Let, $A$ be $(a+b,b+c)$ and $B$ be $(a-b,c-b)$
$\because$ Distance between the points $(x_1,y_1)\text{and}(x_2,y_2)$ is 
$\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
$\therefore AB=\sqrt{(a-b-a-b{)}^2+(c-b-b-c{)}^2}=\sqrt{(-2b{)}^2+(-2b{)}^2}=\sqrt{8b^2}AB=2\sqrt{2}b$
$\therefore$ Distance between the points = $2\sqrt{2}b$

The given points A(-1, 3), B(2, p), C(5, -1) are collinear.

⇒ Area  ΔABC formed by these points should be zero.

⇒ The area of  ΔABC = 0

⇒  $\frac{1}{2}$( $x_1$( $y_2$- $y_3$) +  $x_2$( $y_3$- $y_1$) +  $x_3$( $y_1$ -  $y_2$)) = 0

⇒ -1(p + 1) + 2 (-1 - 3) + 5(3 -p) = 0 

⇒ -p - 1 - 8 + 15 - 5p = 0

⇒ -6p +15 - 9 = 0 ⇒ -6p = -6 ⇒ p = 1

Hence the value of p is 1.

A and B are equidistant from origin and AB is $\sqrt{2}$a.
If we assume the points A and B to be (x,0) and (0,x) respectively,
AB= $\sqrt{x^2+x^2}$ = $\sqrt{2}$x = $\sqrt{2}$a
x=a
So A(0,a) and B(a,0) are the vertices.
Observe that A,B both lie on the line x+y=a and since C also lies on the line y+x = a, they are collinear.
Hence area of the triangle formed by the three points is 0.

Concept : 1 Mark
Application : 1 Mark
Let the coordinates of the third vertex be (x, y). 
We know that the coordinates of the centroid of a triangle whose vertices are
$(x_1,y_1)$,$(x_2,y_2)$ and $(x_3,y_3)$ are
$\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}$
Given points are (3, –5) and (–7, 4) and centroid is (2, –1).
$\Rightarrow \frac{x+3-7}{3}=2$ and $\frac{y-5+4}{3}=-1$
$\Rightarrow x-4=6$ and $y-1=-3$
$\Rightarrow x=10$ and $y=-2$
Thus the coordinates of the third vertex are (10, -2).

Concept : 1 Mark
Application : 1 Mark
Distance between points $A(x_1,y_1)$ and $B(x_2,y_2)$ is,
$D=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
$D=\sqrt{{(4-(-3\left)\right)}^2+{(5-(-3\left)\right)}^2}$
$D=\sqrt{7^2+8^2}$
$D=\sqrt{113}units$

Concept : 1 Mark
Application : 1 Mark
A (1 , -1) , B (5 , 2) and C (9 , 5) are the given points. Then, We have 
Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$
$AB=\sqrt{{(5-1)}^2+{(2+1)}^2}$
So,
$AB=\sqrt{16+9}=5$
$BC=\sqrt{{(5-9)}^2+{(2-5)}^2}$
$BC=\sqrt{16+9}=5$
and , $AC=\sqrt{{(1-9)}^2+{(-1-5)}^2}$
$AC=\sqrt{64+36}=10$
Clearly, AC = AB + BC
Hence A, B, C are collinear points

Concept : 1 Mark
Application : 1 Mark
Let p (x , -1) and Q (3 ,2) be the given points. Then,
PQ = 5 (Given)
Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$
$\Rightarrow \sqrt{{(x-3)}^2+{(-1-2)}^2}=5$
$\Rightarrow {(x-3)}^2+9=5^2$     [Squaring both sides]
$\Rightarrow x^2-6x+18=25$ 
$\Rightarrow x^2-6x-7=0$
$\Rightarrow (x-7)(x+1)=0$
$\Rightarrow x=7orx=-1$

Concept: 1 Mark
Application: 1 Mark
Answer: 1 Mark 
Let the required ratio be $\lambda :1$. Then, the coordinates of the point of division are,
$R\left(\frac{5\lambda +2}{\lambda +1},\frac{6\lambda -3}{\lambda +1}\right)$

But, it is a point on x-axis on which y-coordinates of every point is zero.
$\therefore \frac{6\lambda -3}{\lambda +1}=0\Rightarrow \lambda =\frac{1}{2}$
Thus, the required ratio is $\frac{1}{2}:1$ or 1:2.
Substituting $\lambda =\frac{1}{2}$ in the coordinates of R.
$R\left(\frac{5\lambda +2}{\lambda +1},0\right)$
$\Rightarrow R\left(\frac{\frac{5}{2}+2}{\frac{1}{2}+1},0\right)$
$\Rightarrow R(\frac{\frac{9}{2}}{\frac{3}{2}},0)$
$\Rightarrow R(3,0)$
$\therefore$ the coordinates of R are (3, 0)

Formula: 1 Mark
Concept:1 Mark
Answer: 1 Mark
Let the coordinates of B be $(\alpha ,\beta )$.
It is given that AC:BC = 3:4.

So, the coordinates of C are:
$C(x,y)=\left(\frac{m{x}_2+n{x}_1}{m+n}\right),\left(\frac{m{y}_2+n{y}_1}{m+n}\right)$
$\Rightarrow C(-1,2)=\frac{3\alpha +4\times 2}{3+4},\frac{3\beta +4\times 5}{3+4}$
$\Rightarrow C(-1,2)=\frac{3\alpha +8}{7},\frac{3\beta +20}{7}$
$\Rightarrow \frac{3\alpha +8}{7}=-1and\frac{3\beta +20}{7}=2$
$\Rightarrow \alpha =-5and\beta =-2$
Thus, the coordinates of B are (-5, -2).