Given point P(x, y) is equidistant from the points A(7, 1) and B(3, 5).
So,   AP = BP
\(\Rightarrow\)   AP² = BP²
\(\Rightarrow\) (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²
       \(\left[\because \text { distance }=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\right]\)
\(\Rightarrow\)   x² + 49 - 14x + y² + 1 - 2y
                = x² + 9 - 6x + y² + 25 - 10y
\(\Rightarrow\)  -14x - 2y + 50 = -6x - 10y + 34
\(\Rightarrow\) -6x - 10y + 14x + 2y = 50 - 34
\(\Rightarrow\)              8x - 8y = 16
\(\Rightarrow\)                 x - y = 2
                                [dividing by 8 on both sides]
Hence, the relation between x and y is x - y = 2.

Here, A=(x₁,y₁)=(1, 3) and B=(x₂,y₂)=(a, 7)
Distance between A and B is AB = \(\sqrt { { \left( a-1 \right) }^{ 2 }+{ (7-3) }^{ 2 } } \)  [\(\because \) distance=\(\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }-{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } \)]
\(\Rightarrow \)   5=\(\sqrt { { \left( a-1 \right) }^{ 2 }+{ (4) }^{ 2 } } \)      [\(\because \) distance, AB=5]
\(\Rightarrow \)   5=\(\sqrt { { a }^{ 2 }+1-2a+16 } \)
\(\Rightarrow \)  25=\({ a }^{ 2 }\)-2a+17        [squaring on both sides]
\(\Rightarrow \)  \({ a }^{ 2 }\)-2a-8=0    
\(\Rightarrow \)  \({ a }^{ 2 }\)-4a+2a-8=0
\(\Rightarrow \)  a(a-4)+2(a-4)=0
\(\Rightarrow \)  (a-4)(a+2) =0 \(\Rightarrow \)   a=4, -2
Hence, the possible values of a are 4 and -2.

Given points are A(0,0) and B(36,15) 
Now, using distance formula, we have
AB = \(\sqrt {(36-0)^{2}+(15-0)^{2}}\)
=\(\sqrt{{36}^{2}+{15}^{2}}\) = \(\sqrt{1296+225}\)
=\(\sqrt{1521}\) =39 units

Let us take position of the town A as che origin and the position of the town B as the point B IS) in the coordinate axis as Shown in figure. 
Now, |AB|= \(\sqrt{(36^{2}+(15)^{2}}\)
=\(\sqrt{{36}^{2}+{15}^{2}}\) = \(\sqrt{1296+225}\)
=\(\sqrt{1521}\) =39 units

\(2\sqrt { 2 } \ \ units\)

Mid-points of diagonals of a rectangle coincide

\(\therefore \ \left( \frac { 13+4 }{ 2 } ,\frac { 1-2 }{ 4 } \right) =\left( \frac { x+7 }{ 2 } ,\frac { y+4 }{ 2 } \right) \)
⇒ x=10 and y=-5.
Hence, fourth vertex is D(10,-5)

4

Let the required ratio be k:1.
Then the coordinates of the point of division are \(\left( \frac { 2k+5 }{ k+1 } ,\frac { -3k+6 }{ k+1 } \right) \)
This point lies on the x-axis whose equation is y=0.
\(\therefore \ \frac { -3k+6 }{ k+1 } \ =\ 0\ \Rightarrow \ 3k=6,\ or\ k=2\)
 Line segment joining the two points is divided in the ratio 2:1 internally by x-axis. 

3

Let coordinates of the required point be R(x, y) this means R divides the join of P(4, -3) and Q(8, 5) in the ratio 3:1 internally.

Using the formula for internal division.
\(\left( \frac { { kx }_{ 2 }+{ x }_{ 1 } }{ k+1 } ,\frac { { ky }_{ 2 }+{ y }_{ 1 } }{ k+1 } \right) \)
\(x=\frac { 3(8)+1(4) }{ 3+1 } =\frac { 24+4 }{ 4 } =\frac { 28 }{ 4 } =7\) and \( y=\frac { 3(5)+1(-3) }{ 3+1 } =\frac { 15-3 }{ 4 } =\frac { 12 }{ 4 } =3\)
Thus, the coordinates of R (7, 3) divides PQ in the ratio 3:1.

coordinates of D be \(\left( \alpha ,\beta \right) \) P is mid-point of AC and BD.

