Equations of a line

Consider a triangle $\Delta$ whose two sides lie on the $x$-axis and the line $x + y + 1 = 0$. If the orthocenter of $\Delta$ is $(1, 1)$, then the equation of the circle passing through the vertices of the triangle $\Delta$ is

The area of the region
$$\left\{(x, y) : 0 \le x \le \frac94,\quad 0 \le y \le 1,\quad x \ge 3y,\quad x + y \ge 2\right\}$$
is

Let k be an integer such that the triangle with vertices $(k,-3k)$, $(5,k)$ and $(-k,2)$ has area 28 sq. units. Then the orthocentre of this triangle is at the point :

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 point (2, 1) is translated parallel to the line L : x − y = 4 by $2\sqrt3$ 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 :

In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively $3\hat i+\hat j-\hat k$, $-\hat i+3\hat j+p\hat k$ and $5\hat i+q\hat j-4\hat k$, then the point (p, q) lies on a line :

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 L be the line passing through the point P(1, 2) such that its intercepted segment between the co-ordinate axes is bisected at P. If $L_1$ is the line perpendicular to L and passing through the point (−2, 1), then the point of intersection of L and $L_1$ 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 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 $a$, $b$, $c$ 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 :

Given three points P, Q, R with P(5, 3) and R lies on the $x$-axis. If equation of RQ is $x-2y=2$ and PQ is parallel to the $x$-axis, then the centroid of $\Delta PQR$ lies on the line :

If a line intercepted between the coordinate axes is trisected at a point A(4, 3), which is nearer to $x$-axis, then its equation is :

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 :

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

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

The orthocenter of the triangle PAB is

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

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