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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:

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.

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:

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 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:

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 – 2)^{2} = 0

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

Solution:

(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 – 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} + 2 hxy + 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

Mid point 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 later 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:

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.

### 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} + 8ab xy + (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