Textbook - Coordinate Geometry 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 $39$.

(B) Let the points be $A(1,5), B(2,3),$ and $C(-2,-11)$.
To check if the points are collinear,we calculate the area of the triangle formed by these points using the formula:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the values:
$\text{Area} = \frac{1}{2} |1(3 - (-11)) + 2(-11 - 5) + (-2)(5 - 3)|$
$\text{Area} = \frac{1}{2} |1(14) + 2(-16) + (-2)(2)|$
$\text{Area} = \frac{1}{2} |14 - 32 - 4|$
$\text{Area} = \frac{1}{2} |-22| = 11 \text{ square units}$.
Since the area of the triangle is not $0$,the points are not collinear.

(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 $3x + y - 5 = 0$.

(A) Let $P(x, y)$ be the required point that divides the line segment joining $A(4, -3)$ and $B(8, 5)$ in the ratio $m:n = 3:1$.
Using the section formula,the coordinates $(x, y)$ are given by:
$x = \frac{mx_2 + nx_1}{m + n} = \frac{3(8) + 1(4)}{3 + 1} = \frac{24 + 4}{4} = \frac{28}{4} = 7$
$y = \frac{my_2 + ny_1}{m + n} = \frac{3(5) + 1(-3)}{3 + 1} = \frac{15 - 3}{4} = \frac{12}{4} = 3$
Therefore,the required point is $(7, 3)$.

(A) Let the point $P(-4, 6)$ divide the line segment joining $A(-6, 10)$ and $B(3, -8)$ in the ratio $m_{1}: m_{2}$.
Using the section formula,the coordinates of $P$ are given by:
$P(x, y) = \left( \frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}, \frac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}} \right)$
Substituting the values:
$(-4, 6) = \left( \frac{3m_{1} - 6m_{2}}{m_{1} + m_{2}}, \frac{-8m_{1} + 10m_{2}}{m_{1} + m_{2}} \right)$
Equating the $x$-coordinates:
$-4 = \frac{3m_{1} - 6m_{2}}{m_{1} + m_{2}}$
$-4(m_{1} + m_{2}) = 3m_{1} - 6m_{2}$
$-4m_{1} - 4m_{2} = 3m_{1} - 6m_{2}$
$2m_{2} = 7m_{1}$
$\frac{m_{1}}{m_{2}} = \frac{2}{7}$
Thus,the ratio is $2: 7$.

(N/A) Let $P$ and $Q$ be the points of trisection of $AB$ such that $AP = PQ = QB$.
Point $P$ divides $AB$ internally in the ratio $1:2$. Using the section formula,the coordinates of $P$ are:
$P = \left(\frac{1(-7) + 2(2)}{1 + 2}, \frac{1(4) + 2(-2)}{1 + 2}\right) = \left(\frac{-7 + 4}{3}, \frac{4 - 4}{3}\right) = \left(\frac{-3}{3}, \frac{0}{3}\right) = (-1, 0)$.
Point $Q$ divides $AB$ internally in the ratio $2:1$. Using the section formula,the coordinates of $Q$ are:
$Q = \left(\frac{2(-7) + 1(2)}{2 + 1}, \frac{2(4) + 1(-2)}{2 + 1}\right) = \left(\frac{-14 + 2}{3}, \frac{8 - 2}{3}\right) = \left(\frac{-12}{3}, \frac{6}{3}\right) = (-4, 2)$.
Therefore,the coordinates of the points of trisection are $(-1, 0)$ and $(-4, 2)$.

