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## 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.

Solution:

The given separate equations of the lines are

x – 2y – 3 = 0 and x + y + 5 = 0

∴ The combined equation of the straight lines is

(x – 2y – 3 ) (x + y + 5) = 0

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

x^{2} – xy – 2y^{2} + 2x – 13y – 15 = 0

Question 2.

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

Solution:

Comparing this equation with ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0

we get a = 4, h = \(\frac{4}{2}\) = 2 , b = 1, g = – 3, f = – 3/2, c = – 4

The condition for the lines to be parallel is h^{2} – ab = 0

Now h^{2} – ab = 2^{2} – (4) (1) = 4 – 4 = 0

h^{2} – ab = 0 ⇒ The given equation represents a pair of parallel lines.

Question 3.

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

Solution:

The equation of the given pair of straight lines is

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

Compare this equation with the equation

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

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

The condition for pair of straight lines to be perpendicular is a + b = 0.

2 – 2 = 0

Hence the given pair of lines represents a perpendicular straight lines.

Question 4.

Show that the 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).

Solution:

The given equation represents a pair of straight lines.

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

Solution:

Slope of y = x is m = tan θ = 1

⇒ θ = 45°

The new lines slopes will be

m = tan(45 + α) and m = tan (45 – α)

∴ The equations of the lines passing through the origin is given by

y = tan(45 + α)x and y = tan(45 – α)x

(i.e) y = tan(45 + α)x = 0 and y = tan(45 – α)x = 0

The combined equation is [y – tan (45 + α)x] [y – tan (45 – α)x] = 0

y^{2} + tan(45 + α)tan(45 – α)x^{2} – xy[tan(45 – α) + tan(45 + α)] = 0

Let the equation of lines passes through the origin

So the equations are y = m_{1}x = 0 and y = m_{2}x = 0

So the combined equations is (y – m_{1}x) (y – m_{2}x) = 0

(i.e)y^{2 }– xy(m_{1} + m_{2}) + m_{1}m_{2}x = 0

(i.e) y^{2 }– xy(2sec α) + x^{2}(1) = 0

(i.e) y^{2} – 2xy sec 2α + x^{2} = 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

Solution:

The equation of the given lines are

2x – 3y + 1 = 0 ……….. (1)

5x + y – 3 = 0 ……….. (2)

Equation of any line perpendicular to 2x – 3y + 1 = 0 is

– 3x – 2y + k = 0

3x + 2y – k = 0

This line passes through the point (1, 3)

∴ 3(1) + 2(3) – k = 0

3 + 6 – k = 0 ⇒ k= 9

Substituting the value of k in the above equation we have

3x + 2y – 9 = 0 ………. (3)

Equation of any line perpendicular to 5x + y – 3 = 0 is

x – 5y + k_{1} = 0

This line passes througi the point (1 , 3)

∴ 1 – 5 (3) + k_{1} = 0

1 – 15 + k_{1} ⇒ k_{1} = 14

Substituting the value of k_{1} in the above equation we have

x – 5y + 14 = 0 ……….. (4)

The combined equation of (3) and (4) is

( 3x + 2y – 9) (x – 5y + 14 ) = 0

3x^{2} – 15xy + 42x + 2xy – 10y^{2} + 28y – 9x + 45 y – 126 = 0

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 – 2)^{2} = 0

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

Solution:

(i) 3x^{2} + 2xy – y^{2} = 0

The given equation is

3x^{2} + 2xy – y^{2} = 0 ……. (1)

3x^{2} + 3xy – xy – y^{2} = 0

3x (x + y) – y (x + y) = 0

(3x – y) (x + y) = 0

3x – y = 0 and x + y = 0

∴ The separate equations are

3x – y = 0 and x + y = 0

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

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

(i.e) 6x^{2} – 12x + 6 + 5xy – 10x – 5y + 10 – 4y^{2} + 16y – 16 = 0

(i.e) 6x^{2} + 5xy – 4y^{2} – 22x + 11y = 0

Factorising 6x^{2} + 5xy – 4y^{2} we get

6x^{2} – 3xy + 8xy – 4y^{2} = 3x (2x – y) + 4y (2x – y)

= (3x + 4y)(2x – y)

So, 6x^{2} + 5xy – 4y^{2} – 22x + 11y = (3x + 4y + l )(2x – y + m)

Equating coefficient of x ⇒ 3m + 21 = -22 …….. (1)

Equating coefficient of y ⇒ 4m – l = 11 ……. (2)

Solving (1) and (2) we get l = -11, m = 0

So the separate equations are 3x + 4y – 11 = 0 and 2x – y = 0

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

Factorising 2x^{2} – xy – 3y^{2} we get

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

= 2x(x + y) – 3y(x + y) = (2x – 3y) (x + y)

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

Equating coefficient of x 2m + l = -6 ……. (1)

Equating coefficient of y -3m + l = 19 …….. (2)

Constant term -20 = lm

Solving (1) and (2) we get l = 4 and m = – 5 where lm = – 20.

So the separate equations are 2x – 3y + 4 = 0 and x + y – 5 = 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.

Solution:

ax^{2} + 2hxy + by^{2} = 0

We are given that one slope is twice that of the other.

So let the slopes be m and 2m.

Now sum of the slopes = m + 2m

Question 9.

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

Solution:

Let the slopes be 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.

Solution:

Equation of pair of straight lines is x^{2} – 4xy + y^{2} = 0 ….. (1)

Equation of the given line is x + y – 2 = 0 ⇒ y = 2 – x ……… (2)

On solving (1) and (2) we get x^{2} – 4x (2 – x) + (2 – x)^{2} = 0

(i.e) x^{2} – 8x + 4x^{2} + 4 + x^{2} – 4x = 0

(i.e) 6x^{2} – 12x + 4 = 0

(÷ by 2) 3x^{2} – 6x + 2 = 0

The midpoint of PQ is

Question 11.

Find p and q, ¡f the following equation represents a pair of perpendicular lines 6x^{2} + 5xy – py^{2} + 7x + qy – 50

Solution:

6x^{2} + 5xy – py^{2} + 7x + qy – 50

The given equation represents a pair of perpendicular lines

⇒ coefficient of x^{2} + coefficient of y^{2} = 0

(i.e) 6 – p = 0 ⇒ p = 6

Now comparing the given equation with the general form

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0

we get a = 6, b = -6 and c = -5, f = q/2, g = 7/2 and h = 5/2

The condition for the general form to represent a pair of straight lines is abc + 2fgh – af^{2} – bg^{2} – ch^{2} = 0

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.

Solution:

Comparing the given equation with the general form ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0

we get a = 12, b = -12, c = k, f = 7/2, g = – 1/2, h = 7/2

Here a + b = 0 ⇒ the given equation represents a pair of perpendicular lines

To find k: The condition for the given equation to represent a pair of straight lines is abc + 2fgh – af^{2} – bg^{2} – ch^{2} = 0

Question 13.

For what value of k does the equation 12x^{2} + 2kxy + 2y^{2}, + 11x – 5y + 2 = 0 represent two straight lines.

Solution:

12x^{2} + 2 kxy + 2y^{2} + 11x – 5y + 2 = 0

Comparing this equation with the general form we get

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

4k^{2} + 20k + 35k + 175 = 0

4k(k + 5) + 35(k + 5) = 0

(4k + 35) (k + 5) = 0

k = -5 or -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.

Solution:

Comparing the given equation with ax^{2} + 2kxy + by^{2} = 0 we get a = 9, h = -12, b = 16.

Now h^{2} = (-12)^{2} = 144, ab = (9) (16) = 144

h^{2} = ab ⇒ The given equation represents a pair of parallel lines.

To find their separate equations:

9x^{2} – 24xy + 16y^{2} = (3x – 4y)^{2}

So, 9x^{2} – 24xy +16y^{2} – 12x + 16y – 12 = (3x – 4y + l )(3x – 4y + m)

Here coefficient of x ⇒ 3m + 3l = -12 ⇒ m + l = -4

coefficient of y ⇒ -4m – 4l = 16 ⇒ m + l = -4

Constant term l m = -12

Now l + m = -4 and lm = -12 ⇒ l = -6 and m = 2

So the separate equations are 3x – 4y – 6 = 0 and 3x – 4y + 2 = 0

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.

Solution:

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

a = 4,

b = 1,

h = 4/2 = 2

h^{2} – ab = 2^{2} – (4) (1) = 4 – 4 = 0

⇒ The given equation represents a pair of parallel lines.

To find the separate equations 4x^{2} + 4xy + y^{2} = (2x + y)^{2}

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

Coefficient of x ⇒ 2m + 2l = -6 ⇒ l + m = – 3 ……. (1)

Coefficient of y ⇒ l + m = – 3 ……… (2)

Constant term ⇒ l m = – 4 ……… (3)

Now l + m = -3 and lm = -4 ⇒ l = -4, m = 1

So the separate equations are 2x + y + 1 = 0 and 2x + y – 4 = 0

Question 16.

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

Solution:

Let the slopes be l and m

∵ One line bisects the angle between the coordinate axes ⇒ θ = 45°

So tan θ = 1

The slopes are l and m

Question 17.

