Distance Formula Questions in English

(D) Let the points be $P(3,2), Q(-2,-3)$,and $R(2,3)$.
We use the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ to find the lengths of the sides:
$PQ = \sqrt{(-2-3)^2 + (-3-2)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25+25} = \sqrt{50} = 5\sqrt{2}$.
$QR = \sqrt{(2 - (-2))^2 + (3 - (-3))^2} = \sqrt{4^2 + 6^2} = \sqrt{16+36} = \sqrt{52} = 2\sqrt{13}$.
$PR = \sqrt{(2-3)^2 + (3-2)^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{1+1} = \sqrt{2}$.
Since the sum of any two sides is greater than the third side ($PQ + PR > QR$,$PQ + QR > PR$,and $PR + QR > PQ$),the points form a triangle.
Checking for a right-angled triangle using the converse of the Pythagoras theorem: $PR^2 + PQ^2 = (\sqrt{2})^2 + (\sqrt{50})^2 = 2 + 50 = 52$.
Since $QR^2 = (\sqrt{52})^2 = 52$,we have $PR^2 + PQ^2 = QR^2$.
Thus,the triangle is a right-angled triangle.

(N/A) Let $A(1,7), B(4,2), C(-1,-1)$ and $D(-4,4)$ be the given points.
To prove that $ABCD$ is a square,we must show that all its sides are equal and its diagonals are equal.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB = \sqrt{(4-1)^2 + (2-7)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9+25} = \sqrt{34}$
$BC = \sqrt{(-1-4)^2 + (-1-2)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25+9} = \sqrt{34}$
$CD = \sqrt{(-4 - (-1))^2 + (4 - (-1))^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9+25} = \sqrt{34}$
$DA = \sqrt{(1 - (-4))^2 + (7-4)^2} = \sqrt{5^2 + 3^2} = \sqrt{25+9} = \sqrt{34}$
Now,calculating the diagonals:
$AC = \sqrt{(-1-1)^2 + (-1-7)^2} = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4+64} = \sqrt{68}$
$BD = \sqrt{(-4-4)^2 + (4-2)^2} = \sqrt{(-8)^2 + 2^2} = \sqrt{64+4} = \sqrt{68}$
Since $AB = BC = CD = DA$ and $AC = BD$,all four sides are equal and the diagonals are equal. Therefore,$ABCD$ is a square.

(N/A) Using the distance formula,we have:
$AB = \sqrt{(6-3)^{2} + (4-1)^{2}} = \sqrt{3^{2} + 3^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$
$BC = \sqrt{(8-6)^{2} + (6-4)^{2}} = \sqrt{2^{2} + 2^{2}} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$
$AC = \sqrt{(8-3)^{2} + (6-1)^{2}} = \sqrt{5^{2} + 5^{2}} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$
Since $AB + BC = 3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2} = AC$,the sum of the distances $AB$ and $BC$ is equal to the distance $AC$. Therefore,the points $A$,$B$,and $C$ are collinear,which means they are seated in a line.

(A) Let $P(x, y)$ be equidistant from the points $A(7, 1)$ and $B(3, 5)$.
Given that $AP = BP$,it follows that $AP^2 = BP^2$.
Using the distance formula,we have:
$(x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2$
Expanding both sides:
$(x^2 - 14x + 49) + (y^2 - 2y + 1) = (x^2 - 6x + 9) + (y^2 - 10y + 25)$
Simplifying by cancelling $x^2$ and $y^2$ from both sides:
$-14x - 2y + 50 = -6x - 10y + 34$
Rearranging the terms:
$-14x + 6x - 2y + 10y = 34 - 50$
$-8x + 8y = -16$
Dividing by $-8$:
$x - y = 2$
Thus,the required relation is $x - y = 2$.

(A) We know that any point on the $y$-axis is of the form $(0, y)$. Let the point $P(0, y)$ be equidistant from points $A(6, 5)$ and $B(-4, 3)$.
According to the distance formula,$PA = PB$,so $PA^2 = PB^2$.
$(6 - 0)^2 + (5 - y)^2 = (-4 - 0)^2 + (3 - y)^2$
$36 + (25 - 10y + y^2) = 16 + (9 - 6y + y^2)$
$61 - 10y + y^2 = 25 - 6y + y^2$
Subtracting $y^2$ from both sides:
$61 - 10y = 25 - 6y$
$61 - 25 = 10y - 6y$
$36 = 4y$
$y = 9$
Thus,the required point is $(0, 9)$.

(A) The distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by the distance formula:
$d = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$
Given points are $(x_{1}, y_{1}) = (2, 3)$ and $(x_{2}, y_{2}) = (4, 1)$.
Substituting these values into the formula:
$d = \sqrt{(4-2)^{2} + (1-3)^{2}}$
$d = \sqrt{(2)^{2} + (-2)^{2}}$
$d = \sqrt{4 + 4}$
$d = \sqrt{8}$
$d = 2\sqrt{2}$ units.

(B) The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Given points are $(x_1, y_1) = (-5, 7)$ and $(x_2, y_2) = (-1, 3)$.
Substituting the values into the formula:
$d = \sqrt{(-1 - (-5))^2 + (3 - 7)^2}$
$d = \sqrt{(-1 + 5)^2 + (-4)^2}$
$d = \sqrt{(4)^2 + (-4)^2}$
$d = \sqrt{16 + 16}$
$d = \sqrt{32}$
$d = \sqrt{16 \times 2} = 4\sqrt{2}$ units.