(A) Let the ratio in which the $y$-axis divides the line segment be $k:1$.
By the section formula,the coordinates of the point dividing the segment joining $(x_1, y_1) = (5, -6)$ and $(x_2, y_2) = (-1, -4)$ in the ratio $k:1$ are given by:
$\left( \frac{kx_2 + x_1}{k+1}, \frac{ky_2 + y_1}{k+1} \right) = \left( \frac{k(-1) + 5}{k+1}, \frac{k(-4) + (-6)}{k+1} \right) = \left( \frac{-k+5}{k+1}, \frac{-4k-6}{k+1} \right)$.
Since this point lies on the $y$-axis,its $x$-coordinate (abscissa) must be $0$.
Therefore,$\frac{-k+5}{k+1} = 0$,which implies $-k+5 = 0$,so $k = 5$.
Thus,the ratio is $5:1$.
Now,substitute $k=5$ into the $y$-coordinate expression:
$y = \frac{-4(5) - 6}{5+1} = \frac{-20-6}{6} = \frac{-26}{6} = -\frac{13}{3}$.
Hence,the point of intersection is $(0, -13/3)$.

(D) We know that the diagonals of a parallelogram bisect each other.
Therefore,the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.
The coordinates of the midpoint of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ are given by $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
Midpoint of $AC = \left(\frac{6+9}{2}, \frac{1+4}{2}\right) = \left(\frac{15}{2}, \frac{5}{2}\right)$.
Midpoint of $BD = \left(\frac{8+p}{2}, \frac{2+3}{2}\right) = \left(\frac{8+p}{2}, \frac{5}{2}\right)$.
Equating the midpoints,we get $\frac{15}{2} = \frac{8+p}{2}$.
Multiplying both sides by $2$,we get $15 = 8 + p$.
Thus,$p = 15 - 8 = 7$.

(A) Let $P(x, y)$ be the required point that divides the line segment joining $A(-1, 7)$ and $B(4, -3)$ in the ratio $m:n = 2:3$.
Using the section formula:
$x = \frac{mx_2 + nx_1}{m + n} = \frac{2(4) + 3(-1)}{2 + 3} = \frac{8 - 3}{5} = \frac{5}{5} = 1$
$y = \frac{my_2 + ny_1}{m + n} = \frac{2(-3) + 3(7)}{2 + 3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3$
Therefore,the coordinates of the point are $(1, 3)$.

(A) Let $A = (4, -1)$ and $B = (-2, -3)$. Let $P$ and $Q$ be the points of trisection such that $AP = PQ = QB$.
Point $P$ divides $AB$ in the ratio $1:2$. Using the section formula $\left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)$:
$x_1 = \frac{1(-2) + 2(4)}{1+2} = \frac{-2+8}{3} = \frac{6}{3} = 2$
$y_1 = \frac{1(-3) + 2(-1)}{1+2} = \frac{-3-2}{3} = -\frac{5}{3}$
So,$P = (2, -5/3)$.
Point $Q$ divides $AB$ in the ratio $2:1$. Using the section formula:
$x_2 = \frac{2(-2) + 1(4)}{2+1} = \frac{-4+4}{3} = 0$
$y_2 = \frac{2(-3) + 1(-1)}{2+1} = \frac{-6-1}{3} = -\frac{7}{3}$
So,$Q = (0, -7/3)$.

(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.

(A) Let the ratio in which the line segment joining the points $A(-3, 10)$ and $B(6, -8)$ is divided by the point $P(-1, 6)$ be $k : 1$.
Using the section formula,the coordinates of point $P$ are given by:
$P(x, y) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right)$
Here,$m = k$,$n = 1$,$(x_1, y_1) = (-3, 10)$,and $(x_2, y_2) = (6, -8)$.
Equating the $x$-coordinate:
$-1 = \frac{k(6) + 1(-3)}{k + 1}$
$-1(k + 1) = 6k - 3$
$-k - 1 = 6k - 3$
$-k - 6k = -3 + 1$
$-7k = -2$
$k = \frac{2}{7}$
Thus,the required ratio is $2 : 7$.

