Let the three sides of a triangle be on the lines $4x-7y+10=0$, $x+y=5$ and $7x+4y=15$. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines $x=0$, $y=0$ and $x+y=1$ is
Let the area of $\triangle PQR$ with vertices $P(5,4),\ Q(-2,4)$ and $R(a,b)$ be $35$ square units. If its orthocenter and centroid are $O\!\left(2,\dfrac{14}{5}\right)$ and $C(c,d)$ respectively, then $c+2d$ is equal to:
A ray of light coming from the point (2, $2\sqrt 3 $) is incident at an angle 30o on the line x = 1 at thepoint A. The ray gets reflected on the line x = 1 and meets x-axis at the point B. Then, the line ABpasses through the point :
A straight line $L$ at a distance of $4$ units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of $60^\circ$ with the line $x + y = 0$.
Then an equation of the line $L$ is:
Let $a,b,c$ and $d$ be non-zero numbers. If the point of intersection of the
lines $4ax+2ay+c=0$ and $5bx+2by+d=0$ lies in the fourth quadrant and is
equidistant from the two axes then :
Let $PS$ be the median of the triangle with vertices $P(2,2)$, $Q(6,-1)$ and
$R(7,3)$. The equation of the line passing through $(1,-1)$ and parallel to
$PS$ is :
Let $A$ be the point of intersection of the lines $L_{1}:\ \dfrac{x-7}{1}=\dfrac{y-5}{0}=\dfrac{z-3}{-1}$ and $L_{2}:\ \dfrac{x-1}{3}=\dfrac{y+3}{4}=\dfrac{z+7}{5}$. Let $B$ and $C$ be the points on the lines $L_{1}$ and $L_{2}$ respectively such that $AB=AC=\sqrt{15}$. Then the square of the area of the triangle $ABC$ is:
The straight lines $l_1$ and $l_2$ pass through the origin and trisect the line segment of the line
$L : 9x + 5y = 45$ between the axes.
If $m_1$ and $m_2$ are the slopes of the lines $l_1$ and $l_2$, then the point of intersection of the line
$y = (m_1 + m_2)x$ with $L$ lies on :
Two straight lines through the origin $O$ intersect the line $3x+4y=12$ at points $P$ and $Q$ such that $\triangle OPQ$ is isosceles and $\angle POQ=90^\circ$. If $I=OP^2+PQ^2+QO^2$, then the greatest integer $\le I$ is:
Let $\triangle ABC$ be the triangle such that the equations of lines $AB$ and $AC$ are $3y-x=2$ and $x+y=2$, respectively, and the points $B$ and $C$ lie on the $x$-axis. If $P$ is the orthocentre of $\triangle ABC$, then the area of $\triangle PBC$ is equal to
Let a, b$ \in $R. If the mirror image of the point P(a, 6, 9) with respect to the line ${{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}$ is (20, b, $-$a$-$9), then | a + b |, is equal to :
Let $A(1,1)$, $B(-4,3)$, $C(-2,-5)$ be vertices of a triangle $ABC$, $P$ be a point on side $BC$, and $\Delta_1$ and $\Delta_2$ be the areas of triangles $APB$ and $ABC$, respectively. If $\Delta_1:\Delta_2=4:7$, then the area enclosed by the lines $AP$, $AC$ and the $x$-axis is:
If for $\theta\in\left[-\dfrac{\pi}{3},0\right]$, the points $(x,y)=\big(3\tan(\theta+\tfrac{\pi}{3}),,2\tan(\theta+\tfrac{\pi}{6})\big)$ lie on $xy+\alpha x+\beta y+\gamma=0$, then $\alpha^{2}+\beta^{2}+\gamma^{2}$ is:
Let $d$ be the distance of the point of intersection of the lines $\dfrac{x+6}{3}=\dfrac{y}{2}=\dfrac{z+1}{1}$ and $\dfrac{x-7}{4}=\dfrac{y-9}{3}=\dfrac{z-4}{2}$ from the point $(7,8,9)$. Then $d^{2}+6$ is equal to:
The portion of the line $4x+5y=20$ in the first quadrant is trisected by the lines $L_1$ and $L_2$ passing through the origin.