(B) The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Here,$(x_1, y_1) = (a, b)$ and $(x_2, y_2) = (-a, -b)$.
Substituting these values into the formula:
$d = \sqrt{(-a - a)^2 + (-b - b)^2}$
$d = \sqrt{(-2a)^2 + (-2b)^2}$
$d = \sqrt{4a^2 + 4b^2}$
$d = \sqrt{4(a^2 + b^2)}$
$d = 2\sqrt{a^2 + b^2}$.

(A) The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $(0,0)$ and $(36,15)$.
Substituting the values into the formula:
$d = \sqrt{(36 - 0)^2 + (15 - 0)^2}$
$d = \sqrt{36^2 + 15^2}$
$d = \sqrt{1296 + 225}$
$d = \sqrt{1521}$
$d = 39$.
Therefore,the distance between the points is $39$.

(A) Let the points $A(5,-2), B(6,4),$ and $C(7,-2)$ represent the vertices of the triangle.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB = \sqrt{(6-5)^2 + (4-(-2))^2} = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37}$
$BC = \sqrt{(7-6)^2 + (-2-4)^2} = \sqrt{1^2 + (-6)^2} = \sqrt{1 + 36} = \sqrt{37}$
$CA = \sqrt{(5-7)^2 + (-2-(-2))^2} = \sqrt{(-2)^2 + 0^2} = \sqrt{4} = 2$
Since $AB = BC = \sqrt{37}$,two sides of the triangle are equal in length.
Therefore,the given points are the vertices of an isosceles triangle.

(N/A) The positions of the $4$ friends are $A(3, 4)$,$B(6, 7)$,$C(9, 4)$,and $D(6, 1)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(6 - 3)^2 + (7 - 4)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$
$BC = \sqrt{(9 - 6)^2 + (4 - 7)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$
$CD = \sqrt{(6 - 9)^2 + (1 - 4)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$
$DA = \sqrt{(3 - 6)^2 + (4 - 1)^2} = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$
Now,calculate the diagonals:
$AC = \sqrt{(9 - 3)^2 + (4 - 4)^2} = \sqrt{6^2 + 0^2} = 6$
$BD = \sqrt{(6 - 6)^2 + (1 - 7)^2} = \sqrt{0^2 + (-6)^2} = 6$
Since all sides are equal $(AB = BC = CD = DA = 3\sqrt{2})$ and both diagonals are equal $(AC = BD = 6)$,the quadrilateral $ABCD$ is a square. Therefore,Champa is correct.

(D) Let the points $A(-1,-2), B(1,0), C(-1,2),$ and $D(-3,0)$ be the vertices of the quadrilateral.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB = \sqrt{(1 - (-1))^2 + (0 - (-2))^2} = \sqrt{2^2 + 2^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$
$BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$
$CD = \sqrt{(-3 - (-1))^2 + (0 - 2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$
$DA = \sqrt{(-1 - (-3))^2 + (-2 - 0)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$
Diagonal $AC = \sqrt{(-1 - (-1))^2 + (2 - (-2))^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4$
Diagonal $BD = \sqrt{(-3 - 1)^2 + (0 - 0)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4$
Since all sides are equal $(AB = BC = CD = DA = 2\sqrt{2})$ and both diagonals are equal $(AC = BD = 4)$,the quadrilateral is a square.

(NONE) Let the points $A(-3, 5), B(3, 1), C(0, 3),$ and $D(-1, -4)$ be the vertices of the quadrilateral.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(3 - (-3))^2 + (1 - 5)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$
$BC = \sqrt{(0 - 3)^2 + (3 - 1)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$
$CD = \sqrt{(-1 - 0)^2 + (-4 - 3)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}$
$DA = \sqrt{(-3 - (-1))^2 + (5 - (-4))^2} = \sqrt{(-2)^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85}$
Since all four sides $AB, BC, CD,$ and $DA$ have different lengths,the points do not form any special type of quadrilateral (like a square,rectangle,or parallelogram). It is a general quadrilateral.

(A) Let the points $A(4,5), B(7,6), C(4,3),$ and $D(1,2)$ represent the vertices of the quadrilateral.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB = \sqrt{(7-4)^2 + (6-5)^2} = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}$
$BC = \sqrt{(4-7)^2 + (3-6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$
$CD = \sqrt{(1-4)^2 + (2-3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9+1} = \sqrt{10}$
$DA = \sqrt{(4-1)^2 + (5-2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$
Diagonal $AC = \sqrt{(4-4)^2 + (3-5)^2} = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2$
Diagonal $BD = \sqrt{(1-7)^2 + (2-6)^2} = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$
Since opposite sides are equal ($AB=CD$ and $BC=DA$) and diagonals are unequal $(AC \neq BD)$,the quadrilateral is a parallelogram.

