The equation of the base of an equilateral triangle is $x+y=2$ and one vertex is $(2,-1)$. The length of the side of the triangle is:

The vertices of a triangle are $A(-1, 3)$,$B(-2, 2)$,and $C(3, -1)$. $A$ 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 the origin is:

For an integer $K$,if the point $P(K^2, K+1)$ and the origin $O(0,0)$ lie in the same region between the lines $x+2y-5=0$ and $3x-y+1=0$,then the possible number of such points $P$ is

The line $L$ given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $(13,32)$. The line $K$ is parallel to line $L$ and has the equation $\frac{x}{c}+\frac{y}{3}=1$. Then the distance between $L$ and $K$ is $\qquad$ units.

Let the equation $x(x+2)(12-k)=2$ have equal roots. Then the distance of the point $(k, \frac{k}{2})$ from the line $3x+4y+5=0$ is

$A$ line $L_1$ passing through the point of intersection of the lines $x-2y+3=0$ and $2x-y=0$ is parallel to the line $L_2$. If $L_2$ passes through the origin and also through the point of intersection of the lines $3x-y+2=0$ and $x-3y-2=0$,then the distance between the lines $L_1$ and $L_2$ is