ISC – Maths – XII – Important Q.A.

Pair of Straight Lines.

Q.1. Prove that the following equation represents a pair of straight lines. Find their point of intersection and the angle between them:

2x² + 7xy + 3y² + 2 (4x + 7y + 4) = 0.

Solution : – The equation can be written as :

2x² + 7xy + 3y² + 8x + 14y + 8 = 0.

Then we have, a = 2, b = 3, h = 7/2, g = 4, f = 7, c = 8.

Now,        a b c + 2fgh – af² – bg² – ch²

= (2)(3)(8) + 2(7)(4)(7/2) – (2)(7)² – (3)(4)² – (8)(7/2)²

= 48 + 196 – 98 – 48 – 14 = 244 – 160 = 84.

The given equation does not represent a pair of straight line.

Q.2. Find the value of ‘p’ for which the following equation represents a pair of straight lines :

x² + p x y – 2y² + 3gx + 3fy = 0.

Solution : – We have the equation of straight line :

x² + p x y – 2y² + 3gx + 3fy = 0,

Here, a = 1, b = – 2, h = p/2, g = 3g/2, f = 3f/2, c = 0.

The given equation represents a pair of straight lines,

Hence,                   a b c + 2fgh – af² – bg² – ch² = 0

Or,            (1) (– 2)( 0) + 2(3f/2)(3g/2)(p/2) – (1)(3f/2)² – (– 2)(3g/2)² – (0)(p/2)² = 0

Or,            9pfg/4 – 9f²/4 + 9g²/2 = 0

Or,             pfg – f² + 2g² = 0

Or,             pfg = f² – 2g²

Or,             p = (f² – 2g²)/fg.  [Ans.]

Q.3. Find the equation of a pair of straight lines represented by

8x² + 8xy + 2y² + 26x + 13y + 5 = 0

and show that the angle between them is 0º.

Solution : – We have the equation of straight line :

8x² + 8xy + 2y² + 26x + 13y + 5 = 0

Here, a =  8, b = 2, h = 4, g = 13, f = 13/2, c = 5 and

abc +2fgh – af² – bg² – ch²

= (8)(2)(5) + 2(13/2)(13)(4) – (8)(13/2)² – (2)(13)² – (5)(4)²

= 80 + 676 – 338 – 338 – 80

= 0.

Hence, given equation represents a pair of straight lines.

Now,      8x² + 8xy + 2y² + 26x + 13y + 5 = 0

Or,          8x² + 2x(4y + 13) + (2y² + 13y + 5) = 0

Or,          x = [– 2(4y + 13) ± √{4 (4y + 13)² – 4 × 8 (2y² + 13y + 5)}]/(2 × 8)

= [– 8y – 26 ± √(64y² + 676 + 416y – 64y² – 416y – 160)]/16

= {– 8y – 26 ±√(516)}/16

Or,          16x = – 8y – 26 ± √(516)

Hence, lines are 16x + 8y + 26 + √(516) = 0 and 16x + 8y + 26 – √(516) = 0.

Angle between the lines is given by

tan θ = {2√(h² – ab)}/(a + b)

= [2√{(4)² – 8 ×2}]/(8 + 2) = 0

Therefore,                              θ = 0º [Ans.]

Q.4. Find the angle between the lines represented by the equation :

3x² – y² – √3 x + 3y – 2 = 0

Do yourself.                           [Ans. = 60º]

Q.5. Find the angle between the straight lines represented by the equation :

x² + 2√2 xy + y² + 4x + 2√2 y + 2 = 0.

Do yourself.                          [Ans. = 45º, 135º]

Q.6. Find the equation of the bisectors of the angle between the pair of lines given by:

x² – xy sec² θ – y² tan² θ = 0.

Solution : – We have, x² – xy sec² θ – y² tan² θ = 0

Here,        a = 1, b = – tan² θ, h = – sec² θ.

Therefore, equation of bisectors is:

(x² – y²)/(1 + tan² θ) = xy /(– sec² θ/2)               [(x² – y²)/(a – b) = xy/h]

Or,       (x² – y²)/sec² θ = – 2xy/sec² θ

Or,        x² – y² = – 2xy

Or,        x² + 2xy – y² = 0.  [Ans.]