(A) We need to find a point on the $x$-axis. Therefore,its $y$-coordinate must be $0$.
Let the point on the $x$-axis be $P(x, 0)$.
The distance between $P(x, 0)$ and $A(2, -5)$ is given by the distance formula:
$PA = \sqrt{(x - 2)^2 + (0 - (-5))^2} = \sqrt{(x - 2)^2 + 25}$.
The distance between $P(x, 0)$ and $B(-2, 9)$ is given by:
$PB = \sqrt{(x - (-2))^2 + (0 - 9)^2} = \sqrt{(x + 2)^2 + 81}$.
Since the point is equidistant from $A$ and $B$,we have $PA = PB$,which implies $PA^2 = PB^2$:
$(x - 2)^2 + 25 = (x + 2)^2 + 81$
Expanding the squares:
$x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$
Subtracting $x^2 + 4$ from both sides:
$-4x + 25 = 4x + 81$
Rearranging the terms:
$-8x = 81 - 25$
$-8x = 56$
$x = -7$
Thus,the required point is $(-7, 0)$.

(A) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $P(2, -3)$ and $Q(10, y)$,and the distance $d = 10$.
Substituting the values into the formula:
$10 = \sqrt{(10 - 2)^2 + (y - (-3))^2}$
$10 = \sqrt{(8)^2 + (y + 3)^2}$
Squaring both sides:
$100 = 64 + (y + 3)^2$
$100 - 64 = (y + 3)^2$
$36 = (y + 3)^2$
Taking the square root on both sides:
$y + 3 = \pm 6$
Case $1$: $y + 3 = 6 \implies y = 3$
Case $2$: $y + 3 = -6 \implies y = -9$
Therefore,the possible values of $y$ are $3$ and $-9$.

(X = ±4) Given that $Q(0, 1)$ is equidistant from $P(5, -3)$ and $R(x, 6)$,we have $PQ = QR$.
Using the distance formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$PQ^2 = QR^2$
$(5-0)^2 + (-3-1)^2 = (x-0)^2 + (6-1)^2$
$5^2 + (-4)^2 = x^2 + 5^2$
$25 + 16 = x^2 + 25$
$x^2 = 16 \implies x = \pm 4$.
Case $1$: If $R$ is $(4, 6)$:
$QR = \sqrt{(4-0)^2 + (6-1)^2} = \sqrt{16 + 25} = \sqrt{41}$.
$PR = \sqrt{(4-5)^2 + (6-(-3))^2} = \sqrt{(-1)^2 + 9^2} = \sqrt{1 + 81} = \sqrt{82}$.
Case $2$: If $R$ is $(-4, 6)$:
$QR = \sqrt{(-4-0)^2 + (6-1)^2} = \sqrt{16 + 25} = \sqrt{41}$.
$PR = \sqrt{(-4-5)^2 + (6-(-3))^2} = \sqrt{(-9)^2 + 9^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2}$.

(D) Let the point be $P(x, y)$,$A(3, 6)$,and $B(-3, 4)$.
Since $P$ is equidistant from $A$ and $B$,we have $PA = PB$.
Using the distance formula,$PA^2 = PB^2$.
$(x - 3)^2 + (y - 6)^2 = (x - (-3))^2 + (y - 4)^2$
$(x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2$
Expanding both sides:
$x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16$
Canceling $x^2$ and $y^2$ from both sides:
$-6x - 12y + 45 = 6x - 8y + 25$
Rearranging the terms:
$45 - 25 = 6x + 6x - 8y + 12y$
$20 = 12x + 4y$
Dividing by $4$:
$5 = 3x + y$
Thus,the relation is $3x + y - 5 = 0$.

(N/A) It can be observed that Niharika posted the green flag at $\frac{1}{4}$ of the distance $AD,$ i.e.,$\left(\frac{1}{4} \times 100\right) \, m = 25 \, m$ from the starting point of the $2^{nd}$ line. Therefore,the coordinates of this point $G$ are $(2, 25).$
Preet posted the red flag at $\frac{1}{5}$ of the distance $AD,$ i.e.,$\left(\frac{1}{5} \times 100\right) \, m = 20 \, m$ from the starting point of the $8^{th}$ line. Therefore,the coordinates of this point $R$ are $(8, 20).$
Using the distance formula,the distance between these flags is $GR = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$
$GR = \sqrt{(8 - 2)^2 + (20 - 25)^2} = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61} \, m.$
The point at which Rashmi should post her blue flag is the mid-point of the line segment joining these points. Let this point be $M(x, y).$
$x = \frac{2 + 8}{2} = \frac{10}{2} = 5, \quad y = \frac{25 + 20}{2} = \frac{45}{2} = 22.5.$
Hence,the coordinates of the mid-point are $(5, 22.5).$
Therefore,Rashmi should post her blue flag at $22.5 \, m$ on the $5^{th}$ line.

(B) Let the vertices of the rhombus $ABCD$ be $A(3,0), B(4,5), C(-1,4)$ and $D(-2,-1)$.
The area of a rhombus is given by the formula $\text{Area} = \frac{1}{2} \times d_1 \times d_2$,where $d_1$ and $d_2$ are the lengths of the diagonals.
First,calculate the length of diagonal $AC$ using the distance formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AC = \sqrt{(-1-3)^2 + (4-0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$.
Next,calculate the length of diagonal $BD$:
$BD = \sqrt{(-2-4)^2 + (-1-5)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36+36} = \sqrt{72} = 6\sqrt{2}$.
Now,calculate the area:
$\text{Area} = \frac{1}{2} \times AC \times BD = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 24 \times 2 = 24$ square units.

