Straight Lines and Pair of Straight Lines

Lines are drawn parallel to the line 4x – 3y + 2 = 0, at a distance ${3 \over 5}$ from the origin. Then which one of the following points lies on any of these lines ?

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 :

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 two straight lines drawn from the origin $\mathrm{O}$ intersect the line $3 x+4 y=12$ at the points $\mathrm{P}$ and $\mathrm{Q}$ such that $\triangle \mathrm{OPQ}$ is an isosceles triangle and $\angle \mathrm{POQ}=90^{\circ}$. If $l=\mathrm{OP}^2+\mathrm{PQ}^2+\mathrm{QO}^2$, then the greatest integer less than or equal to $l$ is :

The lines $p\left( {{p^2} + 1} \right)x - y + q = 0$ and $\left( {{p^2} + 1} \right){}^2x + \left( {{p^2} + 1} \right)y + 2q$ $=0$ are perpendicular to a common line for :

A ray of light along $x + \sqrt 3 y = \sqrt 3$ gets reflected upon reaching $X$-axis, the equation of the reflected ray is :

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 :

If the straight line, 2x – 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, $\beta$), then $\beta$ equals :

Suppose that the points (h,k), (1,2) and (–3,4) lie on the line L₁ . If a line L₂ passing through the points (h,k) and (4,3) is perpendicular to L₁ , then $k \over h$ equals :

If the two lines x + (a – 1) y = 1 and 2x + a²y = 1 (a$\in$R – {0, 1}) are perpendicular, then the distance of their point of intersection from the origin is :

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 $\mathrm{L}_1$ and $\mathrm{L}_2$ is :

If the line segment joining the points $(5,2)$ and $(2, a)$ subtends an angle $\frac{\pi}{4}$ at the origin, then the absolute value of the product of all possible values of $a$ is :

Let $P = \left( { - 1,0} \right),\,Q = \left( {0,0} \right)$ and $R = \left( {3,3\sqrt 3 } \right)$ be three point. The equation of the bisector of the angle $PQR$ is :

The lines ${L_1}:y - x = 0$ and ${L_2}:2x + y = 0$ intersect the line ${L_3}:y + 2 = 0$ at $P$ and $Q$ respectively. The bisector of the acute angle between ${L_1}$ and ${L_2}$ intersects ${L_3}$ at $R$. Statement-1: The ratio $PR$ : $RQ$ equals $2\sqrt 2 :\sqrt 5$ Statement-2: In any triangle, bisector of an angle divide the triangle into two similar triangles.

Let ABC be a triangle with A($-$3, 1) and $\angle$ACB = $\theta$, 0 < $\theta$ < ${\pi \over 2}$. If the equation of the median through B is 2x + y $-$ 3 = 0 and the equation of angle bisector of C is 7x $-$ 4y $-$ 1 = 0, then tan$\theta$ is equal to :

The combined equation of the two lines $ax+by+c=0$ and $a'x+b'y+c'=0$ can be written as $(ax+by+c)(a'x+b'y+c')=0$. The equation of the angle bisectors of the lines represented by the equation $2x^2+xy-3y^2=0$ is :

In a $\triangle A B C$, suppose $y=x$ 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 points $A$ and $B$ are respectively $(4,6)$ and $(\alpha, \beta)$, then $\alpha+2 \beta$ is equal to

Let A $\left( {h,k} \right)$, B$\left( {1,1} \right)$ and C $(2, 1)$ be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is $1$ square unit, then the set of values which $'k'$ can take is given by :

The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices $(0, 0)$ $(0, 41)$ and $(41, 0)$ is :