Straight Lines

The circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points $(a^2+1,a^2+1)$ and $(2a,-2a)$, $a\ne0$. Then for any a, the orthocentre of this triangle lies on the line :

If a line L is perpendicular to the line $5x-y=1$, and the area of the triangle formed by the line L and the coordinate axes is 5, then the distance of line L from the line $x+5y=0$ is :

A chord is drawn through the focus of the parabola $y^2=6x$ such that its distance from the vertex of this parabola is $\dfrac{\sqrt5}2$, then its slope can be :

For $a>b>c>0$, the distance between $(1,1)$ and the point of intersection of the lines $ax+by+c=0$ and $bx+ay+c=0$ is less than $2\sqrt2$. Then

$\displaystyle\min_{z\in S}|1-3i-z|=$

If chord $PQ$ subtends an angle $\theta$ at the vertex of $y^2=4ax$, then $\tan\theta=$

The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line $4x-5y=20$ to the circle $x^2+y^2=9$ is

A possible equation of $L$ is

A straight line $L$ through the point $(3,-2)$ is inclined at an angle 60° to the line $\sqrt3x+y=1$. If $L$ also intersects the $x$-axis, then the equation of $L$ is

Let $(x,y)$ be any point on the parabola $y^2=4x$. Let $P$ be the point that divides the line segment from $(0,0)$ to $(x,y)$ in the ratio $1:3$. Then the locus of $P$ is

The straight line $2x-3y=1$ divides the circular region $x^2+y^2\le6$ into two parts. If
$$S=\left\{\left(2,\frac34\right),\left(\frac52,\frac34\right),\left(\frac14,-\frac14\right),\left(\frac18,\frac14\right)\right\},$$
then the number of point(s) in $S$ lying inside the smaller part is

The orthocenter of the triangle PAB is

The equation of the locus of the point whose distances from the point P and the line AB are equal, is

The normal at a point $P$ on the ellipse $x^2+4y^2=16$ meets the $x$-axis at $Q$. If $M$ is the mid point of the line segment $PQ$, then the locus of $M$ intersects the latus rectums of the given ellipse at the points

The locus of the orthocentre of the triangle formed by the lines
$$(1+p)x-py+p(1+p)=0,\qquad(1+q)x-qy+q(1+q)=0,$$
and $y=0$, where $p\ne q$, is

The tangent $PT$ and the normal $PN$ to the parabola $y^2=4ax$ at a point $P$ on it meet its axis at points $T$ and $N$, respectively. The locus of the centroid of the triangle $PTN$ is a parabola whose

Let $a$ and $b$ be non-zero real numbers. Then, the equation
$$(ax^2+by^2+c)(x^2-5xy+6y^2)=0$$
represents

The equation of circle $C$ is

Equations of the sides $QR$, $RP$ are

Consider three points $P=(-\sin(\beta-\alpha),-\cos\beta)$, $Q=(\cos(\beta-\alpha),\sin\beta)$ and $R=(\cos(\beta-\alpha+\theta),\sin(\beta-\theta))$, where $0<\alpha,\beta,\theta<\frac\pi4$. Then,