + 3y + 4 = 0 and a point

(i) (- 2, 4)

(ii) (7, – 3)

Answer:

(i) The distance between the line ax + by + c = 0 and the point (x_{1}, y_{1}) is d = \(\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\)

Here (x_{1}, y_{1}) = (- 2, 4) and the equation of the line is 4x + 3y + 4 = 0

(ii) Given point (x_{1}, y_{1}) = (7, – 3)

Given line 4x + 3y + 4 = 0

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

Answer:

(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.

Answer:

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 of the 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

Answer:

The equation of the straight line passing through the point of intersection of the lines.

4x – y + 3 = 0 and 5x + 2y + 7 = 0 is

( 4x – y + 3) + λ (5x + 2y + 7) = 0 ……… (1)

(i) Through the point (-1,2)

Given that line(1) passes through the point (-1, 2)

(1) ⇒ (4(-1) – 2 + 3) + λ (5(- 1) + 2(2) + 7) = 0

(-4 – 2 + 3) + λ (-5 + 4 + 7) = 0

– 3 + 6 λ = 0 ⇒ λ = \(\frac{3}{6}\) = \(\frac{1}{2}\)

∴ The equation of the required line is

(4x – y + 3) + \(\frac{1}{2}\) (5x + 2y + 7) = 0

2(4x – y + 3) + (5x + 2y + 7) = 0

8x – 2y + 6 + 5x + 2y + 7 = 0

13x +13 = 0 ⇒ x + 1 = 0

(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) Perpendicular to x – 2y + 1 = 0

Given that the line (1) perpendicular to the line

x – 2y + 1 = 0 ………….. (3)

(1) ⇒ (4x – y + 3) + λ (5x + 2y + 7) = 0

4x – y + 3 + 5λx + 2λy + 7λ = 0

(4 + 5λ)x + (2λ – 1 )y + (3 + 7λ) = 0 ……….. (4)

Slope of this line (3) = \(-\frac{4+5 \lambda}{2 \lambda-1}\)

Slope of line (2) = \(-\frac{1}{-2}=\frac{1}{2}\)

Given that line (3) and line (4) are perpendicular

4 + 5λ = 2(2λ – 1 )

4 + 5λ = 4λ – 2

λ = – 6

Substituting the value of λ in equation (1) we have

(4x – y + 3) – 6 (5x + 2y + 7) = 0

4x – y + 3 – 30x – 12y – 42 = 0

-26x – 13y – 39 =0

2x + y + 3 = 0

which is the required equation.

Question 7.

Find the equations of two straight line which are parallel to the line 12x + 5y + 2 = 0 and at a unit distance from the point (1, – 1).

Answer:

The equation of the given line is

12x + 5y + 2 = 0 ……… (1)

Equation of any line parallel to the line (1) is

12x + 5y + k = 0 ………. (2)

Given that line (2) is at a unit distance from the point (1, – 1)

k = – 7 ± 13

k = -7 + 13 or k = – 7 – 13

k = 6 or k = – 20

∴ The equation of the required lines are

12x + 5y + 6 = 0 and 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).

Answer:

The equation of the given line is

3x + 4y – 6 = 0 …………. (1)

The equation of any line perpendicular to line (1) is

4x – 3y + k = 0 …………. (2)

Given that this line is 4 units from the point (2, 1)

± 4 = \(\frac{5+k}{5}\)

5 + k = ± 20

k = ± 20 – 5

k = 20 – 5 or k = -20

k = 15 or k = -25

∴ The equation of the required 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.

Answer:

The equation of the given line is

2x + 3y = 10 ………….. (1)

The equation of any line parallel to (1) is

2x + 3y = k …………. (2)

Given that the sum of the intercepts of the line (2) on the axes is 15

∴ The equation of the required line is 2x + 3y = 18

Question 10.

Find the length of the perpendicular and the coordinates of the foot of the perpendicular from (-10, -2) to the line x + y – 2 = 0.