The tangent of the angle between the lines $L_1$ and $L_2$ is:
Let the lines $3x - 4y - \alpha = 0$, $8x - 11y - 33 = 0$, and $2x - 3y + \lambda = 0$ be concurrent. If the image of the point $(1,2)$ in the line $2x - 3y + \lambda = 0$ is $\left(\dfrac{57}{13}, -\dfrac{40}{13}\right)$, then $|\alpha \lambda|$ is equal to
Let $A(-1,1)$ and $B(2,3)$ be two points and $P$ be a variable point above the line $AB$ such that the area of $\triangle PAB$ is $10$. If the locus of $P$ is $ax+by=15$, then $5a+2b$ is:
The foot of the perpendicular drawn from the origin, on the line, $3x + y = \lambda\ (\lambda \ne 0)$ is $P$. If the line meets $x$-axis at $A$ and $y$-axis at $B$, then the ratio $BP : PA$ is :
The sides of a rhombus $ABCD$ are parallel to the lines, $x - y + 2 = 0$ and $7x - y + 3 = 0$. If the diagonals of the rhombus intersect $P(1,2)$ and the vertex $A$ (different from the origin) is on the $y$-axis, then the coordinate of $A$ is :
Let the area of the triangle with vertices $A(1,\alpha)$, $B(\alpha,0)$ and $C(0,\alpha)$ be $4$ sq. units. If the points $(\alpha,-\alpha)$, $(-\alpha,\alpha)$ and $(\alpha^2,\beta)$ are collinear, then $\beta$ is equal to:
Let the image of the point $(1,0,7)$ in the line $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}$ be the point $(\alpha,\beta,\gamma)$.
Then which one of the following points lies on the line passing through $(\alpha,\beta,\gamma)$ and making angles $\dfrac{2\pi}{3}$ and $\dfrac{3\pi}{4}$ with the $y$-axis and $z$-axis respectively, and an acute angle with the $x$-axis?
If in a parallelogram $ABDC$, the coordinates of $A, B$ and $C$ are respectively $(1,2)$, $(3,4)$ and $(2,5)$, then the equation of the diagonal $AD$ is :
Let the points $\left(\dfrac{11}{2},,\alpha\right)$ lie on or inside the triangle with sides $x+y=11$, $x+2y=16$ and $2x+3y=29$. Then the product of the smallest and the largest values of $\alpha$ is equal to:
Let $R$ be the interior region between the lines $3x - y + 1 = 0$ and $x + 2y - 5 = 0$ containing the origin.
The set of all values of $a$, for which the points $(a^2,\,a+1)$ lie in $R$, is:
Let a variable line of slope $m>0$ passing through $(4,-9)$ intersect the coordinate axes at points $A$ and $B$. The minimum value of the sum of the distances of $A$ and $B$ from the origin is:
A straight line through a fixed point $(2,3)$ intersects the coordinate axes at distinct points $P$ and $Q$. If $O$ is the origin and the rectangle $OPRQ$ is completed, then the locus of $R$ is :
An ordered pair ($\alpha $, $\beta $) for which the system of linear equations
(1 + $\alpha $) x + $\beta $y + z = 2
$\alpha $x + (1 + $\beta $)y + z = 3
$\alpha $x + $\beta $y + 2z = 2
has a unique solution, is :
In $\triangle ABC$, suppose $y=x$ is the equation of the bisector of the angle $B$ and the equation of the side $AC$ is $2x-y=2$. If $2AB=BC$ and the points $A$ and $B$ are respectively $(4,6)$ and $(\alpha,\beta)$, then $\alpha+2\beta$ is equal to:
If a circle of radius $R$ passes through the origin $O$ and intersects the coordinate axes at $A$ and $B$, then the locus of the foot of the perpendicular from $O$ on $AB$ is:
Let $A$ be the point of intersection of the lines $3x+2y=14$ and $5x-y=6$,
and $B$ be the point of intersection of the lines $4x+3y=8$ and $6x+y=5$.
