Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4

Question 1.

Find the combined equation of the straight lines whose separate equations are x – 2y – 3 = 0 and x + y + 5 = 0.

Answer:

Separate equations are x – 2y – 3 = 0; x + y + 5 = 0

So the combined equation is (x – 2y – 3) (x + y + 5) = 0

x^{2} + xy + 5x – 2y^{2} – 2xy – 10y – 3x – 3y – 15 = 0

(i.e) x^{2} – 2y^{2} – xy + 2x – 13y – 15 = 0

Question 2.

Show that 4x^{2} + 4xy + y^{2} – 6x – 3y – 4 = 0 represents a pair of parallel lines.

Answer:

The equation of the given pair of straight lines is

4x^{2} + 4xy + y^{2} – 6x – 3y – 4 = 0 ………. (1)

Compare this equation with the equation

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 ………. (2)

a = 4, 2h = 4, b = 1, 2g = -6,

2f = -3, c = – 4

The condition for parallelism is

h^{2} – ab = 0

2^{2} – (4) (1) = 4 – 4 = 0

∴ The given pair of straight lines represents a pair of parallel straight lines.

Question 3.

Show that 2x^{2} + 3xy – 2y^{2} + 3x + y + 1 = 0 represents a pair of perpendicular straight lines.

Answer:

Comparing the given equation with the general form a = 2,h = 3/2, b = -2,g= 3/2, f = 1/2 and c = 1

Condition for two lines to be perpendicular is a + b = 0. Here a + b = 2 – 2 = 0

⇒ The given equation represents a pair of perpendicular lines.

Question 4.

Show that equation 2x^{2} – xy – 3y^{2} – 6x + 19y – 20 = 0 represents a pair of intersecting lines. Show further that the angle between them is tan^{-1}(5)

Answer:

The equation of the given pair of lines is

2x^{2} – xy – 3y^{2} – 6x + 19y – 20 = 0 —- (1)

Compare this equation with the equation

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 —- (2)

a = 2 , 2h = – 1, b = – 3, 2g = – 6, 2f = 19, c = – 20

h^{2} – ab = \(\left(-\frac{1}{2}\right)^{2}\) – (2)(-3)

= \(\frac{1}{4}\) + 6 ≠ 0

∴ The given line (1) is not parallel.

∴ They are intersecting lines.

Let θ be the angle between the lines.

Taking the acute angle θ = tan^{-1} (5)

Question 5.

Prove that the equation to the straight lines through the origin each of which makes an angle α with the straight line y = x is x^{2} – 2xy sec 2α + y^{2} = 0

Answer:

Let OP be the given line y = x having slope

m = 1 = tan 45°

Given that OA is the line making angle α with the line y = x.

Slope of the line OA = tan (45° – α)

The equation of OA is the equation of the line passing through the point (0, 0) having slope tan (45° – α).

y – 0 = tan(45°- α) (x – 0)

y = x tan (45°- α)

x tan (45° – α) – y = 0

Also given the line OB makes an angle α with the line y = x.

Slope of the line OB = tan(45° + α)

The equation of OB is the equation of the line passing through the point (0, 0) having slope tan(45° + α)

y – 0 = tan (45° + α) (x – 0)

y = x tan (45° + α)

x tan (45° + α) – y = 0 ………. (2)

The combined equation is

(x tan(45°- α) – y) (x tan(45°+ α) – y) = 0

x^{2}tan(45° – α) tan(45° + α) – xy tan (45° – α) – xy tan(45° + α) + yz = 0

Question 6.

Find the equation of the pair of straight lines passing through the poInt (1,3) and

perpendicular to the lines 2x -3y + 1 = 0 and 5x + y – 3 = 0.

Answer:

Equation of a line perpendicular to 2x – 3y + 1 = 0 is of the form 3x + 2y + k = 0.

It passes through (1, 3) ⇒ 3 + 6 + k = 0 ⇒ k = – 9

So the line is 3x + 2y – 9 = 0

The equation of a line perpendicular to 5x + y – 3 = 0 will be of the form x – 5y + k = 0.

It passes through (1, 3) ⇒ 1 – 15 + k = 0 ⇒ k = 14

So the line is x – 5y + 14 = 0.

The equation of the lines is 3x + 2y – 9 = 0 and x – 5y + 14 = 0

Their combined equation is (3x + 2y – 9)(x – 5y + 14) = 0

(i.e) 3x^{2} – 15xy + 42x + 2xy – 10y^{2} + 28y – 9x + 45y – 126 = 0

(i.e) 3x^{2} – 13xy – 10y^{2} + 33x + 73y – 126 = 0

Question 7.

Find the separate equation of the following pair of straight lines.

(i) 3x^{2} + 2xy – y^{2} = 0

(ii) 6(x – 1)^{2} + 5(x – 1) (y – 2) – 4(y – 3)^{2} = 0

(iii) 2x^{2} – xy – 3y^{2} – 6x + 19y – 20 = 0

Answer:

(i) Factorising 3x^{2} + 2xy – y^{2} we get

3x^{2} + 3xy – xy – y^{2} = 3x (x + y) – y (x + y)

= (3 x – y)(x + y)

So 3x^{2} + 2xy – y^{2} = 0 ⇒ (3x – y) (x + y) = 0

⇒ 3x – y = 0 and x + y = 0

(ii) 6(x – 1)^{2} + 5 ( x – 1) (y – 2) – 4(y – 3 )^{2} = 0

Let X = x – 1 and Y = y – 2

∴ The given equation becomes

6X^{2} + 5XY – 4Y^{2} = 0

6X^{2} + 8 XY – 3XY – 4Y^{2} = 0

2X(3X + 4Y) – Y (3X + 4Y) = 0

(2X – Y) (3X + 4Y) = 0

2X – Y = 0 and 3X + 4Y = 0

Substituting for X and Y , we have

2X – Y = 0 ⇒ 2(x – 1) – (y – 2) = 0

⇒ 2x – 2 – y + 2 = 0

⇒ 2x – y = 0

3X + 4Y = 0 ⇒ 3 (x – 1) + 4 ( y – 2 ) = 0

⇒ 3x – 3 + 4y – 8 = 0

⇒ 3x + 4y – 11 = 0

∴ The separate equations are

2x – y = 0 and 3x + 4y – 11 = 0

(iii) 2x^{2} – xy – 3y^{2} – 6x + 19y – 20 = 0

The given equation is

2x^{2} – xy – 3y^{2} – 6x + 19y – 20 = 0 ……… (1)

2x^{2} – xy – 3y^{2} = 2x^{2} – 3xy + 2xy – 3y^{2}

= x (2x – 3y) + y (2x – 3y )

= (x + y) (2x – 3y)

Let the separate equation of the straight lines be

x + y + 1 = 0 and 2x – 3y + m = 0 …….. (A)

(1) ⇒ 2x^{2} – xy – 3y^{2} – 6x + 19y – 20 = (x + y + 1) (2x – 3y + m)

Equating the coeffident of x , y and constant term on both sides, we have

-6 = m + 2l ………. (2)

19 = m – 3l ………. (3)

lm = – 20 ………. (4)

Solving equations (2), (3) and (4) we have

Substituting the values of l and m in equation (A) the required separate equations of the lines are

x + y – 5 = 0 and 2x – 3y + 4 = 0

Question 8.

The slope of one of the straight lines ax^{2} + 2hxy + by^{2} = 0 is twice that of the other, show that 8h^{2} = 9ab.

Answer:

The equation of the given straight line is

ax^{2} + 2hxy + by^{2} = 0 ………… (1)

Given that the slopes of the straight lines are m and 2m

Question 9.

The slope of one of the straight lines ax^{2} + 2h xy + by^{2} = 0 is three times the other, show that 3h^{2} = 4ab.

Answer:

The equation of the given straight line is

ax^{2} + 2hxy + by^{2} = 0 ……….. (1)

Given that the slopes of the lines are m and 3m.

Question 10.

A ∆ OPQ is formed by the pair of straight lines x^{2} – 4xy + y^{2} = 0 and the line PQ. The equation of PQ is x + y – 2 = 0, Find the equation of the median of the triangle ∆ OPQ drawn from the origin O

Answer:

The equation of the given pair of lines is

x^{2} – 4xy + y^{2} = 0 ……….. (1)

The equation of the line PQ is

x + y – 2 = 0

y = 2 – x ……….. (2)

To find the coordinates of P and Q, .

Solve equations (1) and (2)

(1) ⇒ x^{2} – 4x ( 2 – x) + ( 2 – x)^{2} = 0

x^{2} – 8x + 4x^{2} + 4 – 4x + x^{2} = 0

6x^{2} – 12x + 4 = 0

3x^{2} – 6x + 2 = 0

The midpoint of PQ is

The equation of the median drawn from 0 is the equation of the line joining 0 (0, 0) and D (1, 1)

∴ The required equation is x = y

Question 11.

Find p and q, if the following equation represents a pair of perpendicular lines

6x^{2} + 5xy – py^{2} + 7x + qy – 5 = 0

Answer:

The equation of the given pair of straight lines is

6x^{2} + 5xy – py^{2} + 7x + qy – 5 = 0 ……….. (1)

Given that equation (1) represents a pair of perpendicular straight lines.