\(\left( \frac { \alpha +4 }{ 2 } ,\frac { \beta +3 }{ 2 } \right) =\left( \frac { 1+6 }{ 2 } ,\frac { 2+6 }{ 2 } \right) \)
\(\Rightarrow \frac { \alpha +4 }{ 2 } =\frac { 7 }{ 2 } ;\frac { \beta +3 }{ 2 } =\frac { 8 }{ 2 } \)
\(\Rightarrow \alpha +4=7;\beta +3=8\Rightarrow \alpha =3;\beta =5\)
∴ Coordinates of D are (3, 3).

P, Q. R divide AB info four equals parts 

Now, Q is the mid-point of AB.
Coordinates of Q are \(\left( \frac { -2+2 }{ 2 } ,\frac { 2+8 }{ 2 } \right) \) =(0,5)
P is the mid-point of AQ, then the coordinates of P are
\(\frac { -2+0 }{ 2 } ,\frac { 2+5 }{ 2 } =\left( -1,\frac { 7 }{ 2 } \right) \)
R is the mid-point of QB, then the coordinates of R are
\(\left( \frac { 0+2 }{ 2 } ,\frac { 5+8 }{ 2 } \right) =\left( 1,\frac { 13 }{ 2 } \right) \)

Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram.
Since, ABCD is a parallelogram.
\(\therefore\) Diagonals AC and BD will bisect each other. So, the mid-point of AC and mid-point of BD will be same 

Thus mid-point of AC = Mid-point of BD
\(\begin{aligned} \Rightarrow \quad & \left(\frac{1+x}{2}, \frac{2+6}{2}\right)=\left(\frac{4+3}{2}, \frac{y+5}{2}\right) \\ \end{aligned}\)
\(\begin{aligned} & {\left[\because \text { coordinates of mid-point }=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\right] } \end{aligned}\)
On comparing the coordinate from both sides,we get
\(\frac{1+x}{2}=\frac{4+3}{2} \text { and } \frac{2+6}{2}=\frac{5+y}{2}\)
\(\Rightarrow\) 1 + x = 7 and 8 = 5 + y
\(\therefore\)   x = 6 and y = 3

Let P(x,0) divide the line segment AB in the ratio k:1

y-coordinate  \(y=\frac { k\times \left( -4 \right) +1\times \left( 1 \right) }{ k+1 } ,\quad y=\frac { -4k+1 }{ k+1 } \)
on x-axis y=0 ,  \(\therefore \quad 0=\frac { k(5)+1(-5) }{ k-1 } \)
\(0=5k-5\quad \Rightarrow \quad 5k=5\quad \Rightarrow \quad k=1\)
Hence the required ration is 1:1, at point  \(\left( \frac { -3 }{ 2 } ,0 \right) \) .

Let H(x. y) is centre of circle passing through A. B and C. Since AM. BH and CH are radius of circle. 

∴ AH=BH and BH=CH
Also AH²=BH² and BH²=CH²
AH²=(x-5)²+(y+8)²
=x²+25-10x+y²+64+16y
BH=(x-2)²+(y+9)²
=x²+4-4x+y²+81+18y
CH²=(x-2)²+(y-1)²
=x²+4-4x+y²+1-2y
∴ AH²=BH² [Radii of a circle]
∴ x²+25-10x+y²+64+16y
= x²+4-4x+y²+81+18y.
⇒ -10x+4x+16y-18y=-4
⇒ -6x-2y=-4
⇒ 3x+y=2   ---(i)
Also BH²=CH²  --- (i)
∴  x²+4-4x+16y-18y=-4
= x²+4-4x+y²+1-2y
⇒ 18y+2y=1-81
⇒ 20y= -80 ⇒ y=-4
Putting value of y in (i), we get
3x+(-4) =2 ⇒3x=2+4
⇒ 3x=6 ⇒ x=2
∴  Coordinates of centre are (2,-4).