The distance of the point $P(5,-2)$ from the line $AB$ is:
The square of the distance of the point $\left(\dfrac{15}{7},,\dfrac{32}{7},,7\right)$ from the line $\dfrac{x+1}{3}=\dfrac{y+3}{5}=\dfrac{z+5}{7}$ in the direction of the vector $\hat{i}+4\hat{j}+7\hat{k}$ is:
Two equal sides of an isosceles triangle are along $-x+2y=4$ and $x+y=4$. If $m$ is the slope of its third side, then the sum of all possible distinct values of $m$ is:
Let $m_1, m_2$ be the slopes of two adjacent sides of a square of side $a$ such that
$a^{2}+11a+3\left(m_{1}^{2}+m_{2}^{2}\right)=220.$
If one vertex of the square is $\big(10(\cos\alpha-\sin\alpha),\,10(\sin\alpha+\cos\alpha)\big)$, where $\alpha\in(0,\tfrac{\pi}{2})$, and the equation of one diagonal is
$(\cos\alpha-\sin\alpha)x+(\sin\alpha+\cos\alpha)y=10$, then
$
72\left(\sin^{4}\alpha+\cos^{4}\alpha\right)+a^{2}-3a+13
$
is equal to:
If a straight line passing through the point $P(-3,4)$ is such that its intercepted portion between the coordinate axes is bisected at $P$, then its equation is:
Let $P(\alpha,\beta,\gamma)$ be the image of the point $Q(3,-3,1)$ in the line $\dfrac{x-0}{1}=\dfrac{y-3}{1}=\dfrac{z-1}{-1}$ and let $R$ be the point $(2,5,-1)$. If the area of $\triangle PQR$ is $\lambda$ and $\lambda^{2}=14K$, then $K$ is:
If the angle between the lines $\dfrac{x}{2}=\dfrac{y}{2}=\dfrac{z}{1}$ and $\dfrac{5-x}{-2}=\dfrac{7y-14}{p}=\dfrac{z-3}{4}$ is $\cos^{-1}\left(\dfrac{2}{3}\right)$, then $p$ is equal to :
Let $R$ be a rectangle given by the lines $x = 0$, $x = 2$, $y = 0$ and $y = 5$.
Let $A(\alpha, 0)$ and $B(0, \beta)$, $\alpha \in [0, 2]$ and $\beta \in [0, 5]$,
be such that the line segment $AB$ divides the area of the rectangle $R$ in the ratio $4 : 1$.
Then, the mid-point of $AB$ lies on a:
Let a be the length of a side of a square OABC with O being the origin. Its side OA makes an acute angle $\alpha $ with the positive x-axis and the equations of its diagonals are $(\sqrt{3}+1)x+(\sqrt{3}-1)y=0$ and $(\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0$. Then $a$2 is equal to :
A line passing through the point $P(a, 0)$ makes an acute angle $\alpha$ with the positive x-axis. Let this line be rotated about the point $P$ through an angle $\dfrac{\alpha}{2}$ in the clockwise direction. If in the new position, the slope of the line is $2 - \sqrt{3}$ and its distance from the origin is $\dfrac{1}{\sqrt{2}}$, then the value of $3a^2 \tan^2 \alpha - 2\sqrt{3}$ is:
If the foot of the perpendicular from point (4, 3, 8) on the line ${L_1}:{{x - a} \over l} = {{y - 2} \over 3} = {{z - b} \over 4}$, l $\ne$ 0 is (3, 5, 7), then the shortest distance between the line L1 and line ${L_2}:{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}$ is equal to :
A line passing through the point \(A(9,0)\) makes an angle of \(30^\circ\) with the positive
direction of the \(x\)-axis. If this line is rotated about \(A\) through an angle of \(15^\circ\) in
the clockwise direction, then its equation in the new position is:
C
The set of all possible values of $\theta $ in the interval (0, $\pi $) for which the points (1, 2) and (sin $\theta $, cos $\theta $) lie on the same side of the line x + y =1 is :
The equations of two sides $AB$ and $AC$ of a triangle $ABC$ are $4x+y=14$ and $3x-2y=5$, respectively. The point $\left(2,-\frac{4}{3}\right)$ divides the third side $BC$ internally in the ratio $2:1$. The equation of the side $BC$ is