∴ Coefficient of x^{2} + coefficient of y^{2} = 0

6 – p = 0 ⇒ p = 6

6x^{2} + 5xy – 6y^{2} = 6x^{2} + 9xy – 4xy – 6y^{2}

= 3x(2x + 3y) – 2y (2x + 3y)

= (2x + 3y) (3x – 2y)

Let the separate equation of the straight lines be

2x + 3y + 1 = 0 and 3x – 2y + m = 0

6x^{2} + 5xy – 6y^{2} + 7x + qy – 5

= (2x + 3y + 1)(3x – 2y + m)

Comparing the coefficients of x , y and constant terms on both sides

2m + 3l = 7 ………… (2)

3m – 2l = q ……….. (3)

lm = – 5 …….. (4)

Equation (4) ⇒ l = 1, m = – 5

or l = – 1, m = 5

When l = 1, m = – 5 , equation (2) does not satisfy.

∴ l = – 1 , m = 5

Substituting in equation (3)

3 (5) – 2(-1) = q ⇒ q = 17

∴ The required values are p = 6, q = 17

Question 12.

Find the value of k, if the following equation represents a pair of straight lines. Further, find whether these lines are parallel or intersecting

12x^{2} + 7xy – 12y^{2} – x +7y + k = 0

Answer:

The equation of the given pair of straight lines is

12x^{2} + 7xy – 12y^{2} – x + 7y + k = 0 ………. (1)

Compare this equation with the equation

ax^{2} + 2hxy + by^{2} + 2gx + 2f y + c = 0 ……….. (2)

a = 12, 2h = 7 , b = – 12 ,

2g = – 1, 2f = 7, k = c

a = 12, h = \(\frac{7}{2}\), b = – 12,

g = \(-\frac{1}{2}\), f = \(\frac{7}{2}\), c = k

The condition for a second degree equation in x and y to represent a pair of straight lines is

abc + 2fgh – af^{2} – bg^{2} – ch^{2} = 0

Substituting the values

Coefficient of x^{2} + coefficient of y^{2} = 12 – 12 = 0

∴ The given pair of straight lines are perpendicular and hence they are intersecting lines.

Question 13.

For what values of k does the equation

12x^{2} + 2kxy + 2y^{2} +11x – 5y + 2 = 0 represent two straight lines.

Answer:

The given equation of the pair of straight line is

12x^{2} + 2kxy + 2y^{2} + 11x – 5y + 2 = 0 ……… (1)

Compare this equation with the equation

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 ………. (2)

a = 12, 2h = 2k, b = 2,

2g = 11, 2f = – 5 , c = 2 ,

a =12, h = k, b = 2,

g = \(\frac{11}{2}\), f = –\(\frac{5}{2}\), c = 2,

The condition for a second degree equation in x and y to represent a pair of straight lines is

abc + 2fgh – af^{2} – bg^{2} – ch^{2} = 0

96 – 55k – 150 – 121 – 4k^{2} = o

– 4k^{2} – 55k – 175 = 0

4k^{2} + 55k + 175 = 0

∴ The given equation represents a pair of straight lines when

k = – 5 and k = \(\frac{-35}{4}\)

Question 14.

Show that the equation 9x^{2} – 24xy + 16y^{2} – 12x + 16y – 12 = 0 represents a pair of parallel lines. Find the distance between them.

Answer:

The given equation of the pair of straight line is

9x^{2} – 24xy + by^{2} – 12x + 16y – 12 = 0 ………… (1)

9x^{2} – 24xy + 16y^{2} = 9x^{2} – 12xy – 12xy + 16y^{2}

= 3x (3x – 4y) – 4y (3x – 4y)

= (3x – 4y) (3x – 4y)

Let the separate equation of the straight lines be

3x – 4y + 1 = 0 and 3x – 4y + m = 0

9x^{2} – 24xy + 16y^{2} – 12x + 16y – 12

= (3x – 4y + l) ( 3x – 4y + m )

Comparing the coefficients of x , y and constant terms on both sides

3l + 3m = – 12

l + m = – 4 ……….. (2)

– 4l – 4m = 16

l + m = – 4 ………… (3)

lm = – 12 ……….. (4)

(l – m)^{2} = (l + m)^{2} – 4lm

= (- 4)^{2} – 4 × – 12

= 16 + 48 = 64

l – m = √64 = 8

l – m = 8 ………… (5)

Solving equations (2) and (5 ), we have

(2) ⇒ 2 + m = – 4 ⇒ m = – 6

∴ l = 2 and m = – 6

∴ The separate equation of the straight lines are

3x – 4y – 6 = 0 and 3x – 4y + 2 = 0

The distance between the parallel lines is given by

∴ The given pair of straight lines are parallel and the distance between them is \(\frac{8}{5}\) units

Question 15.

Show that the equation 4x^{2} + 4xy + y^{2} – 6x – 3y – 4 = 0 represents a pair of parallel lines. Find the distance between them.

Answer:

The given equation of pair of straight lines is

4x^{2} + 4xy + y^{2} – 6x – 3y – 4 = 0 ………. (1)

4x^{2} + 4xy + y^{2} = (2x + y)^{2}

Let the separate equation of the lines be

2x + y + l = 0 ……….. (2)

2x + y + m = 0 ………. (3)

4x^{2} + 4xy + y^{2} – 6x – 3y – 4 = (2x + y + l) (2x + y + m)

Comparing the coefficients of x , y and constant terms on both sides we have

2l + 2m = – 6

l + m = – 3 ……… (4)

l + m = – 3 ……… (5)

l m = – 4 ……… (6)

(l – m)^{2} = (l + m)^{2} – 4lm

(l – m )^{2} = (- 3)^{2} – 4 × – 4

(l – m)^{2} = 9 + 16 = 25

l – m = 5 ………… (7)

Solving equations (4) and (7)

(4) ⇒ l + m = – 3 ⇒ m = – 4

∴ The separate equation of the straight lines are

2x + y + 1 =0 and 2x + y – 4 = 0

The distance between the parallel lines is

d = \(\frac{5}{\sqrt{5}}\) = \(\sqrt{5}\)

∴ The given equation represents a pair of parallel straight lines and the distance between the parallel lines is \(\sqrt{5}\) units.

Question 16.

Prove that one of the straight lines given by ax^{2} + 2hxy + by^{2} = 0 will bisect the angle between the coordinate axes if (a + b)^{2} = 4h^{2}

Answer:

The equation of the given pair of straight lines is

ax^{2} + 2hxy + by^{2} = 0 ……… (1)

Let m_{1} and m_{2}, be the slopes of the separate straight lines.

Given that one of the straight lines of (1) bisects the angle between the coordinate axes.

∴ The angle made by that line with x-axis 45°.

Slope of that line m_{1} = tan 45°

m_{1} = 1

Squaring on both sides

(a + b)^{2} = (- 2h)^{2}

(a + b)^{2} = 4h^{2}

Question 17.

If the pair of straight lines x^{2} – 2kxy – y^{2} = 0 bisects the angle between the pair of straight lines x^{2} – 2lxy – y^{2} = 0. Show that the later pair also bisects the angle between the former.

Answer:

The equations of the given pair of straight lines are

x^{2} – 2kxy – y^{2} = 0 ………… (1)

x^{2} – 2lxy – y^{2} = 0 ………… (2)

Given that the pair x^{2} – 2kxy – y^{2} = 0 bisects the angle between the pair x^{2} – 2lxy – y^{2} = 0

∴ The equation of the bisector of the pair

x^{2} – 2lxy – y^{2} = 0 is the pair x^{2} – 2kxy – y^{2} = 0

The equation of the bisector of x^{2} – 2lxy – y^{2} = 0 is

Equation (3) and Equation (1) represents the same straight lines. ∴ The coefficients are proportional.

To show that the pair x^{2} – 2lxy – y^{2} = 0 bisects the angle between the pair x^{2} – 2kxy – y^{2} = 0, it is enough to prove the equation of the bisector of x^{2} – 2kxy – y^{2} = 0 is x^{2} – 2lxy – y^{2} = 0

The equation of the bisector of x^{2} – 2kxy – y^{2} = 0 is

x^{2} – y^{2} = 2lxy .

x^{2} – 2lxy – y^{2} = 0

∴ The pair x^{2} – 2lxy – y^{2} = 0 bisects the angle between the pair x^{2} – 2kxy – y^{2} = 0

Question 18.

Prove that the straight lines joining the origin to the points of intersection of 3x^{2} + 5xy – 3y^{2} + 2x + 3y = 0 and 3x – 2y – 1 = 0 are at right angle.

Answer:

Homogenizing the given equations 3x^{2} + 5xy – 3y^{2} + 2x + 3y = 0 and 3x – 2y – 1 = 0

(i.e) 3x – 2y = 1.

We get (3x^{2} + 5xy – 3y^{2}) + (2x + 3y)( 1) = 0

(i.e) (3x^{2} + 5xy – 3y^{2}) + (2x + 3y)(3x – 2y) = 0

3x^{2} + 5xy – 3y^{2} + bx^{2} – 4xy + 9xy – 6y^{2} = 0

9x^{2} + 10xy – 9y^{2} = 0

Coefficient of x^{2} + coefficient of y^{2} = 9 – 9 = 0

⇒ The pair of straight lines are at right angles.