(A) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $(2, -2)$ and $(-1, x)$ with distance $d = 5$.
Substituting the values into the formula:
$5 = \sqrt{(-1 - 2)^2 + (x - (-2))^2}$
$5 = \sqrt{(-3)^2 + (x + 2)^2}$
$5 = \sqrt{9 + (x^2 + 4x + 4)}$
$5 = \sqrt{x^2 + 4x + 13}$
Squaring both sides:
$25 = x^2 + 4x + 13$
$x^2 + 4x - 12 = 0$
Factoring the quadratic equation:
$(x + 6)(x - 2) = 0$
Thus,$x = -6$ or $x = 2$.
Therefore,one of the possible values of $x$ is $2$.

(C) To determine the shape,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
$AB = \sqrt{(9-9)^2 + (6-0)^2} = \sqrt{0^2 + 6^2} = 6$
$BC = \sqrt{(-9-9)^2 + (6-6)^2} = \sqrt{(-18)^2 + 0^2} = 18$
$CD = \sqrt{(-9-(-9))^2 + (0-6)^2} = \sqrt{0^2 + (-6)^2} = 6$
$DA = \sqrt{(9-(-9))^2 + (0-0)^2} = \sqrt{18^2 + 0^2} = 18$
Since opposite sides are equal ($AB = CD = 6$ and $BC = DA = 18$) and the adjacent sides are perpendicular (as the sides are parallel to the coordinate axes),the figure is a rectangle.

(D) We know that for any point $P(x, y)$ in the Cartesian plane:
$1$. The perpendicular distance of the point from the $y$-axis is given by the absolute value of its $x$-coordinate (abscissa),i.e.,$|x|$.
$2$. The perpendicular distance of the point from the $x$-axis is given by the absolute value of its $y$-coordinate (ordinate),i.e.,$|y|$.
For the point $P(2, 3)$,the $x$-coordinate is $2$ and the $y$-coordinate is $3$.
Therefore,the distance of the point $P(2, 3)$ from the $x$-axis is equal to the ordinate of the point,which is $3$ units.

(A) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Given points are $A(0, 6)$ and $B(0, -2)$.
Here,$x_1 = 0, y_1 = 6$ and $x_2 = 0, y_2 = -2$.
Substituting these values into the distance formula:
$AB = \sqrt{(0 - 0)^2 + (-2 - 6)^2}$
$AB = \sqrt{0^2 + (-8)^2}$
$AB = \sqrt{64}$
$AB = 8$
Therefore,the distance between the points $A$ and $B$ is $8$ units.

(B) The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Here,the point $P$ is $(-6, 8)$ and the origin $O$ is $(0, 0)$.
So,$x_1 = -6, y_1 = 8$ and $x_2 = 0, y_2 = 0$.
Substituting these values into the distance formula:
$PO = \sqrt{(0 - (-6))^2 + (0 - 8)^2}$
$PO = \sqrt{(6)^2 + (-8)^2}$
$PO = \sqrt{36 + 64}$
$PO = \sqrt{100}$
$PO = 10$
Thus,the distance of the point $P(-6, 8)$ from the origin is $10$ units.

(C) The distance formula between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by:
$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
Here,$x_{1} = 0, y_{1} = 5$ and $x_{2} = -5, y_{2} = 0$.
Substituting these values into the formula:
$d = \sqrt{(-5 - 0)^{2} + (0 - 5)^{2}}$
$d = \sqrt{(-5)^{2} + (-5)^{2}}$
$d = \sqrt{25 + 25}$
$d = \sqrt{50}$
$d = 5 \sqrt{2}$

(D) In a rectangle $AOBC$,the diagonal is the line segment joining opposite vertices. Here,$AB$ is the diagonal.
The length of the diagonal $AB$ is the distance between the points $A (0, 3)$ and $B (5, 0)$.
The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Here,$x_1 = 0, y_1 = 3$ and $x_2 = 5, y_2 = 0$.
Substituting these values into the distance formula:
$AB = \sqrt{(5 - 0)^2 + (0 - 3)^2}$
$AB = \sqrt{(5)^2 + (-3)^2}$
$AB = \sqrt{25 + 9}$
$AB = \sqrt{34}$
Thus,the length of the diagonal is $\sqrt{34}$.

(A) To find the perimeter of a triangle,we calculate the sum of the lengths of all its sides. Let the vertices of the triangle be $A(0,4)$,$O(0,0)$,and $B(3,0)$.
The perimeter of $\triangle AOB = \text{Sum of the lengths of all its sides} = d(AO) + d(OB) + d(AB)$.
The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Distance $d(AO) = \sqrt{(0-0)^2 + (0-4)^2} = \sqrt{0 + 16} = 4$.
$2$. Distance $d(OB) = \sqrt{(3-0)^2 + (0-0)^2} = \sqrt{9 + 0} = 3$.
$3$. Distance $d(AB) = \sqrt{(3-0)^2 + (0-4)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Perimeter $= 4 + 3 + 5 = 12$.
Thus,the required perimeter of the triangle is $12$.