Let A($-$1, 1), B(3, 4) and C(2, 0) be given three points. A line y = mx, m > 0, intersects lines AC and BC at point P and Q respectively. Let A1 and A2 be the areas of $\Delta$ABC and $\Delta$PQC respectively, such that A1 = 3A2, then the value of m is equal to :
In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4. Let the point B lie on the line x + 3y = 7. If ($\alpha$, $\beta$) is the centroid of $\Delta$ABC, then 15($\alpha$ + $\beta$) is equal to :
If the shortest distance between the lines
$\dfrac{x-\lambda}{2}=\dfrac{y-4}{3}=\dfrac{z-3}{4}$ and
$\dfrac{x-2}{4}=\dfrac{y-4}{6}=\dfrac{z-7}{8}$ is $\dfrac{13}{\sqrt{29}}$, then a value of $\lambda$ is:1
Let the lines
\[
\ell_1:\ \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}
\quad\text{and}\quad
\ell_2:\ 3x+2y+z-2=0\;=\;x-3y+2z-13
\]
be coplanar. If the point $P(a,b,c)$ on $\ell_1$ is nearest to the point $Q(-4,-3,2)$,
then $|a|+|b|+|c|$ is equal to:
If the line segment joining the points $(5,2)$ and $(2,a)$ subtends an angle $\dfrac{\pi}{4}$ at the origin, then the absolute value of the product of all possible values of $a$ is:
Let $A$ be the point of intersection of the lines $3x+2y=14$ and $5x-y=6$, and $B$ be the point of intersection of the lines $4x+3y=8$ and $6x+y=5$. The distance of the point $P(5,-2)$ from the line $AB$ is:
Suppose the points $(h,k)$, $(1,2)$ and $(-3,4)$ lie on the line $L_1$. If a line $L_2$ passing through the points $(h,k)$ and $(4,3)$ is perpendicular to $L_1$, then $\dfrac{k}{h}$ equals:
A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this line on the coordinate axes is $\frac{1}{4}$ . Three stones A, B and C are placed at the points (1, 1), (2, 2) and (4, 4) respectively. Then, which of these stones is / are on the path of the man?
Let $\alpha,\beta,\gamma,\delta\in\mathbb{Z}$ and let
$A(\alpha,\beta),\ B(1,0),\ C(\gamma,\delta),\ D(1,2)$ be the vertices of a parallelogram $ABCD$.
If $AB=\sqrt{10}$ and the points $A$ and $C$ lie on the line $3y=2x+1$, then
$2(\alpha+\beta+\gamma+\delta)$ is equal to:
The equation of one of the straight lines which passes through the point (1, 3) and makes an angles ${\tan ^{ - 1}}\left( {\sqrt 2 } \right)$ with the straight line, y + 1 = 3${\sqrt 2 }$ x is :
Let the vertices $Q$ and $R$ of the triangle $PQR$ lie on the line
$\dfrac{x + 3}{5} = \dfrac{y - 1}{2} = \dfrac{z + 4}{3}$,
$QR = 5$ and the coordinates of the point $P$ be $(0, 2, 3)$.
If the area of the triangle $PQR$ is $\dfrac{m}{n}$, then:
Let $B$ and $C$ be the two points on the line $y+x=0$ such that $B$ and $C$ are symmetric with respect to the origin.
Suppose $A$ is a point on $y-2x=2$ such that $\triangle ABC$ is an equilateral triangle.
Then, the area of the $\triangle ABC$ is:
A variable line $L$ passes through the point $(3,5)$ and intersects the positive coordinate axes at the points $A$ and $B$. The minimum area of the triangle $OAB$, where $O$ is the origin, is:
Let a triangle be bounded by the lines L1 : 2x + 5y = 10; L2 : $-$4x + 3y = 12 and the line L3, which passes through the point P(2, 3), intersects L2 at A and L1 at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to :
Let $(\alpha,\beta)$ be the centroid of the triangle formed by the lines $15x-y=82$, $6x-5y=-4$ and $9x+4y=17$. Then $\alpha+2\beta$ and $2\alpha-\beta$ are the roots of the equation:
A light ray emits from the origin making an angle $30^\circ$ with the positive $x$-axis.
