The origin and the points where the line $L_1$ intersect the $x$ -axis and $y$ -axis are vertices of right angled triangle $T$ whose area is $8$. Also the line $L_1$ is perpendicular to line $L_2$ : $4x -y = 3$, then perimeter of triangle $T$ is -

If the equation of base of an equilateral triangle is $2x - y = 1$ and the vertex is $(-1, 2)$, then the length of the side of the triangle is

$P (x, y)$ moves such that the area of the triangle formed by $P, Q (a , 2 a)$ and $R (- a, - 2 a)$ is equal to the area of the triangle formed by $P, S (a, 2 a)\,\,\, \&\,\, \,T (2 a, 3 a)$. The locus of $'P'$ is a straight line given by :

The line $\frac{x}{a} + \frac{y}{b} = 1$ moves in such a way that $\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{2{c^2}}}$, where $a, b, c \in R_0$ and $c$ is constant, then locus of the foot of the perpendicular from the origin on the given line is -

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

The vertices of a triangle are $\mathrm{A}(-1,3), \mathrm{B}(-2,2)$ and $\mathrm{C}(3,-1)$. $A$ new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :