Distance =\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(0-\sqrt{3})^2+(1-2)^2}\)
=\(\sqrt{(-\sqrt{3})^2+(-1)^2}=\sqrt{3+1}=\sqrt{4}=2\)
BC =\(\sqrt{(0-0)^2+(3-1)^2}\)
=\(\sqrt{0^2+2^2}\)=\(\sqrt{4}\) = 2
AC =\(\sqrt{(0-\sqrt{3})^2+(3-2)^2}\)
=\(\sqrt{(-\sqrt{3})^2+1^2}=\sqrt{3+1}\)
=\(\sqrt{4}\) = 2
AB = BC = AC (Three sides are equal)
ABC is an equilateral triangle.

Radius of the circle = 30 units. The point O is (0, 0).
Let a intersect the x - axis and b intersect the y - axis
∴ The point A is (a, 0) and B is (0, b)
Distance = \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
OA =\(\sqrt { (a-0)^{ 2 }+(0-0)^{ 2 } } \)
30 = \(\sqrt { a^{ 2 } } \)
Squaring on both sides
30² = a²
∴ a = 30
The point A is (30, 0)
OB = \(\sqrt { (0-0)^{ 2 }+(b-0)^{ 2 } } \)
= \(\sqrt { 0+{ b }^{ 2 } } \)
30 = \(\sqrt {b^{ 2 } } \)
Squaring on both sides
30² = b²
∴ b = 30
The point B is (0, 30)
Distance AB =\(\sqrt { (30-0)^{ 2 }+(0-30)^{ 2 } } \)
= \(\sqrt { 30^{ 2 }+30^{ 2 } } =\sqrt { 900+900 } \)
= \(\sqrt { 1800 } =\sqrt { 2\times 900 } =30\sqrt { 2 } \)
∴  Distance between the two points = 30\(\sqrt { 2 } \)

Distance =\(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
OA = \(\sqrt { (11-1)^{ 2 }+(2-2)^{ 2 } } \)
= \(\sqrt { { 10 }^{ 2 }+{ 0 }^{ 2 } } \)
= \(\sqrt { 100 } \) = 10
OB =\(\sqrt { (11-3)^{ 2 }+(2+4)^{ 2 } } \)
= \(\sqrt { { 8 }^{ 2 }+6^{ 2 } } =\sqrt { 64+36 } \)
= \(\sqrt { 100 } \) = 10
OC = \(\sqrt { (11-5)^{ 2 }+(2+6)^{ 2 } } \)
= \(\sqrt { { 6 }^{ 2 }+8^{ 2 } } =\sqrt { 36+64 } \)
= \(\sqrt { 100 } \) = 10
OA = OB = OC = 10 units
O is the centre of the circle passing through A, B and C.

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

Let the point A be (a, a), B is (1, 3)
Distance AB = 10 (Given)
By distance formula \(\sqrt { (a-1)^{ 2 }+(a-3)^{ 2 } } =10\)
Simplifying
2a²- 8a + 10 = 100
a²- 4a - 45 = 0
(a - 9)(a + 5) = 0
⇒ a = -5 ; A = (-5,-5)
a = 9 ; A = (9,9)

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
\(d=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(AB=\sqrt { \left( 11-7 \right) ^{ 2 }+\left( 0-3 \right) ^{ 2 } } \)
\(=\sqrt { \left( 4 \right) ^{ 2 }+\left( -3 \right) ^{ 2 } } =\ \sqrt { 16+9 } =\sqrt { 25 } =5\)​​​​

Let A(–4, –3), B(3, 1), C(3, 6), D(–4, 2) be the four vertices of any quadrilateral ABCD. Using the distance formula.
Let \(d=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(AB=\sqrt { \left( 3+4 \right) ^{ 2 }+\left( 1+3 \right) ^{ 2 } } =\sqrt { 49+16 } =\sqrt { 65 } \)
\(BC=\sqrt { \left( 3-3 \right) ^{ 2 }+\left( 6-1 \right) ^{ 2 } } =\sqrt { 0+25 } =\sqrt { 25 } =5\)
\(CD=\sqrt { \left( -4-3 \right) ^{ 2 }+\left( 2-6 \right) ^{ 2 } } =\sqrt { 49+16 } =\sqrt { 65 } \)
\(AD=\sqrt { \left( -4+4 \right) ^{ 2 }+\left( 2+3 \right) ^{ 2 } } =\sqrt { \left( { 0 }^{ 2 } \right) +\left( { 5 }^{ 2 } \right) } =\sqrt { 25 } =5\)
\(AB=CD=\sqrt { 65 } \) and BC = AD = 5
Here, the opposite sides are equal. Hence ABCD is a parallelogram.

The distance between the points (-4, 3), (2, -3) is
\(d=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 2+4 \right) ^{ 2 }+\left( -3-3 \right) ^{ 2 } } \)
\(=\sqrt { \left( { 6 }^{ 2 }+\left( -6 \right) ^{ 2 } \right) } =\sqrt { \left( 36+36 \right) } \)
\(=\sqrt { 36\times 2 } \)
\(=6\sqrt { 2 } \)