After getting reflected by the line $x+y=1$, if this ray intersects the $x$-axis at $Q$, then the abscissa of $Q$ is:
$y=\tan30^\circ,x=\dfrac{x}{\sqrt3}$ hits the mirror $x+y=1$ at
$P\left(\dfrac{\sqrt3}{\sqrt3+1},,\dfrac{1}{\sqrt3+1}\right)$.
The mirror’s normal is along $(1,1)$, so reflecting the unit direction $u=(\cos30^\circ,\sin30^\circ)=\left(\dfrac{\sqrt3}{2},\dfrac12\right)$ about the line gives
$u'=u-2(u\cdot \hat n)\hat n=\left(-\dfrac12,-\dfrac{\sqrt3}{2}\right)$,
i.e. slope $m'=\sqrt3$.
The reflected ray through $P$ is $y-y_0=\sqrt3(x-x_0)$.
If p and q are the lengths of the perpendiculars from the origin on the lines,:- x cosec $\alpha$ $-$ y sec $\alpha$ = k cot 2$\alpha$ and, x sin$\alpha$ + y cos$\alpha$ = k sin2$\alpha$ respectively, then k2 is equal to :
Two sides of a rhombus are along the lines, $x - y + 1 = 0$ and $7x - y - 5 = 0$.
If its diagonals intersect at $(-1, -2)$, then which one of the following is a vertex of this rhombus?
The shortest distance, between lines $L_1$ and $L_2$, where $L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$ and $L_2$ is the line, passing through the points $\mathrm{A}(-4,4,3), \mathrm{B}(-1,6,3)$ and perpendicular to the line $\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$, is
If the image of the point $P(1, 0, 3)$ in the line joining the points $A(4, 7, 1)$ and $B(3, 5, 3)$ is $Q(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma$ is equal to:
Consider the line $L$ passing through the points $(1,2,3)$ and $(2,3,5)$.
The distance of the point $\left(\dfrac{11}{3},\dfrac{11}{3},\dfrac{19}{3}\right)$ from the line $L$ along the line
$\dfrac{3x-11}{2}=\dfrac{3y-11}{1}=\dfrac{3z-19}{2}$ is equal to:
If the two lines $x+(a-1)y=1$ and $2x+a^{2}y=1$ $(a\in\mathbb{R}\setminus{0,1})$ are perpendicular, then the distance of their point of intersection from the origin is:
The shortest distance between the lines
\[
\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}
\quad\text{and}\quad
\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}
\]
is:
Let a variable line passing through the centre of the circle
$x^{2}+y^{2}-16x-4y=0$ meet the positive coordinate axes at the points $A$ and $B$.
Then the minimum value of $OA+OB$, where $O$ is the origin, is:
If the length of the perpendicular from the point $(\beta,0,\beta)\ (\beta\ne0)$ to the line, $\dfrac{x}{1}=\dfrac{y-1}{0}=\dfrac{z+1}{-1}$ is $\sqrt{\dfrac{3}{2}}$, then $\beta$ is equal to:
The point $(2,1)$ is translated parallel to the line $L : x - y = 4$ by $2\sqrt{3}$ units. If the new point $Q$ lies in the third quadrant, then the equation of the line passing through $Q$ and perpendicular to $L$ is:
If a variable line drawn through the intersection of the lines
$\dfrac{x}{3} + \dfrac{y}{4} = 1$ and $\dfrac{x}{4} + \dfrac{y}{3} = 1$
meets the coordinate axes at $A$ and $B$ $(A \ne B)$, then the locus of the midpoint of $AB$ is:
Let a line passing through the point $(4,1,0)$ intersect the line $\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $\mathrm{L}_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then $\left|\begin{array}{lll}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c\end{array}\right|$ is equal to
Line $L_1$ passes through the point $(1, 2, 3)$ and is parallel to the $z$-axis.
Line $L_2$ passes through the point $(\lambda, 5, 6)$ and is parallel to the $y$-axis.
Let for $\lambda = \lambda_1, \lambda_2,$ $\lambda_2 < \lambda_1,$ the shortest distance between the two lines be $3$.
