Distance Formula

$\left(1,2\right),\left(-1,3\right),\left(3,-1\right)$ and $\left(1,0\right)$ are the vertices of a

A straight road is at a distance of $5\sqrt{2}$ miles from a place. The shortest distance of the road from the place is in the $\mathrm{N}.\mathrm{E}.$ direction. Do the following villages which is $4$ miles East and $3$ miles North from the place, lie on the road or not?

The distance of the point $(1, 2)$ from the line $x+y=0$ measured parallel to the line $3x-y=2$ is

Two particles start from the same point $(2,-1),$ one moving $2$ units along the line $x+y=1$ and the other $5$ units along$x-2 y=4 .$If the particles move towards increasing $y$, find their new positions and the distance between them.

A straight road is at a distance of $5\sqrt{2}$miles from a place. The shortest distance of the road from the place is in the $\mathrm{N}.\mathrm{E}.$ direction. Do the following villages which is $6$ miles East and $4$ miles North from the place, lie on the road or not ?

Let $\mathrm{P}\equiv (a, b), \mathrm{Q}\equiv (c, d)$ and $0<a<b<c<d, \mathrm{L}\equiv (a, 0), \mathrm{M}\equiv (c, 0),$ $R$ lies on $x$ -axis such that $\mathrm{PR} + \mathrm{RQ}$ is minimum, then $\mathrm{R}$ divides $\mathrm{LM}$.
 

A man starts from the point $P (-3, 4)$ and reaches point $Q (0, 1)$ touching $x$ -axis at $R (\alpha , 0)$ such that $\mathrm{PR} + \mathrm{RQ}$ is minimum,
then $\alpha =$
 

Let $A\equiv (-4,0),B\equiv (-1,4).C$ and $D$ are points which are symmetric to points $A$ and $B$ respectively with respect to $y$ -axis, then area of the quadrilateral $ABCD$ is

The distance between two parallel lines is unity. A point $P$ lies between the two lines at a distance $a$ from one of them. $PQR$ is an equilateral triangle having side $2\sqrt{\left(\frac{7}{3}\right)}$  and vertex $Q$ on one of the parallel lines and $\mathit{\text{R}}$ on the other line, then $a=$

$\mathrm{A}(-4,0)$ and $\mathrm{B}(-1,4)$ are two given points.$\mathrm{C}$ and $\mathrm{D}$ are points which are symmetric to the given points $\mathrm{A}$and $\mathrm{B}$ respectively with respect to$y$ -axis. Calculate the perimeter of the trapezium $\mathrm{ABDC}$.

$A(1,-2)$ and $B(2,5)$ are two points. The lines $OA, OB$ are produced to $C$ and $D$ respectively such that $OC = 2OA$ and $OD = 2OB$. Find $CD$.