Family of Lines MCQ Quiz in मल्याळम - Objective Question with Answer for Family of Lines - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Given line is

General equation is

Comparing above equation with , we get

Lines are real and distinct if 0\)

0\)

0\)

(A)

and

So, it represents a pair of lines and

(B)

and

So, it represents a pair of lines and

(C)

which represents parabola

(D)

and

So, it represents a pair of lines and

Hence, only (C) does not represents a pair of lines.

Let ----- equation and ----- equation Multiplying equation with , we get, ----- equation Adding equations and , we get Substituting in the equation , we get Hence, Since the lines are concurrent, should satisfy the equation

Calculation

4a2 + 9b2 – c2 + 12ab = 0
⇒ (2a + 3b)² − c² = 0

⇒ 2a + 3b - c = 0 or 2a + 3b + c = 0

⇒ c = ±(2a + 3b)

∴ ax + by + c = 0

⇒ ax + by ± (2a + 3b) = 0

⇒ a(x ± 2) + b(y ± 3) = 0 (family of lines)

⇒ (-2, -3) or (2, 3)

Hence option 1 is correct 

Calculation

⇒ r²(cosθ.cos ​+ sinθ.sin )² = 2

⇒ 

⇒ (rcosθ + √3rsinθ)² = 8

⇒ (x + y√3)² = 8

⇒ (x + y√3 - 2√2)(x + y√3 + 2√2) = 0

Represents a pair of straight lines

Hence option 4 is correct

Calculation

Locus of point P(x, y) whose distance from  x + 2y + 7 = 0 & 2x – y + 8 = 0 are equal is 

⇒ 

⇒ (x + 2y + 7)² – (2x – y + 8)² = 0 

Combined equation of lines

⇒ (x – 3y + 1) (3x + y + 15) = 0   {  a² - b² = (a + b)(a -  b)}

⇒ 3x² – 3y² – 8xy + 18x – 44y + 15 = 0 

⇒ 

⇒ x² – y² + 2h xy + 2gx 2 + 2fy + c = 0 

⇒ 

⇒ 

Hence option(1) is correct

Concept:

Condition for Pair of straight Lines:

The equation ax² + 2hxy + by² = 0 is a homogenous equation of the second degree, representing a pair of straight lines passing through the origin. But
(i) If h² > ab, then the two straight lines are real and different.
(ii) If h² = ab, then the two straight lines are coincident.
(iii) If h² 

Solution:

(2l − 3)x2 + 2lxy − y2 = 0  . . . (1)

On comparing with  ax² + 2hxy + by² = 0        . . . (2)

we get, a = 2I - 3, h = I, b = - 1

Equation (1) represents a pair of straight lines if 

h² > ab ⇒ I² > (2I - 3)(-1)

⇒ I² >(3 - 2I)

⇒ I² + 2I - 3 > 0

⇒ I² + 3I - I - 3 > 0

⇒ (I + 3)(I - 1) > 0

⇒ I 0

∴ I ∈ R - (- 3, 1)

Hence option (2) is correct.