Then the square of the distance of the point $(\lambda_1, \lambda_2, 7)$ from the line $L_1$ is
Let $\alpha_1, \alpha_2 ; (\alpha_1 < \alpha_2)$ be the values of $\alpha$ for the points $(\alpha, -3), (2, 0)$ and $(1, \alpha)$ to be collinear. Then the equation of the line, passing through $(\alpha_1, \alpha_2)$ and making an angle of $\frac{\pi}{3}$ with the positive direction of the x-axis, is :
A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines $L_1: 2x+y+6=0$ and $L_2: 4x+2y-p=0,; p>0$ at the points $A$ and $B$, respectively. If $|AB|=\dfrac{9}{\sqrt{2}}$ and the foot of the perpendicular from the point $A$ on the line $L_2$ is $M$, then $\dfrac{AM}{BM}$ is equal to
If the shortest distance between the straight lines $3(x - 1) = 6(y - 2) = 2(z - 1)$ and $4(x - 2) = 2(y - \lambda ) = (z - 3),\lambda \in R$ is ${1 \over {\sqrt {38} }}$, then the integral value of $\lambda$ is equal to :
Consider the lines $x(3\lambda+1)+y(7\lambda+2)=17\lambda+5$, $\lambda$ being a parameter, all passing through a point $P$. One of these lines (say $L$) is farthest from the origin. If the distance of $L$ from the point $(3,6)$ is $d$, then the value of $d^2$ is:
Lines are drawn parallel to the line $4x-3y+2=0$, at a distance $\dfrac{3}{5}$ from the origin. Then which one of the following points lies on any of these lines?
A straight line through origin $O$ meets the lines $3y = 10 - 4x$ and $8x + 6y + 5 = 0$ at points $A$ and $B$ respectively.
Then $O$ divides the segment $AB$ in the ratio :
A ray of light is incident along a line which meets another line $7x - y + 1 = 0$ at the point $(0,1)$.
The ray is then reflected from this point along the line $y + 2x = 1$.
Then the equation of the line of incidence of the ray of light is:
Let $\vec a=2\hat{\imath}+\hat{\jmath}+\hat{k}$, and $\vec b,\vec c$ be two nonzero vectors such that $\left\lvert \vec a+\vec b+\vec c \right\rvert=\left\lvert \vec a+\vec b-\vec c \right\rvert$ and $\vec b\cdot\vec c=0$. Consider the statements:
(A) $\left\lvert \vec a+\lambda\vec c \right\rvert \ge \lvert \vec a\rvert \text{ for all } \lambda\in\mathbb{R}$.
(B) $\vec a$ and $\vec c$ are always parallel.
Then,
Let $P$ and $Q$ be the points on the line $\dfrac{x+3}{8}=\dfrac{y-4}{2}=\dfrac{z+1}{2}$ which are at a distance of $6$ units from the point $R(1,2,3)$.
If the centroid of the triangle $PQR$ is $(\alpha,\beta,\gamma)$, then $\alpha^{2}+\beta^{2}+\gamma^{2}$ is:
Let the shortest distance between the lines $\dfrac{x-3}{3}=\dfrac{y-\alpha}{-1}=\dfrac{z-3}{1}$ and $\dfrac{x+3}{-3}=\dfrac{y+7}{2}=\dfrac{z-\beta}{4}$ be $3\sqrt{30}$. Then the positive value of $5\alpha+\beta$ is
A point P moves on the line $2x - 3y + 4 = 0.$
If $Q(1, 4)$ and $R(3, -2)$ are fixed points, then the locus of the centroid of $\triangle PQR$ is a line :
The vertices of a triangle are $A(-1,3)$, $B(-2,2)$ and $C(3,-1)$.
A new triangle is formed by shifting the sides of the triangle by one unit inwards.
Then the equation of the side of the new triangle nearest to the origin is:
A $2,\text{m}$ ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate of $25,\text{cm/sec}$, then the rate (in $\text{cm/sec}$) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is $1,\text{m}$ above the ground is:
Let $A$ and $B$ be two distinct points on the line $L:\ \dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2\sqrt{17}$ from the foot of perpendicular drawn from the point $(1,2,3)$ on the line $L$. If $O$ is the origin, then $\overrightarrow{OA}\cdot\overrightarrow{OB}$ is equal to