A rod of length eight units moves such that its ends $A$ and $B$ always lie on the lines $x-y+2=0$ and $y+2=0$, respectively. If the locus of the point $P$, that divides the rod $A B$ internally in the ratio $2: 1$ is $9\left(x^2+\alpha y^2+\beta x y+\gamma x+28 y\right)-76=0$, then $\alpha-\beta-\gamma$ is equal to :
Let a rectangle $ABCD$ of sides $2$ and $4$ be inscribed in another rectangle $PQRS$ such that the vertices of $ABCD$ lie on the sides of $PQRS$. Let $a$ and $b$ be the sides of $PQRS$ when its area is maximum. Then $(a+b)^2$ is equal to:
Let the line $L$ pass through $(1,1,1)$ and intersect the lines
$\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-1}{4}$ and $\dfrac{x-3}{1} = \dfrac{y-4}{2} = \dfrac{z}{1}$.
Then, which of the following points lies on the line $L$?
If the shortest distance between the lines $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $\dfrac{x}{1}=\dfrac{y}{\alpha}=\dfrac{z-5}{1}$ is $\dfrac{5}{\sqrt6}$, then the sum of all possible values of $\alpha$ is
If the length of the perpendicular drawn from the point $P(a,4,2),;a>0$ on the line $\dfrac{x+1}{2}=\dfrac{y-3}{3}=\dfrac{z-1}{-1}$ is $2\sqrt{6}$ units and $Q(\alpha_{1},\alpha_{2},\alpha_{3})$ is the image of the point $P$ in this line, then $a+\sum_{i=1}^{3}\alpha_{i}$ is equal to:
Let $(\alpha,\beta,\gamma)$ be the image of the point $(8,5,7)$ in the line $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{5}$. Then $\alpha+\beta+\gamma$ is:
Let A be a fixed point (0, 6) and B be a moving point (2t, 0). Let M be the mid-point of AB and the perpendicular bisector of AB meets the y-axis at C. The locus of the mid-point P of MC is :
Let $\binom{n}{r-1}=28$, $\binom{n}{r}=56$ and $\binom{n}{r+1}=70$. Let $A(4\cos t,,4\sin t)$, $B(2\sin t,,-2\cos t)$ and $C(3r-n,,r^{2}-n-1)$ be the vertices of a triangle $ABC$, where $t$ is a parameter. If $(3x-1)^{2}+(3y)^{2}=\alpha$ is the locus of the centroid of triangle $ABC$, then $\alpha$ equals
The locus of the point of intersection of the lines $ \sqrt{2}x - y + 4\sqrt{2}k = 0$ and $\sqrt{2}kx + ky - 4\sqrt{2} = 0$ $(k$ is any non-zero real parameter$)$, is :
A square, of each side $2$, lies above the $x$-axis and has one vertex at the origin.
If one of the sides passing through the origin makes an angle $30^\circ$ with the positive direction of the $x$-axis,
then the sum of the $x$-coordinates of the vertices of the square is:
Let $P_n = \alpha^n + \beta^n$, $n \in \mathbb{N}$.
If $P_{10} = 123$, $P_9 = 76$, $P_8 = 47$ and $P_1 = 1$, then the quadratic equation having roots $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$ is:
The distance between the two points A and A' which lie on y = 2 such that both the line segments AB and A' B (where B is the point (2, 3)) subtend angle ${\pi \over 4}$ at the origin, is equal to :
Let a1, a2, a3, ..... be an A.P. If ${{{a_1} + {a_2} + .... + {a_{10}}} \over {{a_1} + {a_2} + .... + {a_p}}} = {{100} \over {{p^2}}}$, p $\ne$ 10, then ${{{a_{11}}} \over {{a_{10}}}}$ is equal to :
Let the triangle PQR be the image of the triangle with vertices $(1,3),(3,1)$ and $(2,4)$ in the line $x+2 y=2$. If the centroid of $\triangle \mathrm{PQR}$ is the point $(\alpha, \beta)$, then $15(\alpha-\beta)$ is equal to :
The distance of the point $(7,10,11)$ from the line $\dfrac{x-4}{1}=\dfrac{y-4}{0}=\dfrac{z-2}{3}$ along the line $\dfrac{x-9}{2}=\dfrac{y-13}{3}=\dfrac{z-17}{6}$ is
The area (in sq. units) of the region A = {(x, y) : (x – 1)[x] $ \le $ y $ \le $ 2$\sqrt x $, 0 $ \le $ x $ \le $ 2}, where [t] denotes the greatest integer function, is :