(i) The point P (0, -4), Q(0, -1), R(0, 3), S(0, 6) lies on the y-axis.
(ii) Distance = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
(a) PR=\(\sqrt{(0-0)^2+(3+4)^2}\)
\(=\sqrt{0+7^2}=\sqrt{49}=7 \ uint\)
(b) QS =\(\sqrt{(0-0)^2+(6+1)^2}\)
\(=\sqrt{0+7^2}=\sqrt{49}=7 \ uint\)

Distance = \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
AB = \(\sqrt { (2-1)^{ 2 }+(1-1)^{ 2 } } \)
= \(\sqrt { (1)^{ 2 }+(0)^{ 2 } } =\sqrt { 1 } =1\)
BC = \(\sqrt { (2-1)^{ 2 }+(2-1)^{ 2 } } \)
= \(\sqrt { (0)^{ 2 }+(1)^{ 2 } } =\sqrt { 1 } =1\)
CD =\(\sqrt { (1-2)^{ 2 }+(2-2)^{ 2 } } \)
= \(\sqrt { (-1)^{ 2 }+(0)^{ 2 } } =\sqrt { 1 } =1\)
AD =\(\sqrt { (1-1)^{ 2 }+(2-1)^{ 2 } } \)
= \(\sqrt { (0)^{ 2 }+(1)^{ 2 } } =\sqrt { 1 } =1\)
AB = BC = CD = AD =1
All the four sides are equal.
ABCD is a rhombus

Distance =\(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
AB = \(\sqrt { (7-3)^{ 2 }+(6+2)^{ 2 } } =\sqrt { (4)^{ 2 }+(8)^{ 2 } } \)
= \(\sqrt { 16+64 } =\sqrt { 80 } \)
BC = \(\sqrt { (-1-7)^{ 2 }+(2-6)^{ 2 } } =\sqrt { (-8)^{ 2 }+(-4)^{ 2 } } \)
= \(\sqrt { 64+16 } =\sqrt { 80 } \)
CD = \(\sqrt { (-5+1)^{ 2 }+(-6-2)^{ 2 } } =\sqrt { (-4)^{ 2 }+(-8)^{ 2 } } \)
= \(\sqrt { 16+64 } =\sqrt { 80 } \)
AD =\(\sqrt { (-5-3)^{ 2 }+(-6-2)^{ 2 } } =\sqrt { (-8)^{ 2 }+(-4)^{ 2 } } \)
= \(\sqrt { 64+16 } =\sqrt { 80 } \)
AB = BC = CD = AD = \(\sqrt { 80 } \)
All the four sides are equal
∴  ABCD is a rhombus

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)

The two given points are (5,-2), (1, a) and d = 5.
By distance formula
\(d=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 1-5 \right) ^{ 2 }+\left( a+2 \right) ^{ 2 } } =5\)
\(=\sqrt { 16+\left( a+2 \right) ^{ 2 } } =5\)
16 + (a+2)² = 25 (By squaring on both the sides)
(a+2)² = 25–16
(a+2)² = 9
(a+2) = \(\pm \) 3 (By taking the square root on both side)
a = –2 \(\pm \) 3
a = –2 + 3 (or) a = –2 –3
​​​​​​​a = 1 or –5.

Can you find the length, breadth and area of the rectangle?
 

ABCD is a rectangle.

When you join these points, you see that they lie on a line which is parallel to x-axis.

All points lie on x-axis.

(i) To plot (2, 4), draw a vertical line at x = 2 and draw a horizontal line at y = 4. The intersection of these two lines is the position of (2, 4) in the Cartesian plane. Thus, the Point A (2, 4) is located in the I quadrant of Cartesian plane.
(ii) To plot (–3, 5), draw a vertical line at x = –3 and draw a horizontal line at y = 5. The intersection of these two lines is the position of (–3, 5) in the Cartesian plane. Thus, the Point B (–3, 5) is located in the II quadrant of Cartesian plane.
(iii) To plot (–4, –5), draw a vertical line at x = –4 and draw a horizontal line at y =  -5. The intersection of these two lines is the position of (-4, 5) in the Cartesian plane. Thus, the Point C (–4, –5) is located in the III quadrant of Cartesian plane.
(iv) To plot (4, –2), draw a vertical line at x = 4 and draw a horizontal line at y = –2. The Intersection of these two lines is the position of (4,–2) in the Cartesian plane. Thus, the Point D (4,–2) is located in the IV quadrant of Cartesian plane