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Three lines px + qy + r = 0, qx + ry + p = 0 and rx + py + q = 0 are concurrent if (1985  2 Marks)
For concurrency of three lines px + qy + r = 0; qx + ry + p = 0; rx + py + q = 0 We must have,
⇒ (p + q + r) (pq – q^{2 }– rp + rq – r^{2} + pr + pr – p^{2}) = 0
⇒ (p + q + r) (p^{2} + q^{2} + r^{2} – pq – pr – rq ) = 0
⇒ p^{3} + q^{3} + r^{3} – 3pqr = 0
It is clear that a, b, c are correct options.
The points (1, 3) and (82, 30) are vertices of
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.
All points lying inside the triangle formed by the points (1, 3), (5, 0) and (–1, 2) satisfy (1986  2 Marks)
Substituting the coordinates 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 coordinates 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 coordinates 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.
A vector has components 2p and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to the new system , ar has components p + 1 and 1, then (1986  2 Marks)
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.
If (P(1, 2), Q(4, 6), R(5, 7) and S(a, b) are the vertices of a parallelogram PQRS, then (1998  2 Marks)
PQRS will represent a parallelogram if and only if the midpoint of PR is same as that of the midpoint of QS.
That is, if and only if
⇒ a = 2 and b = 3.
The diagonals of a parallelogram PQRS are along the lines x +3y = 4 and 6x – 2y = 7. Then PQRS must be a. (1998  2 Marks)
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.
If the vertices P, Q, R of a triangle PQR are rational points, which of the following points of the triangle PQR is (are) always rational point(s)? (1998  2 Marks)
Since the coordinates of in the centre depend on lengths of side of Δ.
∴ it can have irrational coordinates
Let L_{1} be a strainght line passing through the origin and L_{2} be the straight line x +y = 1. If the intercepts made by the circle x^{2} + y^{2}  x + 3y= 0 on L_{1} and L_{2} are equal, then which of the following equations can represent L_{1}? (1999  3 Marks)
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 x^{2 }+ y^{2} – x + 3y = 0 with centre
and radius =
L_{1} : y = mx (any line through origin)
L_{2} : x + y – 1 = 0 (given line)
ATQ circle makes equal intercepts on L_{1} and L_{2}
⇒ m^{2 }+ 6m + 9 = 8m^{2} + 8 ⇒ 7m^{2} – 6m – 1 = 0
⇒ 7m^{2} – 7m + m – 1 = 0 ⇒ (7m + 1) (m – 1) = 0
⇒ m = 1, – 1/7
∴ The required line L_{1} is y = x or y =
i.e., x – y = 0 or x + 7y = 0.
For a > b > c > 0, the distance between (1, 1) and the point of intersection of the lines ax + by + c = 0 and bx + ay + c = 0 is less than . Then (JEE Adv. 2013)
ANSWER : a
Solution : ax+by+c=0 ;equation(i)
bx+ay+c=0 ;equation(ii)
equation(i)×a−equation(ii)×b
a^{2}x+aby+ac−b^{2}x−aby+bc=0
⇒x(a^{2}−b^{2})+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}]/t^{2}<4
⇒(t+c)^{2}<4t^{2}
⇒4t^{2}−(t+c)^{2}>0
⇒4t^{2}−t^{2}−c^{2}−2tc>0
⇒3t^{2}−2tc−c^{2}>0
⇒3t^{2}−3tc+tc−c^{2} >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
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