A.T.Q. Diameter = 10\(\sqrt {2}\)
OA²=50
⇒ (2a-11)²+(a-7+9)²=50
⇒​ 4a²+121-44a+(a+2)²=50
⇒ 4a²-44a+121+a²+4+4a-50=0
⇒ 5a²-40a+75=0,
⇒ a²-8a+15=0
⇒ (a-5)(a-3)=0 ⇒ a=5,3​​​​​​

OA = 10 units
\(\Rightarrow \ OA=\sqrt { { (2\alpha -1+3 })^{ 2 }+(7+1)^{ 2 } } \)
\(\Rightarrow \ 10=\sqrt { 4{ \alpha }^{ 2 }+4+8\alpha +64 } \)
Squaring \(100\ =\ { 4\alpha }^{ 2 }+8\alpha +68\)
\(\Rightarrow \ { 4\alpha }^{ 2 }+8\alpha -32=0\ \Rightarrow \ { \alpha }^{ 2 }+2\alpha -8=0\)
\(\Rightarrow \ { \alpha }^{ 2 }+4\alpha -2\alpha -8=0\ \Rightarrow \alpha (\alpha +4)-2(\alpha +4)=0\ \Rightarrow \ (\alpha +4)(\alpha -2)=0\)
\(\\ \alpha =-4\ ,\ \alpha =2\)

Let A(7,6), B(-3,4)) be the given points and P(x,0) be the required point.
Since, P is equidistant from A and B, therefore,
AP = BP ⇒  AP²=BP²
⇒ (x-7)²+(0-6)²=(x+3)²+(0-4)² ⇒ x²+49-14x+36 = x²+9+6x+16
⇒ -14x-6x = 25-85    ⇒ -20x=-60
\(\Rightarrow x=\frac { -60 }{ -20 } =3\)
∴ Required point on axis is (3,0).

Let point on y-axis be (0,a)
Now distance of this point from (5,-2) equal to distance from point (-3,2)
i.e.,  \(\Rightarrow \sqrt { 5^{ 2 }+\left( -2-a \right) ^{ 2 } } =\sqrt { \left( 3 \right) ^{ 2 }+\left( a-2 \right) ^{ 2 } } \)
Squaring and simplifying, we get
25+4+a²+4a = 9+a²+4-4a ⇒ 8a=-16  ⇒ a=-2

(i) We know the diagonals of a rhombus bisect each other.
Thus, the point of intersection of both the diagonals is the mid-point of P(2,-3) and R(-2,1).
So, \(\left(\frac{2-2}{2}, \frac{-3+1}{2}\right)=(0,-1)\)
(ii) Finds the mid-point of Q(6,5) and S(x, y) as \(\left(\frac{6+x}{2}, \frac{5+y}{2}\right)\), where x and y are the coordinates of the fourth vertex S.
Uses the above steps and equates the respective coordinates of the mid-points to get the following relationships:
(a) \(0=\frac{6+x}{2} \Rightarrow x=-6\)
(b) \(-1=\frac{5+y}{2} \Rightarrow y=-7\)
Concludes that the coordinates of the fourth vertex S are (-6,-7).

Given points are A (2, -2), B (-3, 8) and C (-1, 4).
Now area of \(\Delta \)ABC = \(\frac { 1 }{ 2 } \)[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
=\(\frac { 1 }{ 2 } \)[2(8-4)-3(4+2)-1(-2-8)]
=\(\frac { 1 }{ 2 } \)[8-18+1
=\(\frac { 1 }{ 2 } \)(18-18)=0
Since, the area of  \(\Delta \)ABC is zero.
Hence, given points A, B and C are collinear.   
Hence proved.

Let P and Q be the points of trisection as shown below

Then, AP: PB=1.2 and AQ: QB=2:1
(i) Let P divides AB in the ratio 1: 2.
Then,        \(\frac { { m }_{ 1 } }{ { m }_{ 2 } } =\frac { 1 }{ 2 } \)

Here, A(x₁, y₁) = (2, -3) and B(x₂, y₂) = (4, -1)
Now,  \(P(x,\quad y)=P\left( \frac { { m }_{ 1 }{ x }_{ 2 }+{ m }_{ 2 }{ x }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } ,\frac { { m }_{ 1 }{ y }_{ 2 }+{ m }_{ 2 }{ y }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) \)
\( =P\left( \frac { 1\times 4+2\times 2 }{ 1+2 } ,\frac { 1\times (-1)+2\times (-3) }{1+2 } \right) \\ =P\left( \frac { 4+4 }{3 } ,\frac { -1-6 }{3} \right) =P\left( \frac { 8 }{ 3 } ,\frac { -7 }{ 3 } \right) \)
(ii) Let Q divides AB in the ratio 2: 1, then \(\frac { { m }_{ 1 } }{ { m }_{ 2 } } =\frac { 2 }{ 1 } \)
Here, A(x₁, y₁) = (2, -3) and B(x₂, y₂) = (4, -1)
 