(A) Let the ratio in which the line segment joining $A(1, -5)$ and $B(-4, 5)$ is divided by the $x$-axis be $k: 1$.
Using the section formula,the coordinates of the point of division are $\left(\frac{-4k + 1}{k + 1}, \frac{5k - 5}{k + 1}\right)$.
Since the point lies on the $x$-axis,its $y$-coordinate must be $0$.
Therefore,$\frac{5k - 5}{k + 1} = 0$.
This implies $5k - 5 = 0$,so $k = 1$.
The ratio is $1: 1$.
Substituting $k = 1$ into the $x$-coordinate expression: $\frac{-4(1) + 1}{1 + 1} = \frac{-3}{2}$.
Thus,the point of division is $\left(\frac{-3}{2}, 0\right)$.

(N/A) Let the vertices of the parallelogram $ABCD$ be $A(1, 2), B(4, y), C(x, 6),$ and $D(3, 5).$
In a parallelogram,the diagonals bisect each other at the same point $O.$
Therefore,$O$ is the midpoint of both diagonals $AC$ and $BD.$
The midpoint of diagonal $AC$ is given by $\left(\frac{1+x}{2}, \frac{2+6}{2}\right) = \left(\frac{x+1}{2}, 4\right).$
The midpoint of diagonal $BD$ is given by $\left(\frac{4+3}{2}, \frac{y+5}{2}\right) = \left(\frac{7}{2}, \frac{y+5}{2}\right).$
Since both represent the same point $O,$ we equate their coordinates:
$\frac{x+1}{2} = \frac{7}{2} \implies x+1 = 7 \implies x = 6.$
$4 = \frac{y+5}{2} \implies 8 = y+5 \implies y = 3.$
Thus,$x = 6$ and $y = 3.$

(A) Let the coordinates of point $A$ be $(x, y).$
Since $AB$ is the diameter of the circle,the center of the circle is the midpoint of the diameter $AB.$
Given that the center is $(2, -3)$ and point $B$ is $(1, 4).$
Using the midpoint formula,the coordinates of the center are $\left(\frac{x+1}{2}, \frac{y+4}{2}\right).$
Equating this to the given center $(2, -3),$
$\frac{x+1}{2} = 2 \Rightarrow x+1 = 4 \Rightarrow x = 3.$
$\frac{y+4}{2} = -3 \Rightarrow y+4 = -6 \Rightarrow y = -10.$
Therefore,the coordinates of point $A$ are $(3, -10).$

(N/A) The coordinates of points $A$ and $B$ are $(-2, -2)$ and $(2, -4)$ respectively.
Since $AP = \frac{3}{7} AB$,it implies that $AP : AB = 3 : 7$.
Therefore,$AP : PB = 3 : (7 - 3) = 3 : 4$.
Point $P$ divides the line segment $AB$ in the ratio $m : n = 3 : 4$.
Using the section formula,the coordinates of $P$ are given by:
$P = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right)$
$P = \left( \frac{3(2) + 4(-2)}{3 + 4}, \frac{3(-4) + 4(-2)}{3 + 4} \right)$
$P = \left( \frac{6 - 8}{7}, \frac{-12 - 8}{7} \right)$
$P = \left( -\frac{2}{7}, -\frac{20}{7} \right)$

(N/A) Let the points dividing the line segment $AB$ into four equal parts be $X, Y, Z$. These points divide the segment in the ratios $1:3, 1:1, 3:1$ respectively.
Coordinates of $X = \left(\frac{1 \times 2 + 3 \times (-2)}{1 + 3}, \frac{1 \times 8 + 3 \times 2}{1 + 3}\right) = \left(\frac{2 - 6}{4}, \frac{8 + 6}{4}\right) = \left(-1, \frac{14}{4}\right) = \left(-1, 3.5\right)$.
Coordinates of $Y$ (midpoint of $AB$) $= \left(\frac{-2 + 2}{2}, \frac{2 + 8}{2}\right) = \left(0, \frac{10}{2}\right) = (0, 5)$.
Coordinates of $Z = \left(\frac{3 \times 2 + 1 \times (-2)}{3 + 1}, \frac{3 \times 8 + 1 \times 2}{3 + 1}\right) = \left(\frac{6 - 2}{4}, \frac{24 + 2}{4}\right) = \left(\frac{4}{4}, \frac{26}{4}\right) = (1, 6.5)$.