(C) Let the vertices be $A(-4,0), B(4,0),$ and $C(0,3)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(4 - (-4))^2 + (0 - 0)^2} = \sqrt{8^2 + 0^2} = 8$
$BC = \sqrt{(0 - 4)^2 + (3 - 0)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$
$AC = \sqrt{(0 - (-4))^2 + (3 - 0)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$
Since $BC = AC = 5$,two sides of the triangle are equal.
Therefore,$\triangle ABC$ is an isosceles triangle.

(A) We know that the perpendicular bisector of any line segment passes through the midpoint of that line segment.
The midpoint of the line segment joining the points $A(-2, -5)$ and $B(2, 5)$ is calculated using the midpoint formula:
Midpoint $= \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$
Substituting the given coordinates:
Midpoint $= \left( \frac{-2 + 2}{2}, \frac{-5 + 5}{2} \right) = \left( \frac{0}{2}, \frac{0}{2} \right) = (0, 0)$
Since the perpendicular bisector must pass through the midpoint of the segment,the point $(0, 0)$ lies on the perpendicular bisector.

(C) The center of the circle is $(0,0)$ and it passes through the point $\left(\frac{13}{2}, 0\right)$.
Therefore,the radius of the circle $r$ is the distance between $(0,0)$ and $\left(\frac{13}{2}, 0\right)$.
$r = \sqrt{\left(\frac{13}{2}-0\right)^{2} + (0-0)^{2}} = \sqrt{\left(\frac{13}{2}\right)^{2}} = \frac{13}{2} = 6.5$.
$A$ point $(x, y)$ lies in the interior of the circle if its distance from the center is less than $r$,on the circle if the distance is equal to $r$,and outside if the distance is greater than $r$.
$(a)$ Distance of $\left(-\frac{3}{4}, 1\right)$ from $(0,0) = \sqrt{\left(-\frac{3}{4}\right)^{2} + 1^{2}} = \sqrt{\frac{9}{16} + 1} = \sqrt{\frac{25}{16}} = 1.25 < 6.5$. (Interior)
$(b)$ Distance of $\left(2, \frac{7}{3}\right)$ from $(0,0) = \sqrt{2^{2} + \left(\frac{7}{3}\right)^{2}} = \sqrt{4 + \frac{49}{9}} = \sqrt{\frac{85}{9}} \approx 3.07 < 6.5$. (Interior)
$(c)$ Distance of $\left(-6, \frac{5}{2}\right)$ from $(0,0) = \sqrt{(-6)^{2} + \left(\frac{5}{2}\right)^{2}} = \sqrt{36 + \frac{25}{4}} = \sqrt{\frac{144+25}{4}} = \sqrt{\frac{169}{4}} = \frac{13}{2} = 6.5$. (On the circle)
$(d)$ Distance of $\left(5, -\frac{1}{2}\right)$ from $(0,0) = \sqrt{5^{2} + \left(-\frac{1}{2}\right)^{2}} = \sqrt{25 + \frac{1}{4}} = \sqrt{\frac{101}{4}} \approx 5.02 < 6.5$. (Interior)
Thus,the point $\left(-6, \frac{5}{2}\right)$ does not lie in the interior of the circle.

(B) According to the distance formula,the distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $(4, p)$ and $(1, 0)$ with distance $d = 5$.
Substituting the values into the formula:
$5 = \sqrt{(1 - 4)^2 + (0 - p)^2}$
$5 = \sqrt{(-3)^2 + (-p)^2}$
$5 = \sqrt{9 + p^2}$
Squaring both sides:
$25 = 9 + p^2$
$p^2 = 25 - 9$
$p^2 = 16$
$p = \pm \sqrt{16}$
$p = \pm 4$
Thus,the value of $p$ is $\pm 4$.

(TRUE) The statement is True.
$1$. $A$ point on the $y$-axis has coordinates $(0, y)$. Since $P (0, -7)$ lies on the $y$-axis,it satisfies the first condition.
$2$. For $P$ to be on the perpendicular bisector of $AB$,it must be equidistant from $A (-1, 0)$ and $B (7, -6)$.
$3$. Calculate the distance $PA = \sqrt{(0 - (-1))^2 + (-7 - 0)^2} = \sqrt{1^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50}$ units.
$4$. Calculate the distance $PB = \sqrt{(0 - 7)^2 + (-7 - (-6))^2} = \sqrt{(-7)^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50}$ units.
$5$. Since $PA = PB$,point $P$ lies on the perpendicular bisector of $AB$. Thus,the statement is true.

(B) False.
We know that any point lying on the perpendicular bisector of a line segment is equidistant from the endpoints of that segment.
Let us calculate the distance of point $P(0, 2)$ from points $A(-1, 1)$ and $B(3, 3)$ using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$PA = \sqrt{(0 - (-1))^2 + (2 - 1)^2} = \sqrt{(1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$.
$PB = \sqrt{(0 - 3)^2 + (2 - 3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$.
Since $PA \neq PB$,the point $P(0, 2)$ is not equidistant from $A$ and $B$.
Therefore,the point $P(0, 2)$ does not lie on the perpendicular bisector of the line segment $AB$.