Now, \(Q(x,\quad y)=Q\left( \frac { { m }_{ 1 }{ x }_{ 2 }+{ m }_{ 2 }{ x }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } ,\frac { { m }_{ 1 }{ y }_{ 2 }+{ m }_{ 2 }{ y }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) \)
\(=Q\left( \frac { 2\times 4+1\times 2 }{2+1 } ,\frac { 2\times (-1)+1\times (-3) }{2+1} \right) \\ =Q\left( \frac { 8+2 }{3 } ,\frac { -2-3 }{3} \right) =Q\left( \frac { 10 }{ 3 } ,\frac { -5 }{ 3 } \right) \)

Let P(x, y) be the required point.
Then, P divides AB internally in the ratio 3: 2.

Here, \(\frac { { m }_{ 1 } }{ { m }_{ 2 } } =\frac { 3 }{ 2 } \) and \(\left( { x }_{ 1 },\quad { y }_{ 1 } \right) =(4,\quad -3);\quad \left( { x }_{ 2 },\quad { y }_{ 2 } \right) =(9,\quad 7)\)
Then, \(P(x,\quad y)=P\left( \frac { { m }_{ 1 }{ x }_{ 2 }+{ m }_{ 2 }{ x }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } ,\frac { { m }_{ 1 }{ y }_{ 2 }+{ m }_{ 2 }{ y }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) \)
\( =P\left( \frac { 3\times 9+2\times 4 }{ 3+2 } ,\frac { 3\times 7+2\times (-3) }{ 3+2 } \right) \\ \ =P\left( \frac { 27+8 }{ 5 } ,\frac { 21-6 }{ 5 } \right) =P\left( \frac { 35 }{ 5 } ,\frac { 15 }{ 5 } \right) \\=P(7, 3)\)
Therefore, (7, 3) is the required point.

Let p \(\left( \frac { 3 }{ 4 } ,\frac { 5 }{ 12 } \right) \)divides the line segment joining points A\(\left( \frac { 1 }{ 2 } ,\frac { 3 }{ 2 } \right) \) and B (2,-5)in the ration k:1
\(\therefore \) Coordinates of P are
\(P\left( \frac { 2k+\frac { 1 }{ 2 } }{ k+1 } \frac { -5k+\frac { 3 }{ 2 } }{ k+1 } \right) =P\left( \frac { 3 }{ 4 } ,\frac { 5 }{ 12 } \right) \)

\(\Rightarrow \) \(\frac { 2k+\frac { 1 }{ 2 } }{ k+1 } =\frac { 3 }{ 4 } and\quad \frac { -5k+\frac { 3 }{ 2 } }{ k+1 } =\frac { 5 }{ 2 } \)
\(\Rightarrow \) 8k+2=3k+3 and -60k+18=5k+5
\(\Rightarrow \) 5k=1 and -65k=-13
\(\Rightarrow \) \(k=\frac { 1 }{ 5 } \)   and\(k=\frac { 1 }{ 5 } \)
Hence the required ratio is 1:5

Since A(6-1), B(k,-6) and C(0,-7) are collinear.
∴ Area of △ABC = 0

=\(\frac{1}{2}\)|-36-7k-0+k-0+42|=0
⇒ \(\frac{1}{2}\)|-6k+6|=0
⇒ -6k+6=0
⇒ 6k=6
⇒ k=1

 Let P(0,2) is equidistant from A(3,k) and B(k,5).
∴ |PA| = |PB|
⇒ PA² = PB²
⇒ (3-0)²+(k-2)² = (k-0)²+(5-2)²
⇒ 9+k²+4-4k = k²+9
⇒ 4-4k=0
⇒ k=1

A(1,2), B(5,4), C(3,8) and D(-1, 6)
\(AB=\sqrt{4^2+2^2}=\sqrt{16+4}=\sqrt{20};\ BC=\sqrt{(-2)^2+(4)^2}=\sqrt{4+16}=\sqrt{20}\)
\(CD=\sqrt{(-4)^2+(-2)^2}=\sqrt{16+4}=\sqrt{20};\ DA=\sqrt{(-2)^2+(4)^2}=\sqrt{4+16}=\sqrt{20}\)
Here AB=BC=CA=DA
\(AC=\sqrt{2^2+6^2}=\sqrt{40}\ and\ BD=\sqrt{(-6)^2+(2)^2}=\sqrt{36+4}=\sqrt{40}\)
All sides of quadrilateral are equal and diagonals are equal.
ABCD is square.