JEE Advanced (Fill in the Blanks): Straight Lines & Pair of Straight Lines | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. The area enclosed within the curve | x | + | y | = 1 is ................... (1981 - 2 Marks) 

Ans. 2 sq. units.

Sol.  | x | + | y | = 1

 The curve represents four lines
x + y = 1, x – y = 1, – x + y = 1,  – x – y = 1
which enclose a square of side = distance between opp. sides x + y = 1 and x + y = – 1

Side

∴ Req. area = (side)² = 2 sq. units.

Q.2. y = 10^x is the reflection of y = log10 ^x in the line whose equation is ................ (1982 - 2 Marks) 

Ans. y = x

Sol.  As y = log₁₀ x  can be obtained by replacing x by y and y by x in y = 10^x

∴ The line of reflection is y = x.

Q.3. The set of lines ax+by+c = 0, where 3a + 2b + 4c = 0 is concurrent at the point ................... (1982 - 2 Marks) 

Ans. (3/4, 1/2)

Sol.  Given that 3a + 2b + 4c = 0

⇒ The set of lines ax + by + c = 0 passes through the point (3/4, 1/2).

Q.4. Given the points A (0, 4) and B (0, –4), the equation of the locus of the point P(x, y) such that | AP – BP | = 6 is ................... (1983 - 1 Mark)

Ans. 

Sol.   AP – BP | = 6
We know that locus of a point, difference of whose distances from two fixed points is constant, is hyperbola with the fixed points as focii and the difference of distances as length of transverse axis.
Thus, ae = 4 and 2a = 6   ⇒ a = 3, e = 4 /3

   ∴  Equation is  

(foci being on y-axis, it is vertical hyperbola)

Q.5. If a, b and c are in A.P., then the straight line ax + by + c = 0 will always pass through a fixed point whose coordinates are ................... (1984 - 2 Marks) 

Ans. (1, – 2)

Sol.  If a, b, c are in A.P. then a + c = 2b ⇒ a – 2b + c = 0

⇒ ax + by + c = 0 passes through (1,– 2).

Q.6. The orthocentre of the triangle for med by the lines x + y = 1, 2x + 3y = 6 and 4x – y + 4 = 0 lies in quadrant number ................... . (1985 - 2 Marks) 

Ans. first quadrant

Sol.  
The equations of sides of triangle ABC are
AB : x + y = 1
BC : 2x + 3y = 6
CA : 4x – y = – 4
Solving these pairwise we get the vertices of D as follows A (– 3/5, 8/5) B (– 3, 4) C (– 3/7, 16/7)
Now AD is line ⊥ ^lar to BC and passes through A. Any line perpendicular to BC is 3x – 2y + l = 0 As it passes through A(– 3/5, 8/5)

∴ Equation of altitude AD is 3x – 2y + 5 = 0         ...(1)

Any line perpendicular to side AC is  x + 4y + µ = 0

As it passes through point B (– 3, 4)

∴ – 3 + 16 + µ = 0 ⇒ µ  = – 13

∴ Equation of altitude BE is x + 4y –13 = 0      ...(2)

Now orthocentre is the point of intersection of equations (1) and (2) (AD and BE)

Solving (1) and (2), we get x = 3/7, y = 22/7

As both the co-ordinates are positive, orthocentre lies in first quadrant.

Q.7. Let the algebraic sum of the perpendicular distances from the points (2, 0), (0, 2) and (1, 1) to a variable straight line be zero; then the line passes through a fixed point whose cordinates are ................... . (1991 -  2 Marks)

Ans. (1, 1)

Sol.  Let the variable line be ax + by + c = 0 .....… (1)  

Then ⊥^lar distance of line from (0, 2) =  

⊥^lar  distance of line from (1, 1) =  

ATQ  p₁ + p₂ + p₃ = 0

⇒ 3a + 3b + 3c = 0

⇒ a + b + c = 0 .....… (2)

From (1) and (2), we can say variable line (1) passes through the fixed point (1, 1).

Q.8. The vertices of a triangle are A (–1, –7), B (5, 1) and C (1, 4).
 The equation of th e bisector of the an gle ∠ABC is ................... . (1993 -  2 Marks)

Ans. x - 7y + 2 = 0

Sol.  Let BD be the bisector of ∠ABC.
NOTE THIS STEP :

Then AD : DC = AB : BC

And

AB =

∴ By section formula

Therefore equation of BD is

⇒ 7y – 7 = x – 5 ⇒ x – 7y + 2 = 0