Let the three sides of a triangle be on the lines $4x-7y+10=0$, $x+y=5$ and $7x+4y=15$. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines $x=0$, $y=0$ and $x+y=1$ is
Let the area of $\triangle PQR$ with vertices $P(5,4),\ Q(-2,4)$ and $R(a,b)$ be $35$ square units. If its orthocenter and centroid are $O\!\left(2,\dfrac{14}{5}\right)$ and $C(c,d)$ respectively, then $c+2d$ is equal to:
A ray of light coming from the point (2, $2\sqrt 3 $) is incident at an angle 30o on the line x = 1 at thepoint A. The ray gets reflected on the line x = 1 and meets x-axis at the point B. Then, the line ABpasses through the point :
A straight line $L$ at a distance of $4$ units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of $60^\circ$ with the line $x + y = 0$.
Then an equation of the line $L$ 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 :
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 $A$ be the point of intersection of the lines $L_{1}:\ \dfrac{x-7}{1}=\dfrac{y-5}{0}=\dfrac{z-3}{-1}$ and $L_{2}:\ \dfrac{x-1}{3}=\dfrac{y+3}{4}=\dfrac{z+7}{5}$. Let $B$ and $C$ be the points on the lines $L_{1}$ and $L_{2}$ respectively such that $AB=AC=\sqrt{15}$. Then the square of the area of the triangle $ABC$ is: