Distance = \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-{ y }_{ 1 })^{ 2 } } \)
AB= \(\sqrt { (7-3)^{ 2 }+(2-2)^{ 2 } } \)
= \(\sqrt { { 4 }^{ 2 }+0^{ 2 } } \)
= \(\sqrt { 16 } =4\)
BC = \(\sqrt { (7-7)^{ 2 }+(5-2)^{ 2 } } \)
= \(\sqrt { 0+(3)^{ 2 } } \)
= \(\sqrt { 9 } =3\)
AC = \(\sqrt { (7-3)^{ 2 }+(5-2)^{ 2 } } \)
= \(\sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 } } \)
= \(\sqrt { 16+9 } =\sqrt { 25 } =5\)
Perimeter of ΔABC = AB + BC + AC
= 4 + 3 + 5 =12 units

Given points are A(2, 3) and B(2, -4)
The point P lines on the x-axis.
∴ The point P is (x, 0)
AP  = \(\frac { 3 }{ 7 } \) AB
\(\frac { AP }{ AB } \) = \(\frac { 3 }{ 7 } \)
\(\frac { AP }{ PB } \) = \(\frac { 3 }{ 7 } \)  ........(1)
Distance =\(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
AP = \(\sqrt { (x-2)^{ 2 }+(0-3)^{ 2 } } \)
= \(\sqrt { { x }^{ 2 }-4x+4+9 } \)
= \(\sqrt { { x }^{ 2 }-4x+13 } \)
BP =\(\sqrt { (x-2)^{ 2 }+(0+4)^{ 2 } } \)
= \(\sqrt { x^{ 2 }-4x+4+16 } \)
= \(\sqrt { x^{ 2 }-4x+20 } \)
From (1) we get
\(\frac { AP }{ PB } \) = \(\frac { 3 }{4 } \)
\(\frac { \sqrt { { x }^{ 2 }-4x+13 } }{ \sqrt { { x }^{ 2 }-4x+20 } } =\frac { 3 }{ 4 } \)    (Squaring on both sides)
\(\frac { { x }^{ 2 }-4x+13 }{ { x }^{ 2 }-4x+20 } =\frac { 9 }{ 16 } \)
16x²- 64x + 208 = 9x² - 36x + 180
7x² - 28x + 28 = 0
x²- 4x + 4 = 0
(x-2)²= 0
x-2 =0
x = 2
∴ The point P is (2, 0)

Let the point O be (x, y), A be (3, 4) and B be (-5, 6).
Distance = \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
Given OA = OB
\(\sqrt { (x-3)^{ 2 }+(y-4)^{ 2 } } =\sqrt { (x+5)^{ 2 }+(y-6)^{ 2 } } \)
Squaring on both sides
(x-3)² + (y-4)² = (x+5)²+(y-6)²
x²-6x+9+y²-8y+16 = x²+10x+25+y²-12y+36
x²+y²-6x-8y+25 = x²+y²+10x-12y+61
-6x-10x-8y+12y = 61-25
⇒ -4x + y = 9
The relation between x and y is y = 4x + 9

Since B is on the x-axis, the y-coordinate of B is 0.
So, the coordinates of the point B is (11, 0)
By the distance formula the distance between the points A (7, 3), B (11, 0) is
\(d=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(AB=\sqrt { \left( 11-7 \right) ^{ 2 }+\left( 0-3 \right) ^{ 2 } } \)
\(=\sqrt { \left( 4 \right) ^{ 2 }+\left( -3 \right) ^{ 2 } } =\ \sqrt { 16+9 } =\sqrt { 25 } =5\)​​​​

Using the distance formula, we have
\(AB=\sqrt { \left( 6-3 \right) ^{ 2 }+\left( 4-1 \right) ^{ 2 } } =\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
\(BC=\sqrt { \left( 8-6 \right) ^{ 2 }+\left( 6-4 \right) ^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 } \)
\(AC=\sqrt { \left( 8-3 \right) ^{ 2 }+\left( 6-1 \right) ^{ 2 } } =\sqrt { 25+25 } =\sqrt { 50 } =5\sqrt { 2 } \)
\(AB+BC=3\sqrt { 2 } +3\sqrt { 2 } =5\sqrt { 2 } =AC\)
Therefore the points lie on a straight line.

Take a graph sheet and plot all the points and join the points to form a rectangle.
Take a mirror and see the image of the rectangle formed.

Distance between the two points (3,- 9) and (-2, 3)
= \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+({ y }_{ 2 }-{ y }_{ 1 })^{ 2 } } \)
= \(\sqrt { (-2-3)^{ 2 }+(3+9)^{ 2 } } =\sqrt { (-5)^{ 2 }+(12)^{ 2 } } \)
=\(\sqrt { 25+144 } =\sqrt { 169 } \) = 13 units

The shape of the geometrical figure Trapezium.

The geometrical shape of the figure is square

The line is on the y - axis

Distance = \(\sqrt{(x_2-x_2)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(6-3)^2+(5-7)^2}\)
\(\sqrt{3^2+(-2)^2}=\sqrt{9+4}=\sqrt{13}\)
BC=\(\sqrt{(15-6)^2+(-1-5)^2}=\sqrt{(9)^2+(-6)^2}\)
\(\sqrt{81+36}=\sqrt{117}\)
\(\sqrt{9\times 13}-3\sqrt{13}\)
AC =\(\sqrt{(15-3)^2+(-1-7)^2}\)
\(\sqrt{12^2+(-8)^2}=\sqrt{144+64}\)
\(=\sqrt{208}=\sqrt{16\times 13}=4\sqrt{13}\)

AB + BC =AC\(\Rightarrow \sqrt{13}+3\sqrt{13}=4\sqrt{13}\) 
\(\therefore\)The points A,B,C are collinear.

Distance = \(\sqrt{(x_2+x_1)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(3-3)^2+(2+2)^2}\)
=\(\sqrt{0+4^2}=\sqrt{16}=4\)
BC =\(\sqrt{(-1-3)^2+(2-2)^2}\)
\(\sqrt{(-4)^2+0}=\sqrt{16}=4\)
CD =\(\sqrt{(-1-1)^2+(2-2)^2}\)
\(\sqrt{0+(-4)^2}=\sqrt{16}=4\)
AD =\(\sqrt{(-1-3)^2+(-2+2)^2}\)
\(\sqrt{(-4)^2+0}=\sqrt{16}=4\)
AB = BC = CD = DA = 4. All the four sides are equal.

\(\therefore \)ABCD is a Rhombus .................(1)
Diagonal AC =\(\sqrt{(3+1)^2+(-2-2)^2}\)
\(=\sqrt{4^2+(-4)^2}=\sqrt{16+16}=\sqrt{32}\)
Diagonal BD =\(\sqrt{(3+1)^2+(2+2)^2}\)
\(=\sqrt{4^2+4^2}=\sqrt{16+16}=\sqrt{32}\)
Diagonal AC = Diagonal BD =\(\sqrt{32}\) ................ (2)
From (1) and (2) we getABCD is a square.