Let $A B C D$ be a square of side length $1$ . Let $P, Q, R, S$ be points in the interiors of the sides $A D, B C, A B, C D$ respectively, such that $P Q$ and $R S$ intersect at right angles. If $P Q=\frac{3 \sqrt{3}}{4}$, then $R S$ equals
$\frac{2}{\sqrt{3}}$
$\frac{3 \sqrt{3}}{4}$
$\frac{\sqrt{2}+1}{2}$
$4-2 \sqrt{2}$
Area of the parallelogram formed by the lines ${a_1}x + {b_1}y + {c_1} = 0$,${a_1}x + {b_1}y + {d_1} = 0$and ${a_2}x + {b_2}y + {c_2} = 0$, ${a_2}x + {b_2}y + {d_2} = 0$is
Let $A B C$ be an isosceles triangle in which $A$ is at $(-1,0), \angle A=\frac{2 \pi}{3}, A B=A C$ and $B$ is on the positive $\mathrm{x}$-axis. If $\mathrm{BC}=4 \sqrt{3}$ and the line $\mathrm{BC}$ intersects the line $y=x+3$ at $(\alpha, \beta)$, then $\frac{\beta^4}{\alpha^2}$ is :
lf 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
Let $B$ and $C$ be the two points on the line $y+x=0$ such that $B$ and $C$ are symmetric with respect to the origin. Suppose $A$ is a point on $y -2 x =2$ such that $\triangle ABC$ is an equilateral triangle. Then, the area of the $\triangle ABC$ is
Co-ordinates of the orthocentre of the triangle whose vertices are $A(0, 0) , B(3, 4)$ and $C(4, 0)$ is