Test: MCQs (One or More Correct Option): Straight Lines and Pair of Straight Lines | JEE Advanced - JEE MCQ

For concurrency of three lines px + qy + r = 0; qx + ry + p = 0; rx + py + q = 0 We must have,

⇒ (p + q + r) (pq – q² – rp + rq – r² + pr + pr – p²) = 0

⇒ (p + q + r) (p² + q² + r² – pq – pr – rq ) = 0

⇒ p³ + q³ + r³ – 3pqr = 0

It is clear that a, b, c are correct options.

Let A (0, 8/3), B (1, 3) and C (82, 30).

Now, slope of line AB = 

Slope of line BC = 

⇒ AB || BC and B is common point.
⇒ A, B, C are collinear.

Substituting the co-ordinates of the points (1, 3), (5, 0) and (– 1, 2) in 3x + 2y, we obtain the value 8, 15 and 1 which are all +ve. Therefore, all the points lying inside the triangle formed by given points satisfy 3x + 2y ≥ 0.
Hence (a) is correct answer.
Substituting the co-ordinates of the given points in 2x + y – 13, we find the values – 8, – 3 and – 13 which are all –ve.
So, (b) is not correct.
Again substituting the given points in 2x – 3y – 12 we get – 19, – 2, – 20 which are all –ve.
It follows that all points lying inside the triangle formed by given points satisfy 2x – 3y – 12 ≤ 0.
So, (c) is the correct answer.
Finally substituting the co-ordinates of the given points in – 2x + y, we get 1, – 10 and 4 which are not all +ve.
So, (d) is not correct.
Hence, (a) and (c) are the correct answers.

Consider  with respect to original axes and a = ( p + 1) i+j with respect to new axes.
Now, as length of vector will remain the same

⇒ p2 + 2p + 2 = 4p2 + 1

⇒ 3p2 – 2p – 1 = 0

⇒ p = 1 or – 1/3

∴ (b) is the correct answer.

PQRS will represent a parallelogram if and only if the mid-point of PR is same as that of the mid-point of QS.
That is, if and only if

⇒ a = 2 and b = 3.

Slope of x + 3y = 4 is – 1/3 and slope of  6x – 2y = 7 is  3.
Therefore, these two lines are perpendicular which shows that both diagonals are perpendicular. Hence PQRS must be a rhombus.

Since the co-ordinates of in the centre depend on lengths of side of Δ.

∴ it  can have irrational coordinates

We know that length of intercept made by a circle on a line is given by =

where p = ⊥ distance of line from the centre of the circle.

Here circle is x² + y² – x + 3y = 0 with centre 

and radius = 

L₁ : y = mx (any line through origin)

L₂ : x + y – 1 = 0 (given line)

ATQ circle makes equal intercepts on L₁ and L₂

⇒ m² + 6m + 9 = 8m² + 8   ⇒ 7m² – 6m – 1 = 0

⇒ 7m² – 7m + m – 1 = 0  ⇒ (7m + 1) (m – 1) = 0

⇒ m = 1, – 1/7

∴ The required line L₁ is y = x  or y =

i.e., x – y = 0 or x + 7y = 0.

ANSWER :- a

Solution :- ax+by+c=0 ;equation(i)

bx+ay+c=0 ;equation(ii)

equation(i)×a−equation(ii)×b

a²x+aby+ac−b²x−aby+bc=0

⇒x(a²−b²)+c(a−b)=0

⇒x(a−b)(a+b)+c(a−b)=0

Dividing both sides by (a−b), we get,

x(a+b)+c=0

⇒x = −c/(a+b)

Similarly equation(ii)×a−equation(i)×b, yields

y = −c/(a+b)

⇒(x,y)=(−c/(a+b), − c/(a+b))

Given that distance between (x,y) and (1,1) is less than 2√2

√{1−(−c/(a+b))}² + {1−(−c/(a+b))}²

√((a+b+c)/a+b)² + ((a+b+c)/a+b)²  < 2√2

= √2((a+b+c)/a+b)² < 2√2

Squaring both sides, we get,

2((a+b+c)/a+b)²<8

⇒((a+b+c)/a+b)²<4

⇒Let a+b=t

⇒ [(t+c)²]/t²<4

⇒(t+c)²<4t²

⇒4t²−(t+c)²>0

⇒4t²−t²−c²−2tc>0

⇒3t²−2tc−c²>0

⇒3t²−3tc+tc−c² >0

⇒3t(t−c)+c(t−c)>0

⇒(3t+c)(t−c)>0

As t=a+b

⇒(3a+3b+c)(a+b−c)>0

Given that a>b>c>0

⇒3a+3b+c>0

Hence for product of (3a+3b+c) and (a+b−c) to be positive, (a+b−c) should also be

 positive.

⇒a+b−c>0