Let the point $P(\alpha, \beta)$ be at a unit distance from each of the two lines $L_1: 3x - 4y + 12 = 0$, and $L_2: 8x + 6y + 11 = 0$. If $P$ lies below $L_1$ and above $L_2$, then $100(\alpha + \beta)$ is equal to
(1) $-14$
(2) 42
(3) $-22$
(4) 14

The equations of the sides $A B , B C$ and $C A$ of a triangle $A B C$ are $2 x + y = 0 , x + p y = 39$ and $x - y = 3$ respectively and $P ( 2,3 )$ is its circumcentre. Then which of the following is NOT true
(1) $( A C ) ^ { 2 } = 9 p$
(2) $( A C ) ^ { 2 } + p ^ { 2 } = 136$
(3) $32 <$ area $( \triangle A B C ) < 36$
(4) $34 <$ area $( \triangle A B C ) < 38$

Let $m_1, m_2$ be the slopes of two adjacent sides of a square of side $a$ such that $a^2 + 11a + 3(m_1^2 + m_2^2) = 220$. If one vertex of the square is $(10\cos\alpha - \sin\alpha, 10\sin\alpha + \cos\alpha)$, where $\alpha \in \left(0, \frac{\pi}{2}\right)$ and the equation of one diagonal is $(\cos\alpha - \sin\alpha)x + (\sin\alpha + \cos\alpha)y = 10$, then $72(\sin^4\alpha + \cos^4\alpha) + a^2 - 3a + 13$ is equal to
(1) 119
(2) 128
(3) 145
(4) 155

Let a circle $C$ of radius 5 lie below the $x$-axis. The line $L _ { 1 } : 4 x + 3 y + 2 = 0$ passes through the centre $P$ of the circle $C$ and intersects the line $L _ { 2 } : 3 x - 4 y - 11 = 0$ at $Q$. The line $L _ { 2 }$ touches $C$ at the point $Q$. Then the distance of $P$ from the line $5 x - 12 y + 51 = 0$ is

The equations of the sides $AB , BC$ and $CA$ of a triangle $ABC$ are $2 x + y = 0 , x + p y = 15 a$ and $x - y = 3$ respectively. If its orthocentre is $( 2 , a ) , - \frac { 1 } { 2 } < a < 2$, then $p$ is equal to $\_\_\_\_$.

Let $( \alpha , \beta )$ be the centroid of the triangle formed by the lines $15 x - y = 82$, $6 x - 5 y = - 4$ and $9 x + 4 y = 17$. Then $\alpha + 2 \beta$ and $2 \alpha - \beta$ are the roots of the equation
(1) $x ^ { 2 } - 7 x + 12 = 0$
(2) $x ^ { 2 } - 14 x + 48 = 0$
(3) $x ^ { 2 } - 13 x + 42 = 0$
(4) $x ^ { 2 } - 10 x + 25 = 0$

If $( \alpha , \beta )$ is the orthocenter of the triangle $ABC$ with vertices $A ( 3 , - 7 ) , B ( - 1,2 )$ and $C ( 4,5 )$, then $9 \alpha - 6 \beta + 60$ is equal to
(1) 25
(2) 35
(3) 30
(4) 40

If the orthocentre of the triangle, whose vertices are $(1,2)$, $(2,3)$ and $(3,1)$ is $(\alpha, \beta)$, then the quadratic equation whose roots are $\alpha + 4\beta$ and $4\alpha + \beta$, is
(1) $x^2 - 19x + 90 = 0$
(2) $x^2 - 18x + 80 = 0$
(3) $x^2 - 22x + 120 = 0$
(4) $x^2 - 20x + 99 = 0$

The equations of the sides $AB , BC \& CA$ of a triangle $ABC$ are $2x + y = 0 , x + py = 21a ( a \neq 0 )$ and $x - y = 3$ respectively. Let $P ( 2 , a )$ be the centroid of the triangle $ABC$, then $( BC ) ^ { 2 }$ is equal to

The distance of the point $( 6 , - 2 \sqrt { 2 } )$ from the common tangent $y = m x + c , m > 0$, of the curves $x = 2 y ^ { 2 }$ and $x = 1 + y ^ { 2 }$ is
(1) $\frac { 1 } { 3 }$
(2) 5
(3) $\frac { 14 } { 3 }$
(4) $5 \sqrt { 3 }$

A light ray emits from the origin making an angle $30 ^ { \circ }$ with the positive $x$-axis. After getting reflected by the line $\mathrm { x } + \mathrm { y } = 1$, if this ray intersects x-axis at Q , then the abscissa of Q is
(1) $\frac { 2 } { ( \sqrt { 3 } - 1 ) }$
(2) $\frac { 2 } { 3 + \sqrt { 3 } }$
(3) $\frac { 2 } { 3 - \sqrt { 3 } }$
(4) $\frac { \sqrt { 3 } } { 2 ( \sqrt { 3 } + 1 ) }$

The equations of sides $AB$ and $AC$ of a triangle $ABC$ are $( \lambda + 1 ) x + \lambda y = 4$ and $\lambda x + ( 1 - \lambda ) y + \lambda = 0$ respectively. Its vertex $A$ is on the $y$-axis and its orthocentre is $( 1,2 )$. The length of the tangent from the point $C$ to the part of the parabola $y ^ { 2 } = 6 x$ in the first quadrant is
(1) $\sqrt { 6 }$
(2) $2 \sqrt { 2 }$
(3) 2
(4) 4

A triangle is formed by $X$-axis, $Y$-axis and the line $3 x + 4 y = 60$. Then the number of points $P ( a , b )$ which lie strictly inside the triangle, where $a$ is an integer and $b$ is a multiple of $a$, is $\_\_\_\_$.

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 $\mathrm { y } - 2 \mathrm { x } = 2$ such that $\triangle A B C$ is an equilateral triangle. Then, the area of the $\triangle A B C$ is
(1) $3 \sqrt { 3 }$
(2) $2 \sqrt { 3 }$
(3) $\frac { 8 } { \sqrt { 3 } }$
(4) $\frac { 10 } { \sqrt { 3 } }$

A line passing through the point $A ( 9,0 )$ makes an angle of $30 ^ { \circ }$ with the positive direction of $x$-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
(1) $\frac { y } { \sqrt { 3 } - 2 } + x = 9$
(2) $\frac { x } { \sqrt { 3 } - 2 } + y = 9$
(3) $\frac { x } { \sqrt { 3 } + 2 } + y = 9$
(4) $\frac { y } { \sqrt { 3 } + 2 } + x = 9$

Let $\alpha , \quad \beta , \quad \gamma , \quad \delta \in Z$ and let $A(\alpha , \beta)$, $B(1, 0)$, $C(\gamma , \delta)$ and $D(1, 2)$ be the vertices of a parallelogram $ABCD$. If $AB = \sqrt { 10 }$ and the points $A$ and $C$ lie on the line $3 y = 2 x + 1$, then $2 \alpha + \beta + \gamma + \delta$ is equal to
(1) 10
(2) 5
(3) 12
(4) 8

Let a variable line of slope $m > 0$ passing through the point $( 4 , - 9 )$ intersect the coordinate axes at the points $A$ and $B$. The minimum value of the sum of the distances of $A$ and $B$ from the origin is
(1) 30
(2) 25
(3) 15
(4) 10

For $\alpha, \beta \in \left(0, \frac{\pi}{2}\right)$ let $3\sin(\alpha + \beta) = 2\sin(\alpha - \beta)$ and a real number $k$ be such that $\tan\alpha = k\tan\beta$. Then the value of $k$ is equal to
(1) $-5$
(2) $5$
(3) $\frac{2}{3}$
(4) $-\frac{2}{3}$

The portion of the line $4 x + 5 y = 20$ in the first quadrant is trisected by the lines $\mathrm { L } _ { 1 }$ and $\mathrm { L } _ { 2 }$ passing through the origin. The tangent of an angle between the lines $L _ { 1 }$ and $L _ { 2 }$ is:
(1) $\frac { 8 } { 5 }$
(2) $\frac { 25 } { 41 }$
(3) $\frac { 2 } { 5 }$
(4) $\frac { 30 } { 41 }$

Let $A ( - 1,1 )$ and $B ( 2,3 )$ be two points and $P$ be a variable point above the line $A B$ such that the area of $\triangle \mathrm { PAB }$ is 10 . If the locus of P is $\mathrm { a } x + \mathrm { b } y = 15$, then $5 \mathrm { a } + 2 \mathrm {~b}$ is :
(1) 6
(2) $- \frac { 6 } { 5 }$
(3) 4
(4) $- \frac { 12 } { 5 }$

The equations of two sides AB and AC of a triangle ABC are $4 x + y = 14$ and $3 x - 2 y = 5$, 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
(1) $x + 3 y + 2 = 0$
(2) $x - 6 y - 10 = 0$
(3) $x - 3 y - 6 = 0$
(4) $x + 6 y + 6 = 0$

Two vertices of a triangle ABC are $\mathrm { A } ( 3 , - 1 )$ and $\mathrm { B } ( - 2,3 )$, and its orthocentre is $\mathrm { P } ( 1,1 )$. If the coordinates of the point C are $( \alpha , \beta )$ and the centre of the of the circle circumscribing the triangle PAB is $( \mathrm { h } , \mathrm { k } )$, then the value of $( \alpha + \beta ) + 2 ( \mathrm {~h} + \mathrm { k } )$ equals
(1) 5
(2) 81
(3) 15
(4) 51

Let $A$ be the point of intersection of the lines $3 x + 2 y = 14,5 x - y = 6$ and $B$ be the point of intersection of the lines $4 x + 3 y = 8,6 x + y = 5$. The distance of the point $P ( 5 , - 2 )$ from the line $A B$ is
(1) $\frac { 13 } { 2 }$
(2) 8
(3) $\frac { 5 } { 2 }$
(4) 6

Let $A(a, b)$, $B(3, 4)$ and $(-6, -8)$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $P(2a+3, 7b+5)$ from the line $2x + 3y - 4 = 0$ measured parallel to the line $x - 2y - 1 = 0$ is
(1) $\dfrac{15\sqrt{5}}{7}$
(2) $\dfrac{17\sqrt{5}}{6}$
(3) $\dfrac{17\sqrt{5}}{7}$
(4) $\dfrac{\sqrt{5}}{17}$

In a $\triangle \mathrm { ABC }$, suppose $\mathrm { y } = \mathrm { x }$ is the equation of the bisector of the angle $B$ and the equation of the side $A C$ is $2 x - y = 2$. If $2 A B = B C$ and the point $A$ and $B$ are respectively $( 4,6 )$ and $( \alpha , \beta )$, then $\alpha + 2 \beta$ is equal to
(1) - 4
(2) 42
(3) 2
(4) - 1