Q.7. A pair of straight lines x² – 2pxy – y² = 0 and x² – 2qxy – y² = 0 be such that each pair bisects the angle between the other pair, prove that p q = – 1.

Solution : – The first pair of lines is : x² – 2pxy – y² = 0 ,

Here ,  a= 1, b = – 1, h = – p.

Equation of bisectors is :

(x² – y²)/(1 – (– 1)) = xy /– p      [(x² – y²)/(a – b) = xy/h]

Or,                – px² – 2xy + py² = 0        ——————— (1)

As per question, (1) is identical with x² – 2qxy – y² = 0

Thus comparing the coefficients we get,

–        p /1 = – 2 /– 2q

Or,                                        – 2pq = 2

Or,                                             pq = –1.    [Proved.]

Q.8. Show that the second degree equation x² – 5xy + 4y² + x + 2y – 2 = 0 represents a pair of straight lines. Find the equation of the individual lines and their point of intersection.

Solution : – We have x² – 5xy + 4y² + x + 2y – 2 = 0 ———————- (1)

Comparing it with  ax² + 2hxy + by² + 2gx + 2fy +c = 0 ,

We get,  a = 1 , b = 4, c = – 2, h = – 5/2, g = 1/2 , f = 1.

And  abc + 2fgh – af² – bg² – ch² = –8 + 2 × 1 × 1/2 × (–5/2) –1(1)² –4(1/2)² + 2(–5/2)²

= – 8 – 5/2 – 1 – 1 + 25/2 = 0

Hence the given equation represents a pair of straight lines.      [Proved.]

Writing (1) as a function of x , we get

x² + x(–5 y + 1) + (4y2 + 2y – 2) = 0

Or,                   x = [– (–5y + 1 ) ± √{(–5y + 1)2 – 4 (4y2 + 2y – 2 )}]/2

= [5y – 1 ± √{25y2 + 1 – 10y – 16 y2 – 8y +8}]/2

= [5y – 1 ± √{9y2 – 18y + 9}]/2

= [5y – 1 ± 3 (y – 1)]/2

Or,          2x = 5y – 1 ± (3y – 3)

Or,          2x – 8y + 4 = 0     &       2x – 2y – 2 = 0      [Ans.]

Q.9. Find the two straight lines represented by the equation:

x² – y² – 4x + 4 = 0

and hence or otherwise, find the angle between them.

Do yourself.                               [Ans. x + y = 2; x – y = 2; 90º]

Q.10. Find the value of ‘k’ so that the second degree equation  12x² – 10xy + ky² + 14x – 5y + 2 = 0 may represent a pair of straight lines. For this value of ‘k’ find the angle between the lines represented by the above equation.

Solution : – We have ,     12x² – 10xy + ky² + 14x – 5y + 2 = 0 ,

Comparing with ax² + 2hxy + by² + 2gx + 2fy + c = 0 , we get

a = 12, h = – 5, b = k, g = 7, f = –5/2, c = 2 .

The condition is                            abc + 2fgh – af² – bg² – ch² = 0

Or,                              12×k×2 + 2×(– 5/2)×7×(– 5) – 12 (– 5/2)² – k(7)² – 2 (– 5)² = 0

Or,                                           k = 2 . [Ans.]

Q.11. Find the value of λso that the equation :

x² + 4xy + 4y² + λx + 10y + 4 = 0

may represent a pair of straight lines and find both the lines separately.

Do yourself.                                  [Ans. 5; x + 2y + 1 = 0; x + 2y + 4 = 0]

Q.12. Show that the bisectors of the angles between the pair of straight lines 11x² – 16xy – y² = 0 are parallel and perpendicular to x + 2y + 5 = 0

Solution : – Equation of bisector is given by :

(x² – y²)(a – b) = xy/h    —————- (1)

Hence,  a = 11, h = – 8 , b = – 1 .

Putting in (1) we get (x² – y²)(11 + 1) xy /– 8

Or,                                8x² + 12xy – 8y² = 0

Or,                                 2x² + 3xy – 2y² = 0

Or,                             (2x – y)(x + 2y) = 0

Or,                               2x – y = 0  and x + 2y = 0

Slopes are 2 and – 1/2 .

Slope of x + 2y + 5 = 0 is – 1/2 .

