Test: Single Correct MCQs: Straight Lines and Pair of Straight Lines | JEE Advanced

The given points are A (– a, – b), B (0, 0), C (a, b) and D (a², ab).
Slope of   Slope of BC = slope of BD

∴ A, B, C, D are collinear.

Reflection about the line y = x, changes the point (4, 1) to (1, 4).
On translation of (1, 4) through a distance of 2 units along +ve direction of x-axis the point becomes (1 + 2, 4), i.e., (3, 4).

On rotation about origin through an angle p/4 the point P takes the position P' such that OP = OP'

Also OP = 5 = OP' and  

Now, x = OP' 

Solving the given equations of lines pairwise, we get the vertices of Δ as A (– 2, 2) B (2, – 2), C (1, 1)

Then AB  = 

BC = 

CA =            ∴   Δ is isosceles.

We have P = (1, 0), Q = (– 1, 0), R = (2, 0)
Let   S = (x, y) ATQ   SQ² + SR² = 2SP²
⇒ (x + 1)² + y² + (x – 2)² + y² = 2[(x – 1)² + y²]
⇒ 2x² + 2y² – 2x + 5 = 2x² + 2y² – 4x + 2
⇒ 2x + 3 = 0 ⇒ x = – 3/2
Which is a straight line parallel to y-axis.

As L has intercepts a and b on axes, equation of L is

     .....… (1)
Let x and y axes be rotated through an angle q in anticlockwise direction.
In new system intercepts are p and q, therefore equation of L becomes

.....… (2)
KEY CONCEPT :  As the origin is fixed in rotation, the distance of line from origin in both the cases should be same.

∴ We get  d = 

∴ (b) is the correct answer.

Let the two perpendicular lines be the co-ordinate axes.
Let (x, y) be the point sum of whose distances from two axes is 1 then we must have | x | + | y | = 1  or ± x ± y = 1
These are the four lines x + y = 1, x – y = 1, – x + y = 1, – x – y = 1

Any two adjacent sides are perpendicular to each other.
Also each line is equidistant from origin. Therefore figure formed is a square.

If variable point is P and S (– 2, 0) then  PS =   PM where PM is the perpendicular distance of point P from given line x = – 9/2

∴ By definition P describes an ellipse.  

The sides of parallelogram are x = 2, x = 3, y = 1, y = 5.

∴ Diagonal AC is  or y = 4x – 7

Equation diagonal BD is or 4x + y = 13

The lines by which Δ is formed are x = 0, y = 0 and x + y = 1.
Clearly, it is right Δ and we know that in a right Δ orthocentre coincides with the vertex at which right ∠ is formed.
∴ Orthocentre is (0, 0).

Let m be the slope of PQ then

⇒ m + 2 = 1 – 2m or – 1 + 2m = m + 2
⇒ m = – 1/3 or m = 3

As PR also makes ∠ 45° with RQ.
∴ The above two values of m are for PQ and PR.

∴ Equation of PQ, 

⇒ 3y – 3 = – x + 2 ⇒ x + 3y – 5 = 0

and equation of PR is ⇒ 3x – y – 5 = 0
∴ Combined equation of PQ and PR is (x – 3y – 5) (3x – y – 5) = 0
⇒ 3x² – 3y² + 8xy – 20x – 10y + 25 = 0

x₂ = x₁r, x₃ = x₁r² and so is y₂ = y₁r, y³ = y₁r²

Hence the points lie on a line, i.e., they are collinear.

S is the midpoint of Q and R

Therefore,  

Now slope of  PS = m 

Now equation of the line passing through (1, – 1) and parallel to PS is

y + 1 =      or 2x + 9y + 7 = 0

Here AB = BC = CA = 2. So, it is an equilaterial triangle and the incentre coincides with centroid.
Therefore,

Incentre = Centroid = 

Intersection of 3x + 4y = 9 and y = mx + 1.
For x co-ordinate 3x + 4 (mx + 1) = 9 ⇒ (3 + 4m) x = 5

For x to be an integer 3 + 4m should be a divisor of 5 i.e., 1, – 1, 5 or – 5. 3 + 4m = 1
⇒ m = – 1/2 (not integer) 3 + 4m = – 1 ⇒ m = – 1 (integer) 3 + 4m =  5
⇒ m = 1/2 (not an integer) 3 + 4m = – 5 ⇒ m = – 2 (integer)
∴ There are 2 integral values of m.
∴ (a) is the correct alternative.

The vertices, 

⇒ bisector will have slope tan 120°
⇒ equation of bisector is

Clearly OP = OQ = 1 and   

The bisector of ∠QOP will be a perpendicular to PQ and also bisect it. Hence Q is reflection of P in the line OM which makes an angle ∠MOP + ∠POX with x- axis,

So that slope of OM is tan a/2.

The given lines are 2x + y = 9/2 .....… (1)
and 2x + y = – 6 .....… (2)
Signs of constants on R.H.S. show that two lines lie on opp. sides of origin. Let any line through origin meets these lines in P and Q respectively then req. ratio is OP : OQ

Now in DOPA and DOQC, Ð POA = Ð QOC (ver. opp. Ð' s) Ð PAO = Ð OCQ (alt. int. Ð' s) ∠
∴DOPA ~ DOQC (by AA similarly)

∴ Req. ratio is 3 : 4.

Total no. of points within the square OABC = 20 × 20 = 400

Points on line AB = 20 ((1, 1), (2, 2), ....… (20, 20))
∴ Points within DOBC and DABC = 400 – 20 = 380
By symmetry points within  

We know that or thocen tre is the meeting point of altitudes of a Δ.

Equation of alt. AD
⇒ line parallel to y-axis through (3, 4)
⇒ x = 3 ......... (1)
Similarly eqⁿ of OE ⊥ AB is

⇒ y = x/4 ......... (2)

Solving (1) and (2), we get orthocentre as (3, 3/4)

x² – y² + 2y = 1  ⇒ x = ± (y – 1)

Bisectors of above lines are x = 0 and y = 1.

So area between x = 0, y = 1 and x + y = 3 is shaded region shown in figure.

Area  =   sq. units.

∵ Ar (ΔOPR) = Ar (ΔPQR) = Ar (ΔOQR)

∴ By simply geometry R should be the centroid of ΔPQO

Let the slope of line L be m.

∵ L intersects x-axis,      

∴ Equation of L is y + 2 =    (x – 3)

or