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

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 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

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 AB = \(\sqrt { (a-a)^{ 2 }+(3+2)^{ 2 } } \)
= \(\sqrt { 0+{ 5 }^{ 2 } } =\sqrt { 25 } =5\)
BC = \(\sqrt { (a-a)^{ 2 }+(0-3)^{ 2 } } \)
= \(\sqrt { 0+9 } =\sqrt { 9 } =3\)
AC = \(\sqrt { (a-a)^{ 2 }+(0+2)^{ 2 } } \)
= \(\sqrt { 0+2^{ 2 } } =\sqrt { 4 } =2\)
AC + BC = AB ⇒ 2 + 3 = 5
∴ The given three points are collinear

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.

AD = 3 units
DA = 3 units (From graph)
Yes, Distance AD = Distance DA
The students are asked to do the activity with different points and verify the result.

(i) The point P (-2, 0) , Q (2, 0), and R (3, 0) lies on the x-axis,
(ii) Distance =\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
(a) PR =\(\sqrt{(3+2)^2+(0-0)^2}\)
\(=\sqrt{5^2+0}=\sqrt{25}=5units\)
(b) QR =\(\sqrt{(3-2)^2+(0-0)^2}\)
\(=\sqrt{1^2+0}=\sqrt{1}=1 \ units\)

Let P, Q, R be the points on the line segment joining A(−11, 4) and B(9, 8) such that AP = PQ = QR = RB .
Here Q is the mid-point of AB, P is the mid-point of AQ and R is the mid-point of QB.
Q is the mid-point of AB =\(\left( \frac { -11+9 }{ 2 } ,\frac { 4+8 }{ 2 } \right) =\left( \frac { -2 }{ 2 } ,\frac { 12 }{ 2 } \right) \)= (-1, 6)
P is the mid-point of AQ =\(\left( \frac { -11-1 }{ 2 } ,\frac { 4+6 }{ 2 } \right) =\left( \frac { -12 }{ 2 } ,\frac { 10 }{ 2 } \right) \) = (-6, 5)
R is the mid-point of QB =\(\left( \frac { -1+9 }{ 2 } ,\frac { 6+8 }{ 2 } \right) =\left( \frac { 8 }{ 2 } ,\frac { 14 }{ 2 } \right) \) = (4, 7)
Hence the points which divides AB into four equal parts are P(–6, 5), Q(–1, 6) and R(4, 7).

The centroid G(x, y) of a triangle whose vertices are (x₁, y₁), (x₂ , y₂ ) and (x₃ , y₃) is given by

G(x,y)=G\(\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }3 }{ } \right) \)
We have (x₁, y₁) = (6, −1); (x₂ , y₂ ) = (8, 3); (x₃ , y₃) = (10, −5)
The centroid of the triangle
G(x, y) =G\(\left( \frac { 6+8+10 }{ 3 } ,\frac { -1+3-5 }{ 3 } \right) \)
= G \((\frac{24}{3},\frac{-3}{3})\)= G(8, -1)