Distance = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
\(PQ=\sqrt{(1-6)^2(3+1)^2}\)
\(=\sqrt{(-5)^2+4^2}=\sqrt{25+16}\)
\(=\sqrt{41}\)
\(QR=\sqrt{(x-1)^2+(8-3)^2}\)
\(=\sqrt{(x-1)^2+5^2}\)
\(=\sqrt{(x-1)^2+25}\)
But PQ = QR
\(\sqrt{(x-1)^2+25}=\sqrt{41}\)
Squaring on both sides
(x - 1)²- 25 = 41
(x-1)² = 41-25 = 16
x-1 =\(\sqrt{16}\) = ±4
x-1 = 4 (or) x-1 = - 4
x = 5 (or) x = - 4 + 1 = - 3
The value of x = 5 or - 3

Let the point A  (-1, -1), (1, 1) and (\(-\sqrt{13},\sqrt{13}\))
Distance = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(1+1)^2+(1+1)^2}\)
\(\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}\)
BC =\(\sqrt{(-\sqrt{3}+1)^2+(\sqrt{3}-1)^2}\)
\(\sqrt{(\sqrt{3}+1)^2+(\sqrt{3}-1)^2}\)
\(\sqrt{3+1+2\sqrt{3}+3+1-2\sqrt{3}}\)
\(\sqrt{3+1+3+1}=\sqrt{8}\)
AC =\(\sqrt{(-\sqrt{3}+1)^2+(\sqrt{3}+1)^2}\)
\(=\sqrt{3+1-2\sqrt{3}+3+1+2\sqrt{3}}\)
\(=\sqrt{4+4}=\sqrt{8}\)

AB= BC=AC= \(\sqrt{8}\)
\(\therefore\)ABC is an equilateral triangle.

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.

Let A(1, 1),B(5, 4) and C(-2, 5)
Distance =\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(5-1)^2+(4-1)^2}\)
\(=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\)
BC =\(\sqrt{(-2-5)^2+(5-4)^2}\)
\(=\sqrt{(-7)^2+1^2}=\sqrt{49+1}=\sqrt{50}\)
AC =\(\sqrt{(-2-1)^2+(5-1)^2}\)
\(=\sqrt{(-3)^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\)

AB = 5, AC = 5,
\(\therefore\) ABC is an isosceles triangle .................(1)
BC² = AB² + AC²
50 = 25 + 25\(\Rightarrow\) 50 = 50
\(\therefore\) \(\angle\)A = 90° ...................(2)
From (1) and (2) we get ABC is an isosceles right angle triangle.

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)

Let POQ be the right angled triangle and O be placed at the origin. Let OQ = a units and OP be b units. Let us name the coordinates of P as (0,b) and Q as (a,0).
By mid-point formula, if M is the mid-point of the hypotenuse PQ [PM=MQ], then M is

\(\left( \frac { a+0 }{ 2 } ,\frac { b+0 }{ 2 } \right) =\left( \frac { a }{ 2 } ,\frac { b }{ 2 } \right) \)
We now use the distance formula and find that
OM=\(\sqrt { { \left( \frac { a }{ 2 } -0 \right) }^{ 2 }{ +\left( \frac { b }{ 2 } -0 \right) }^{ 2 } } =\sqrt { \frac { { a }^{ 2 } }{ 4 } +\frac { { b }^{ 2 } }{ 4 } } \) which is the same value as
QM=\(\sqrt { { \left( a-\frac { a }{ 2 } \right) }^{ 2 }{ +\left( 0-\frac { b }{ 2 } \right) }^{ 2 } } =\sqrt { \frac { { a }^{ 2 } }{ 4 } +\frac { { b }^{ 2 } }{ 4 } } \) and similarly PM=\(\sqrt { \frac { { a }^{ 2 } }{ 4 } +\frac { { b }^{ 2 } }{ 4 } } \)
This shows OM = QM = PM, which we desired to prove.

(3, 5)

P (- 4,4), Q (8, 2), R (6, -3) and S (-2, -2)

A (- 6, 4), B (- 3, -3) and C (2, 2)

\((10,-1)=\left( \frac { 3+5+x }{ 3 } ,\frac { 2-11+y }{ 3 } \right) \)
\(10=\frac { 3+5+x }{ 3 } ,\)
30 = 3 + 5 + x
22 = x
\(-1=\frac { 2-11+y }{ 3 } \)
-3 = 2 - 11 + y
-3 + 9 = y
y = 6
(22, 6)

\(G\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) =\left( \frac { 2+5+9 }{ 3 } ,\frac { -5+11+9 }{ 9 } \right) =\left( \frac { 16 }{ 3 } ,5 \right) \)

\(120=\frac { 8(50)+4(y) }{ 12 } \)
1440 = 400 + 4y
4y = 1040
\(y=\frac { 1040 }{ 4 } =260\ km\)

\(\left( \frac { 5(3)+3(5) }{ 5+3 } ,\frac { 5(1)+3(13) }{ 5+3 } \right) =\left( \frac { 15+15 }{ 8 } ,\frac { 5+39 }{ 8 } \right) =\left( \frac { 30 }{ 8 } ,\frac { 44 }{ 8 } \right) =\left( \frac { 15 }{ 4 } ,\frac { 11 }{ 2 } \right) \)

Centre of the circle \(=\left( \frac { 10+2 }{ 2 } ,\frac { 11+3 }{ 2 } \right) =(6,7)\)