(B) False.
To determine if the points form a parallelogram,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$AB = \sqrt{(6-4)^2 + (4-3)^2} = \sqrt{2^2 + 1^2} = \sqrt{5}$
$BC = \sqrt{(5-6)^2 + (-6-4)^2} = \sqrt{(-1)^2 + (-10)^2} = \sqrt{1 + 100} = \sqrt{101}$
$CD = \sqrt{(-3-5)^2 + (5 - (-6))^2} = \sqrt{(-8)^2 + 11^2} = \sqrt{64 + 121} = \sqrt{185}$
$DA = \sqrt{(4 - (-3))^2 + (3-5)^2} = \sqrt{7^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53}$
In a parallelogram,opposite sides must be equal ($AB = CD$ and $BC = DA$). Since all sides are unequal,the given points do not form a parallelogram.

(B) False.
If a point $A$ lies on the perpendicular bisector of a line segment $PQ$,then it must be equidistant from the endpoints $P$ and $Q$,i.e.,$AP = AQ$.
First,calculate the distance $AP$ using the distance formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AP = \sqrt{(6-2)^2 + (5-7)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20}$.
Next,calculate the distance $AQ$:
$AQ = \sqrt{(0-2)^2 + (-4-7)^2} = \sqrt{(-2)^2 + (-11)^2} = \sqrt{4 + 121} = \sqrt{125}$.
Since $AP \neq AQ$ (as $\sqrt{20} \neq \sqrt{125}$),the point $A (2,7)$ does not lie on the perpendicular bisector of the line segment $PQ$.

(A) True.
First,we check if the points are collinear by calculating the area of the triangle formed by them.
Area of triangle $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Here,$x_1 = -6, x_2 = -4, x_3 = 3$ and $y_1 = 10, y_2 = 6, y_3 = -8$.
Area $= \frac{1}{2} | -6(6 - (-8)) + (-4)(-8 - 10) + 3(10 - 6) |$.
Area $= \frac{1}{2} | -6(14) + (-4)(-18) + 3(4) |$.
Area $= \frac{1}{2} | -84 + 72 + 12 | = \frac{1}{2} | 0 | = 0$.
Since the area is $0$,the points are collinear.
Next,we calculate the distances $AB$ and $AC$ using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$AB = \sqrt{(-4 - (-6))^2 + (6 - 10)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$.
$AC = \sqrt{(3 - (-6))^2 + (-8 - 10)^2} = \sqrt{9^2 + (-18)^2} = \sqrt{81 + 324} = \sqrt{405} = 9\sqrt{5}$.
Now,check the relation $AB = \frac{2}{9} AC$:
$\frac{2}{9} AC = \frac{2}{9} \times 9\sqrt{5} = 2\sqrt{5}$.
Since $AB = 2\sqrt{5}$,the relation holds true.

(B) False.
$A$ point lies on a circle if and only if the distance between the point and the centre of the circle is equal to the radius of the circle.
Let the centre be $C(3, 5)$ and the point be $P(-2, 4)$. The distance $CP$ is calculated using the distance formula:
$CP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
$CP = \sqrt{(3 - (-2))^2 + (5 - 4)^2}$
$CP = \sqrt{(3 + 2)^2 + (1)^2}$
$CP = \sqrt{5^2 + 1^2}$
$CP = \sqrt{25 + 1} = \sqrt{26}$
Since $\sqrt{26} \neq 6$,the distance between the point $P$ and the centre $C$ is not equal to the radius of the circle.
Therefore,the point $P(-2, 4)$ does not lie on the circle.

(A) Let $O(x, y)$ be the centre of the circle. Let the points $A(6, -6), B(3, -7),$ and $C(3, 3)$ lie on the circumference.
Since the distance from the centre to any point on the circle is equal (radius $r$):
$OA^2 = OB^2 = OC^2$
$OA^2 = (x - 6)^2 + (y + 6)^2$
$OB^2 = (x - 3)^2 + (y + 7)^2$
$OC^2 = (x - 3)^2 + (y - 3)^2$
Equating $OB^2 = OC^2$:
$(x - 3)^2 + (y + 7)^2 = (x - 3)^2 + (y - 3)^2$
$(y + 7)^2 = (y - 3)^2$
$y^2 + 14y + 49 = y^2 - 6y + 9$
$20y = -40 \Rightarrow y = -2$
Equating $OA^2 = OC^2$:
$(x - 6)^2 + (y + 6)^2 = (x - 3)^2 + (y - 3)^2$
Substitute $y = -2$:
$(x - 6)^2 + (-2 + 6)^2 = (x - 3)^2 + (-2 - 3)^2$
$(x - 6)^2 + 16 = (x - 3)^2 + 25$
$x^2 - 12x + 36 + 16 = x^2 - 6x + 9 + 25$
$-6x = -18 \Rightarrow x = 3$
Thus,the centre is $(3, -2)$.

