Straight lines practice · Mixed difficulty

Let $m_1, m_2$ be the slopes of two adjacent sides of a square of side a such that $a^2+11 a+3\left(m_1^2+m_2^2\right)$ $=220$. 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 $72\left(\sin ^4 \alpha+\cos ^4 \alpha\right)$ $+a^2-3 a+13$ is equal to :

Normal form of line $x+y+1=0$ is

Consider the following statements.

I. The equation of a straight line making intercepts $a$ and $b$ on $x$ and $y$-axis respectively is given by $\frac{x}{a}+\frac{y}{b}=1$

II. If $a x+b y+c_1=0$ and $a x+b y+c_2=0$ be two parallel lines, then distance between two parallel lines, $d=\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right|$.

Choose the correct option.

A line has slope m and y-intercept 4. The distance between the origin and the line is equal to

If $\left( { - 2,6} \right)$ is the image of the point (4,2) with respect to the line L=0, then L=

The straight line $y=m x+c$ cuts the circle $x^2+y^2=a^2$ at the real points if

If the three distinct lines $x+2 a y+a=0, x+3 b y+b=0$ and $x+4 a y+a=0$ are concurrent, then the point $(a, b)$ lies on a:

If $A$ and $B$ are two points having coordinates (3, 4) and $\left( {5, - 2} \right)$ respectively and $P$ is a point such that $PA = PB$ and area of triangle $PAB = 10$ sq unit, then the coordinates of $P$ are

If sum of distances of a point from the origin and lines $x=2$ is 4 , then its locus is

The image of the point (3,5) in the line $x - y + 1 = 0$, lies on

If two vertices of a triangle are $(5,-1)$ and $(-2,3)$, and if its orthocentre lies at the origin, then the coordinates of the third vertex, are

If $O$ is the origin and $P\left( {{x_1},{y_1}} \right),Q\left( {{x_2},{y_2}} \right)$ are two points, then $OP.OQ\sin \angle POQ =$

The points $(p+1,1),(2 p+1,3)$ and $(2 p+2,2 p)$ are collinear, if

The number of integral points (i.e., the points whose $x$ and $y$-coordinates are both integers) within the area (not including the boundary) enclosed by the right branch of hyperbola $x^2-y^2=1$ and the line $x=100$ is

Let $C$ be the centroid of the triangle with vertices (3,-1) and (1,3) and (2,4). Let $P$ be the point of intersection of the lines $x + 3y - 1 = 0$ and $3x - y + 1 = 0.$ Then the line passing through the points $C$ and $P$ also passes through the point

If points $A\left( {{x_1},{y_1}} \right),\;B\left( {{x_2},{y_2}} \right)$ and $C\left( {{x_3},{y_3}} \right)$ are such that $C\left( {{x_3},{y_3}} \right)$ and ${y_1},{y_2},{y_3}$ are in AP, then

The distance between $4 x+3 y=11$ and $8 x+6 y=15$, is

Ratio in which the join of $(2,1)$ and $(-1,2)$ is divided by the line $x+3 y+5=0$ is -

The angle between the lines represented by the equation $a x^2+x y+b y^2=0$ will be $45^{\circ}$, if

The equation of the bisector of that angle between the lines $x+2 y-11=0$, $3 x-6 y-5=0$ which contains the point $(1,-3)$ is

If the straight line through the point $P(3,4)$ makes an angle $\frac{\pi}{6}$ with the $x$-axis and meets the line $12 x+5 y+10=0$ at $Q$, then the length $P Q$ is $\frac{132}{A \sqrt{3}+B}$. Find $A+B$.

If the point $(p, 5)$ lies on the line parallel to $y$-axis and passing through the intersection of the lines $2\left(a^2+1\right) x+b y+4\left(a^3+a\right)=0$ and $\left(a^2+1\right) x-6 b y+2\left(a^3+a\right)=0$, then $p$ is equal to

The equation of line through $(1,2)$ and parallel to $3 x-y-4=0$ is-

A line passing through the point $A(9,0)$ makes 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 :

If the angle between the lines represented by the equation $y^2+k x y-x^2 \tan ^2 A=0$ be $2 A$, then $k=$

$P\left( {3,\;1} \right),Q\left( {6,5} \right)$ and $R\left( {x,y} \right)$ are three points such that the angle PRQ is a right angle and the area of $\Delta \;RQP = 7,$ then the number of such points R is

The equation of the straight line equally inclined to the coordinate axes and equidistant from the point $(1,-2)$ and $(3,4)$, is

A line with positive rational slope, which passes through the point $(6,0)$ and at a distance of 5 units from the point $(1,3)$ is equal to $\frac{p}{q}$ (where $p, q$ are co-prime), then find the value of $(p+q)$.

A straight line through P(1,2) is such that its intercept between the axes is bisected at P. Its equation is

$A \equiv(-4,0), B \equiv(4,0), M$ and $N$ are variable points on $Y$-axis such that $M$ lies below $N$ and $M N=4$. If the line joining $A M$ and $B N$ intersect at $P$, then locus of $P$ is