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 = \(\sqrt{(x_2-x_2)^2+(y_2-y_1)^2}\)
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 get ABCD 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.

Given P (6, –1), Q (1, 3) and R (a, 8)
\(PQ=\sqrt { \left( 1-6 \right) ^{ 2 }+\left( 3+1 \right) ^{ 2 } } =\sqrt { \left( -5 \right) ^{ 2 }+\left( 4 \right) ^{ 2 } } = \sqrt { 41 } \)
\(QR=\sqrt { \left( a-1 \right) ^{ 2 }+\left( 8-3 \right) ^{ 2 } } =\sqrt { \left( a-1 \right) ^{ 2 }+\left( 5 \right) ^{ 2 } } \)
Given PQ = QR
Therefore \(\sqrt { 41 } =\sqrt { \left( a-1 \right) ^{ 2 }+\left( 5 \right) ^{ 2 } } \)
41= (a –1)² + 25 [Squaring both sides]
(a–1)² + 25 = 41
(a–1)² = 41 - 25
(a–1)² = 16
(a–1) = \(\pm \) 4 [taking square root on both sides]
a = 1  \(\pm \)4
a = 1 + 4 or a = 1 – 4
a = 5, –3

Let P(4, 3), A(9, 3), B(7, –1) and C(–1, 3)
If P is the centre of the circle which passes through the points A, B, and C, then P is equidistant from A, B and C (i.e.) PA = PB = PC
By distance formula,
\(d=\sqrt{(x_2 -x_1)^2 +(y_2 -y_1)^2}\)
\(AP=PA =\sqrt{(4-9)^2+(3-3)^2}=\sqrt{(-5)^2+0}=\sqrt{25}=5\)
\(BP=PB =\sqrt{(4-7)^2+(3+1)^2}=\sqrt{(3)^2+(4)^2}=\sqrt{9+16}=\sqrt{25}=5\)
\(CP=CB =\sqrt{(4+1)^2+(3-3)^2}=\sqrt{(5)^2+0}=\sqrt{25}=5\)
PA = PB = PC, Radius = 5
Therefore P is the centre of the circle, passing through A, B and C

Given points are A(2, 2) and B(8, –4) and let P = (x, 0) [P lies on x axis]
By the distance formula
\(d=\sqrt{(x_2 -x_1)^2 +(y_2 -y_1)^2}\)
\(AP=\sqrt{(x-2)^2+(0-2)^2}=\sqrt{X^2-4x+4+4}=\sqrt{x^2 -4x+8}\)
\(BAP=\sqrt{(x-8)^2+(0+4)^2}=\sqrt{X^2-16x+64+16}=\sqrt{x^2 -16x+80}\)
Given, AP : PB = 1 : 2
i.e \(\frac{AP}{BP}=\frac{1}{2} (\therefore BP=PB)\)
2AP = BP
squaring on both sides
4AP² = BP^{2
\(4(x^2 -16x+8)=(x^2-16x+80)\)}
\(4x^2 -16x+32=x^2-16x+80\)
3x²- 48 = 0
3x²= 48
x² = 16
x =\(\pm\)4
As the point P lies on x-axis (positive side), its x- coordinate cannot be –4.
Hence the coordinates of P is (4, 0)

Let A(3, 5), B(6, 2), C(3, -1), and D(0, 2) be the vertices of any quadrilateral ABCD
\(d=\sqrt{(x_2 -x_1)^2 +(y_2 -y_1)^2}\)
By using the distance formula we get,
\(AB=\sqrt{(6-3)^2+(2-5)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\)
\(BC=\sqrt{(3-6)^2 +(-1-2)^2}=\sqrt{(-3)^2 +(-3)^2}=\sqrt{9+9}\sqrt{18}=3\sqrt{2}\)
\(CD=\sqrt{(0-3)^2 +(2+1)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\)
\(AD=\sqrt{(0-3)^2 +(2-5)^2}=\sqrt{(-3)^2 +(-3)^2}=\sqrt{9+9}=\sqrt{18}=\sqrt3{2}\)
From the above results, we see that AB = BC = CD = DA = \(3\sqrt{2}\)
(i.e.) All the four sides are equal.
Further, A(3, 5), C(3, –1)
Diagonal AC =\(\sqrt{(3-3)^2 +(-1-5)^2}=\sqrt{(0)^2 +(-6)^2}=\sqrt{36}=6\)
Diagonal BD= \(\sqrt{(0-6)^2 +(2-2)^2}=\sqrt{(-6)^2 +(0)^2}=\sqrt{36}=6\)
From the above we see that AB = CD = 6
Hence ABCD is a square

Here A = (7, 10), B = (-2, 5), C = (3,-4)
\(AB=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( -2-7 \right) ^{ 2 }+\left( 5-10 \right) ^{ 2 } } \)
\(=\sqrt { \left( -9 \right) ^{ 2 }+\left( -5 \right) ^{ 2 } } \)
\(=\sqrt { \left( 81+25 \right) } \)
\(=\sqrt { 106 } \)
\(A{ B }^{ 2 }=106\)............1
\(BC=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 3-\left( -2 \right) \right) ^{ 2 }+\left( -4-5 \right) ^{ 2 } } \)
\(=\sqrt { \left( 5 \right) ^{ 2 }+\left( -9 \right) ^{ 2 } } \)
\(=\sqrt { \left( 81+25 \right) } \)
\(=\sqrt { 106 } \)
\({ BC }^{ 2 }=106\)...............2
\(AC=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 3-7 \right) ^{ 2 }+\left( -4-10 \right) ^{ 2 } } \)
\(=\sqrt { \left( -4 \right) ^{ 2 }+\left( -14 \right) ^{ 2 } } \)
\(=\sqrt { \left( 16+196 \right) } \)
\(=\sqrt { 212 } \)
\({ AC }^{ 2 }=212\)..........3
From (1), (2) & (3) we get
AB²+BC² = 106 +106 = 212 = AC²
AB² + BC² = AC² 
ΔABC is a right angled triangle, right angled at B.