(A) Let $A(-1, 2)$ and $C(3, 2)$ be the given opposite vertices of a square $ABCD$. Let the other two vertices be $B(x, y)$ and $D(x_1, y_1)$.
Since $ABCD$ is a square,all sides are equal,so $AB = BC$.
Using the distance formula: $\sqrt{(x+1)^2 + (y-2)^2} = \sqrt{(x-3)^2 + (y-2)^2}$.
Squaring both sides: $(x+1)^2 + (y-2)^2 = (x-3)^2 + (y-2)^2$.
$x^2 + 2x + 1 = x^2 - 6x + 9 \Rightarrow 8x = 8 \Rightarrow x = 1$.
In $\triangle ABC$,$\angle B = 90^\circ$,so by Pythagoras theorem: $AB^2 + BC^2 = AC^2$.
$AC^2 = (3 - (-1))^2 + (2 - 2)^2 = 4^2 + 0^2 = 16$.
Since $AB = BC$,$2AB^2 = 16 \Rightarrow AB^2 = 8$.
$(x+1)^2 + (y-2)^2 = 8$. Substituting $x=1$: $(1+1)^2 + (y-2)^2 = 8 \Rightarrow 4 + (y-2)^2 = 8 \Rightarrow (y-2)^2 = 4$.
$y-2 = \pm 2 \Rightarrow y = 4$ or $y = 0$.
Thus,$B$ is $(1, 4)$ and $D$ is $(1, 0)$ (or vice versa).
The coordinates of the other two vertices are $(1, 0)$ and $(1, 4)$.

(N/A) Taking $A$ as the origin $(0,0)$,we consider $AD$ as the $x$-axis and $AB$ as the $y$-axis.
By observing the grid provided in the figure:
- The point $P$ is located at $4$ units along the $x$-axis and $6$ units along the $y$-axis,so its coordinates are $P(4, 6)$.
- The point $Q$ is located at $3$ units along the $x$-axis and $2$ units along the $y$-axis,so its coordinates are $Q(3, 2)$.
- The point $R$ is located at $6$ units along the $x$-axis and $5$ units along the $y$-axis,so its coordinates are $R(6, 5)$.
Thus,the coordinates of the vertices of the triangle are $P(4, 6)$,$Q(3, 2)$,and $R(6, 5)$.

(A) Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$PQ = \sqrt{(-\sqrt{2} - \sqrt{2})^2 + (-\sqrt{2} - \sqrt{2})^2} = \sqrt{(-2\sqrt{2})^2 + (-2\sqrt{2})^2} = \sqrt{8 + 8} = \sqrt{16} = 4$
$PR = \sqrt{(-\sqrt{6} - \sqrt{2})^2 + (\sqrt{6} - \sqrt{2})^2} = \sqrt{(6 + 2 + 2\sqrt{12}) + (6 + 2 - 2\sqrt{12})} = \sqrt{8 + 8} = \sqrt{16} = 4$
$RQ = \sqrt{(-\sqrt{6} - (-\sqrt{2}))^2 + (\sqrt{6} - (-\sqrt{2}))^2} = \sqrt{(-\sqrt{6} + \sqrt{2})^2 + (\sqrt{6} + \sqrt{2})^2} = \sqrt{(6 + 2 - 2\sqrt{12}) + (6 + 2 + 2\sqrt{12})} = \sqrt{8 + 8} = \sqrt{16} = 4$
Since $PQ = PR = RQ = 4$,all three sides are equal. Therefore,the triangle $PQR$ is an equilateral triangle.

(C) To determine the type of triangle,we calculate the lengths of all three sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of side $AB$:
$AB = \sqrt{(-4 - (-5))^2 + (-2 - 6)^2} = \sqrt{(1)^2 + (-8)^2} = \sqrt{1 + 64} = \sqrt{65}$ units.
$2$. Length of side $BC$:
$BC = \sqrt{(7 - (-4))^2 + (5 - (-2))^2} = \sqrt{(11)^2 + (7)^2} = \sqrt{121 + 49} = \sqrt{170}$ units.
$3$. Length of side $CA$:
$CA = \sqrt{(-5 - 7)^2 + (6 - 5)^2} = \sqrt{(-12)^2 + (1)^2} = \sqrt{144 + 1} = \sqrt{145}$ units.
Since $AB \neq BC \neq CA$,all three sides have different lengths.
Therefore,the triangle formed by these points is a scalene triangle.

(A) We know that every point on the $x$-axis is of the form $(x, 0)$. Let $P(x, 0)$ be a point on the $x$-axis at a distance of $2\sqrt{5}$ from the point $Q(7, -4)$.
Using the distance formula,$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given $PQ = 2\sqrt{5}$,so $(PQ)^2 = (2\sqrt{5})^2 = 4 \times 5 = 20$.
Substituting the coordinates:
$(x - 7)^2 + (0 - (-4))^2 = 20$
$(x - 7)^2 + (4)^2 = 20$
$x^2 - 14x + 49 + 16 = 20$
$x^2 - 14x + 65 = 20$
$x^2 - 14x + 45 = 0$
Factoring the quadratic equation:
$x^2 - 9x - 5x + 45 = 0$
$x(x - 9) - 5(x - 9) = 0$
$(x - 9)(x - 5) = 0$
Thus,$x = 5$ or $x = 9$.
The points are $(5, 0)$ and $(9, 0)$.
There are $2$ such points.

(B) According to the distance formula,the distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(-3, -14)$ and $B(a, -5)$ with distance $d = 9$.
Substituting the values into the formula:
$9 = \sqrt{(a - (-3))^2 + (-5 - (-14))^2}$
$9 = \sqrt{(a + 3)^2 + (-5 + 14)^2}$
$9 = \sqrt{(a + 3)^2 + (9)^2}$
Squaring both sides:
$81 = (a + 3)^2 + 81$
Subtracting $81$ from both sides:
$(a + 3)^2 = 0$
Taking the square root:
$a + 3 = 0$
$a = -3$.
Thus,the required value of $a$ is $-3$.