Therefore , 2x – y is perpendicular to x + 2y + 5 = 0

and x + 2y = 0 is parallel to x + 2y + 5 = 0 . [Proved.]

Q.13. Show that the equation x² – 3xy + 2y² + 3x – 5y + 2 = 0 represents two straight lines. Find the angle between them.

Solution : –

We have , x² – 3xy + 2y² + 3x – 5y + 2 = 0 ,

Comparing with ax² + 2hxy + by² + 2gx + 2fy + c = 0 , we get

a = 1, h = – 3/2 , b = 2, g = 3/2, f = – 5/2 , c = 2.

Then,  abc + 2fgh – af² – bg² – ch²

= 1×2×2 + 2×(– 5/2)×(3/2)×(– 3/2) – 1 (– 5/2)² – 2(3/2)² – 2(– 3/2)²

= 4 + 45/4 – 25/4 – 9/2 – 9/2

= (16 + 45 – 25 – 18 – 18)/4 = 0 .

Therefore, given equation represents pair of straight lines.

Let angle between the lines be θ then tan θ = [2√(h² – ab)]/(a + b)

Or,                                                       tan θ = [2√{(– 3/2)² – 1×2}]/(1 + 2)

= [2√{(9/4) – 2 }]/3

= [2×(1/2)]/3 = 1/3 .

Therefore,                                            θ = tan ^–1 (1/3) . [Ans.]

Q.14. Find the equation of the bisectors of the angle between the pair of lines given by

(4 ax + 3 by)² = 3 (3 b x – 4 ay)².

Solution : –

We have ,  (4ax + 3by)² = 3 (3bx – 4ay)²

Or,            16a²x² + 9b²y² + 24abxy = 3 (9b²x² + 16a²y² – 24abxy)

Or,             16a²x² + 9b²y² + 24abxy – 27b²x² – 48a²y² + 72abxy = 0

Or,              (16a² – 27b²)x² + 96abxy + (9b² – 48a²)y² = 0 ————— (1)

Therefore, Equation of bisector of (1) is

(x² – y²)/[(16a² – 27b²) – (9b² – 48a²)] = xy/48ab [Using : (x² – y²)/(a – b) = xy/h]

Or,                         (x² – y²)/(64a² – 36b²) = xy/48ab

Or,                      12ab(x² – y²) = (16a² – 9b²)x y .  [Ans.]

Q.15. The equation 2x² – 3xy – py² + x + q y = 0 represents two perpendicular lines. Find the value of p and q.

Solution : –

Comparing  2x² – 3xy – py² + x + q y = 0 , with

ax² + 2hxy + by² + 2gx + 2fy + c = 0

we get   a = 2, h = – 3/2, f = q/2, b = – p, g = 1/2 , c = 0 .

As line are perpendicular then  a + b = 0 => 2 – p = 0  = > p = 2 . [Ans.]

As lines are straight lines then

abc + 2fgh – af² – bg² – ch² = 0

Or,                     0 + 2(q/2)(1/2)(–3/2) – 2 (q²/4) + p (1/4) = 0

Or,                                                        (–3q)/2 – (q²)/2 + 2/4 = 0

Or,                                                               – 3q – 2q² + 2 = 0

Or,                                                               (2q – 1)(q + 2) = 0

Or,                                                                       q = 1/2 or – 2 . [Ans.]

Q.16. Find the angle and the equation of the bisectors of the angle between the pair of lines given by :

2x² – 3xy + 7y² = 0 .

Solution : – We have   2x² – 3xy + 7y² = 0

Here  a = 2, h = – 7/2, b = 3.

Then    angle      = θ = tan ^–1 [{2√(h² –  ab)}/(a +b)]

= tan ^–1 [{2√(49/4 – 6)}/(5)]

= tan ^–1 [√(25)/5]

= tan ^–1 (1) = 45º. [Ans.]

Equation of bisectors is , h(x2 – y2) – xy(a – b) = 0

Or,                         – 7/2 (x² – y²) – x y(2 – 3) = 0

Or,                        – 7/2 (x² – y²) + x y = 0

Or,                        – 7 (x² – y²) + 2xy = 0

Or,                           7x² – 2xy – 7y² = 0 . [Ans.]