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

Solution:

Given that x^{2} – 2kxy – y^{2} = 0 …….. (1)

Bisect the angle between the lines x^{2} – 11xy – y^{2} = 0 …… (2)

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 angles.

Solution:

The equation of the pair of straight lines is

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

The given line is 3x – 2y – 1 = 0

3x – 2y = 1 ……….. (2)

The equation of the straight lines joining the origin to the points of intersection of the pair of lines (1) and the line (2) is obtained by homogeneous using equation (1) by using equation (2)

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

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

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

9x^{2} + 10xy – 9y^{2} = 0 ………. (3)

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

∴ The pair of straight line (3) represents a perpendicular straight lines.

### Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 Additional Questions Solved

Question 1.

Find the angle between the pair of straight lines given by (a^{2} – 3b^{2})x^{2} + 8abxy + (b^{2} – 3a^{2})y^{2} = 0.

Solution:

Question 2.

Show that 9x^{2} + 24xy + 16y^{2} + 21x + 28y + 6 = 0 represents a pair of parallel straight lines and find the distance between them.

Solution:

9x^{2} + 24xy + 16y^{2} + 21x + 28y + 6 = 0

Here a = 9.6,

b = 16,

g = \(\frac{21}{2}\),

f = 14,

c = 6,

h = 12

h^{2} – ab = (12)^{2} – 9(16) = 144 – 144 = 0

∴ The lines are parallel.

9x^{2} + 24xy + 16y^{2} = (3x + 4y)(3x + 4y)

Let 9x^{2} + 24xy + 16y^{2} + 21x + 28y + 6 = (3x + 4y + l)(3x + 4y + m)

Equating the coefficients of x and constant term

3l + 3m = 21

lm = 6

Solving we get, l = 1 or 6

m = 6 or 1

∴ The separate equations are 3x + 4y + 1 = 0 and 3x + 4y + 6 = 0

Question 3.

If the equation 12x^{2} – 10xy + 2y^{2} + 14x – 5y + c = 0 represents a pair of straight lines, find the value of c. Find the separate equations of the straight lines and also the angle

between them.

Solution:

12x^{2} – 10xy – 2y^{2} + 14x – 5y + c = 0

ax^{2} + 2hxy + by^{2} +2gx + 2fy – c = 0

Here a = 12,

b = 2,

g = 7,

f = 5/2,

c = c,

h = -5

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

The equation is 12x^{2} – 10y + 2y^{2} + 14x – 5y + 2 = 0

12x^{2} – 10xy + 2y = (3x – y)(4x – 2y)

Let 12x^{2} – 10y + 2y^{2} + 14x – 5y + 2(3x – y + l)(4x – 2y + m)

So that 4l + 3m = 14 , -2l – m = -5

Question 4.

For what value of k does 12x^{2} + 7xy + ky^{2} + 13x – y + 3 = 0 represents a pair of straight lines? Also write the separate equations.

Solution:

12x^{2} + 7xy + ky^{2} + 13x – y + 3 = 0

a = 12,

h = \(\frac{7}{2}\),

f = \(-\frac{1}{2}\) ,

c = 3

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

⇒ 12 + 169k + 147 – 144k + 91 = 0

25k = – 250 ⇒ k = -10

The equation is 12x^{2} + 7xy – 10y^{2} + 13x – y + 3 = 0

To find separate equations: 12x^{2} + 7xy – 10y^{2} = (3x – 2y)(4x + 5y)

Let 12x^{2} + 7xy – 10y^{2} + 13x – y + 3 = 0(3x – 2y + l)(4x + 5y + m)

Equating the coefficient of x ⇒ 4l + 3m = 13 …… (1)

Equating the coefficient of y ⇒ 5l – 2m = -1 …… (2)

(1) × 2 ⇒ 8l + 6m = 26

(2) × 3 ⇒ 15l – 6m = -3

23l = 23 ⇒ l = 1

4 + 3 m = 13

3 m = 9 ⇒ m = 3

The separate equations are 3x – 2y + 1 = 0 and 4x + 5y + 3 = 0

Question 5.

Show that 3x^{2} + 10xy + 8y^{2} + 14x + 22y + 15 = 0 represents a pair of straight lines and the angle between them is tan^{-1}\(\left(\frac{2}{11}\right)\)

Solution:

3x^{2} + 10xy + 8y^{2} + 14x + 22y + 15 = 0

a = 3,

A = 5,

b = 8,

g = 7,

f = 11,

c = 15

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

3(11)^{2} + 8(7)^{2} + 15 (5)^{2} – (3)(8)(15) – 2(11)(7)(5) = 363 + 392 + 375 – 360 – 770 = 0

The angle between the pair of straight line is given by