Answer:

The coordinate of the foot of the perpendicular from the point (x_{1}, y_{1}) on the line ax + by + c = 0 is

∴ The coordinate of the foot of the perpendicular from the point (- 10, – 2) on the line x + y – 2 = 0 is

x + 10 = y + 2 = \(\frac{14}{2}\)

x + 10 = y + 2 = 7

x + 10 = 7, y + 2 = 7

x = – 3, y = 5

∴ The required foot of the perpendicular is (- 3, 5).

Length of the perpendicular

Question 11.

If p_{1} and p_{2} are the lengths of the perpendiculars from the origin to the straight lines x 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}.

Answer:

Given P_{1} is the length of the perpendicular from the origin to the straight line

x sec θ + y cosec θ – 2a = 0

Also given P_{2} is the length of the perpendicular from the origin to the straight line

x cos θ – y sin θ – a cos 2θ = 0

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

Answer:

(i) 12x + 5y = 7 and 12x + 5y + 7 = 0

The equation of the given lines are

12x + 5y – 7 = 0 ……….. (1)

12x + 5y + 7 = 0 ……….. (2)

The distance between the parallel lines

ax + by + c_{1} = 0 and ax + by + c_{2} = 0 is

The equation of any line parallel to (1) is

The distance cannot be negative

∴ Required distance = \(\frac{14}{13}\)

(ii) 3x – 4y + 5 = 0 and 6x – 8y -15 = 0

The equation of the given lines are

3x – 4y + 5 = 0 ……….. (1)

6x – 8y – 15 = 0

3x – 4y – 1 = 0 ………… (2)

The distance between the parallel lines (1) and (2) is

Question 13.

Find the family of straight lines

(i) Perpendicular

(ii) Parallel to 3x + 4y – 12

Answer:

(i) Equation of lines perpendicular to 3x + 4y – 12 = 0 will be of the form 4x – 3y + k = 0, k ∈ R

(ii) 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 an anticlockwise direction through an angle of 15° , then find the equation of the line in the new position.

Answer:

Slope of the line AB

m = tan θ = \(\frac{1-0}{3-2}\)

tan θ = 1

θ = 45°

∴ The line AB makes an angle 45° with x-axis.

Given that the line AB is rotated through an angle of 15° about the point A in the anticlockwise direction.

∴ The angle made by the new line AB’ is 45° + 15° = 60°

Slope of the new line AB’ is m_{1} = tan 60° = √3

∴ The equation of the new line AB’ is the equation of the straight line passing through the point A (2, 0) and having slope m_{1} = √3

y – 0 = √3 (x – 2)

y = √3x – 2√3

√3x – y – 2√3 = 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 coordinates of point A.

Answer:

Let P(1, 2) and (5, 3) are the given points.

By the property of reflection,

∠XAB = ∠OAP = θ

(Angle of incidence = Angle of reflection)

Slope of the line OA (x – axis) m_{1} = 0

Slope of the line joining the points P (1, 2) and A (x, 0)

Slope of AP, m_{2} = \(\frac{2-0}{1-x}=\frac{2}{1-x}\)

Slope of the line joining the points B (5, 3 ) and A (x, 0)

Slope of AP, m_{3} = \(\frac{3-0}{5-x}=\frac{3}{5-x}\)

From equations (1) and (2)

\(\frac{3}{5-x}\) = –\(\frac{2}{1-x}\)

3(1 – x) = – 2 (5 – x)

3 – 3x = – 10 + 2x

2x + 3x = 10 + 3

5x = 13 ⇒ x = \(\frac{13}{5}\)

∴ The required point A is \(\left(\frac{13}{5}, 0\right)\)

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.

Answer:

Let the given line be PQ whose equation is

5x = y + 7

5x – y – 7 = 0 ——- (1)

Let AB be the line perpendicular to the line PQ such that the area of the triangle OAB is 10 units.

The equation of the line AB is

– x – 5y + k = 0

x + 5y – k = 0

x + 5y = k ——- (2)

∴ A is (k, 0) and B is (0, \(\frac{\mathrm{k}}{5}\))

OA = k and OB = \(\frac{\mathrm{k}}{5}\)

Area of ∆OAB = \(\frac{1}{2}\) × OA × OB

= \(\frac{1}{2}\) × k × \(\frac{\mathrm{k}}{5}\)

Given area of ∆ OAB = 10

∴ \(\frac{\mathrm{k}^{2}}{10}\) = 10

k^{2} = 100 ⇒ k = ±10

∴ The required equation of the straight line is

x + 5y = ±10

Question 17.