(D) Let $P(h, k)$ be a point equidistant from $A(-5, 4)$ and $B(-1, 6)$.
By the distance formula,$PA = PB$,which implies $PA^2 = PB^2$.
$(-5 - h)^2 + (4 - k)^2 = (-1 - h)^2 + (6 - k)^2$
Expanding both sides:
$25 + h^2 + 10h + 16 + k^2 - 8k = 1 + h^2 + 2h + 36 + k^2 - 12k$
Simplifying the equation:
$10h - 2h - 8k + 12k + 41 - 37 = 0$
$8h + 4k + 4 = 0$
Dividing by $4$,we get:
$2h + k + 1 = 0$
This is the equation of the perpendicular bisector of the line segment $AB$. Any point lying on this line will be equidistant from $A$ and $B$. Since a line contains an infinite number of points,there are infinitely many such points.
Replacing $h$ and $k$ with $x$ and $y$,the locus of all such points is $2x + y + 1 = 0$.

(N/A) According to the question,point $A (2, -4)$ is equidistant from $P (3, 8)$ and $Q (-10, y)$.
This means $PA = QA$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$,we have:
$\sqrt{(2 - 3)^2 + (-4 - 8)^2} = \sqrt{(2 - (-10))^2 + (-4 - y)^2}$
$\sqrt{(-1)^2 + (-12)^2} = \sqrt{(12)^2 + (-(4 + y))^2}$
$\sqrt{1 + 144} = \sqrt{144 + (4 + y)^2}$
$\sqrt{145} = \sqrt{144 + 16 + 8y + y^2}$
Squaring both sides:
$145 = 160 + 8y + y^2$
$y^2 + 8y + 15 = 0$
$(y + 5)(y + 3) = 0$
So,$y = -5$ or $y = -3$.
Now,calculating distance $PQ = \sqrt{(-10 - 3)^2 + (y - 8)^2} = \sqrt{(-13)^2 + (y - 8)^2} = \sqrt{169 + (y - 8)^2}$.
For $y = -3$,$PQ = \sqrt{169 + (-3 - 8)^2} = \sqrt{169 + 121} = \sqrt{290}$.
For $y = -5$,$PQ = \sqrt{169 + (-5 - 8)^2} = \sqrt{169 + 169} = \sqrt{338} = 13\sqrt{2}$.

(A) The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Distance from House $(2,4)$ to Bank $(5,8)$:
$d_1 = \sqrt{(5-2)^2 + (8-4)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5 \, km$.
$2$. Distance from Bank $(5,8)$ to School $(13,14)$:
$d_2 = \sqrt{(13-5)^2 + (14-8)^2} = \sqrt{8^2 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10 \, km$.
$3$. Distance from School $(13,14)$ to Office $(13,26)$:
$d_3 = \sqrt{(13-13)^2 + (26-14)^2} = \sqrt{0^2 + 12^2} = 12 \, km$.
Total distance travelled by Ayush $= 5 + 10 + 12 = 27 \, km$.
$4$. Direct distance from House $(2,4)$ to Office $(13,26)$:
$d_{direct} = \sqrt{(13-2)^2 + (26-4)^2} = \sqrt{11^2 + 22^2} = \sqrt{121 + 484} = \sqrt{605} \approx 24.6 \, km$.
Extra distance travelled $= 27 - 24.6 = 2.4 \, km$.

(B) The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(5, 8)$ and $B(1, 2)$.
Here,$x_1 = 5, y_1 = 8, x_2 = 1, y_2 = 2$.
Substituting these values into the formula:
$AB = \sqrt{(1 - 5)^2 + (2 - 8)^2}$
$AB = \sqrt{(-4)^2 + (-6)^2}$
$AB = \sqrt{16 + 36}$
$AB = \sqrt{52}$
$AB = \sqrt{4 \times 13}$
$AB = 2 \sqrt{13}$.

(A) Let the vertices be $A(2,2), B(5,2), C(5,5)$ and $D(2,5)$.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB^2 = (5-2)^2 + (2-2)^2 = 3^2 + 0^2 = 9 \implies AB = 3$.
$BC^2 = (5-5)^2 + (5-2)^2 = 0^2 + 3^2 = 9 \implies BC = 3$.
$CD^2 = (2-5)^2 + (5-5)^2 = (-3)^2 + 0^2 = 9 \implies CD = 3$.
$DA^2 = (2-2)^2 + (2-5)^2 = 0^2 + (-3)^2 = 9 \implies DA = 3$.
Since $AB = BC = CD = DA = 3$,the quadrilateral is a rhombus.
Now,check the diagonals:
$AC^2 = (5-2)^2 + (5-2)^2 = 3^2 + 3^2 = 9 + 9 = 18 \implies AC = \sqrt{18} = 3\sqrt{2}$.
$BD^2 = (2-5)^2 + (5-2)^2 = (-3)^2 + 3^2 = 9 + 9 = 18 \implies BD = \sqrt{18} = 3\sqrt{2}$.
Since all sides are equal and the diagonals are equal $(AC = BD)$,the quadrilateral $ABCD$ is a square.