Find the image of the point (- 2, 3) about the line x + 2y – 9 = 0.

Answer:

The image of the point (x_{1}, y_{1}) about the line ax + by + c = 0 is

∴ The image of the point (- 2, 3) about the line x + 2y – 9 = 0 is

∴ The required point is (0, 7)

Question 18.

A photocopy store charges ₹ 1.50 per copy for the first 10 copies and ₹ 1.00 per copy after the 10^{th} copy. Let x be the number of copies, and y be the total cost of photo coping.

(i) Draw graph the cost as x goes from 0 to 50 copies

(ii) Find the cost of making 40 copies.

Answer:

(i) Draw graph of the cost as x goes from 0 to 50 copies:

Let x represent the number of copies and y represent the cost of photocopying.

Given photo copying charge for 1 copy is Rs. 1.50 for the first 10 copies and Rs. 1.00 per copy after the 10^{th} copy.

∴ The relation connecting the number of copies and cost of photocopying charge is given by

y = 1.50x, 0 ≤ x ≤ 10

y = 10(1.50) + (x – 10) (1)

y = 15 + x – 10

y = x + 5 ………… (1)

The graph for 0 to 50 copies:

When x = 10, y = 1.50 × x

⇒ y = 1.50 × 10 = 15

The corresponding point is (10 , 15)

When x = 20, y = x + 5

⇒ y = 20 + 5 = 25

The corresponding point is (20 , 25)

When x = 30, y = x + 5

⇒ y = 30 + 5 = 35

The corresponding point is (30, 35)

When x = 40, y = 40 + 5 = 45

The corresponding point is (40, 45 )

When x = 50, y = 50 + 5 = 55

The corresponding point is (50, 55 )

The cost of 40 copies is the value of y

When x = 40 , y = 40 + 5 = 45 rupees

Cost of 40 copies = 45 rupees

Question 19.

Find at least two equations of the straight lines in the family of file 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.

Answer:

The equations of the given straight lines are

y = 5x + b ……….. (1)

3x – 4y = 6 ……….. (2)

To find atleast two equations from the family y = 5x + b for which b is an integer and x – coordinate of the point of intersection of (1) and (2) is an integer. Solving (1)and (2) using equation (1) inequation (2) (2) ⇒ 3x – 4 (5x + b) = 6

3x – 20x – 4b = 6

-17x = 6 + 4b

The corresponding equation of the line is = 5x + 7

When b = – 10, we have x = \(\frac{6-40}{-17}\)

= \(\frac{-34}{-17}\) = 2

The corresponding equation of the line is y = 5x – 10

Thus y = 5x + 7 and y = 5x – 10 are the two straight lines belonging to the family such that b is an integer and the x – coordinate of the point of intersection with the line (2) is an integer.

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.

Answer:

The equations of the given lines are

y = mx – 3 ………. (1)

x – y = 6 ………. (2)

Solving equations (1) and (2)

(2) ⇒ x – (mx – 3 ) = 6

x – mx + 3 = 6

x (1 – m) = 3

x = \(\frac{3}{1-\mathrm{m}}\) …….. (3)

From equation (3) let us find the values of x and m for which they are integers. The only values of m for which , x is an integer are m = 0, 2, -2

When m = 0, x = \(\frac{3}{1-0}\) = 3

The corresponding equation is

y = 0 . x – 3

y + 3 = 0

When m = 2, x = \(\frac{3}{1-2}\) = \(\frac{3}{-1}\) = – 3

The corresponding equation is y = -2x + 3

2x + y – 3 = 0

When m = – 2, x = \(\frac{3}{1+2}\) = \(\frac{3}{3}\) = 1

The corresponding equation is

y = – 2 x + 3

2x + y – 3 = 0

∴ The required equations of the lines are

y + 3 = 0 , 2x – y – 3 = 0 and

2x + y – 3 = 0