(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.

(C) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Given vertices are $A(1, -1), B(-4, 6)$ and $C(-3, -5)$.
Substituting these values into the formula:
Area $= \frac{1}{2} |1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6)|$
Area $= \frac{1}{2} |1(11) + (-4)(-4) + (-3)(-7)|$
Area $= \frac{1}{2} |11 + 16 + 21|$
Area $= \frac{1}{2} |48| = 24$
Thus,the area of the triangle is $24$ square units.

(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Given vertices are $A(5, 2)$,$B(4, 7)$,and $C(7, -4)$.
Substituting the values into the formula:
Area $= \frac{1}{2} |5(7 - (-4)) + 4(-4 - 2) + 7(2 - 7)|$
Area $= \frac{1}{2} |5(11) + 4(-6) + 7(-5)|$
Area $= \frac{1}{2} |55 - 24 - 35|$
Area $= \frac{1}{2} |55 - 59|$
Area $= \frac{1}{2} |-4|$
Area $= \frac{1}{2} \times 4 = 2$ square units.

(A) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the given coordinates $P(-1.5, 3)$,$Q(6, -2)$,and $R(-3, 4)$:
Area $= \frac{1}{2} |-1.5(-2 - 4) + 6(4 - 3) + (-3)(3 - (-2))|$
Area $= \frac{1}{2} |-1.5(-6) + 6(1) - 3(5)|$
Area $= \frac{1}{2} |9 + 6 - 15|$
Area $= \frac{1}{2} |0| = 0$ square units.
Since the area is $0$,the points $P$,$Q$,and $R$ are collinear,meaning they lie on the same straight line and do not form a triangle.

(B) Since the given points $A(2, 3)$,$B(4, k)$,and $C(6, -3)$ are collinear,the area of the triangle formed by these points must be $0$.
The formula for the area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$
Substituting the given coordinates:
$\frac{1}{2} |2(k - (-3)) + 4(-3 - 3) + 6(3 - k)| = 0$
$\frac{1}{2} |2(k + 3) + 4(-6) + 6(3 - k)| = 0$
$\frac{1}{2} |2k + 6 - 24 + 18 - 6k| = 0$
$\frac{1}{2} |-4k| = 0$
$-2k = 0$
Therefore,$k = 0$.

(C) To find the area of the quadrilateral $ABCD$,we can divide it into two triangles by drawing a diagonal $BD$.
The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
$1$. Area of $\Delta ABD$ with vertices $A(-5, 7)$,$B(-4, -5)$,and $D(4, 5)$:
$\text{Area} = \frac{1}{2} |-5(-5 - 5) + (-4)(5 - 7) + 4(7 - (-5))|$
$= \frac{1}{2} |-5(-10) - 4(-2) + 4(12)|$
$= \frac{1}{2} |50 + 8 + 48| = \frac{106}{2} = 53$ square units.
$2$. Area of $\Delta BCD$ with vertices $B(-4, -5)$,$C(-1, -6)$,and $D(4, 5)$:
$\text{Area} = \frac{1}{2} |-4(-6 - 5) + (-1)(5 - (-5)) + 4(-5 - (-6))|$
$= \frac{1}{2} |-4(-11) - 1(10) + 4(1)|$
$= \frac{1}{2} |44 - 10 + 4| = \frac{38}{2} = 19$ square units.
Total area of quadrilateral $ABCD = \text{Area}(\Delta ABD) + \text{Area}(\Delta BCD) = 53 + 19 = 72$ square units.

(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Given vertices are $(x_1, y_1) = (2, 3), (x_2, y_2) = (-1, 0),$ and $(x_3, y_3) = (2, -4)$.
Substituting these values into the formula:
Area $= \frac{1}{2} |2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0)|$
Area $= \frac{1}{2} |2(4) + (-1)(-7) + 2(3)|$
Area $= \frac{1}{2} |8 + 7 + 6|$
Area $= \frac{1}{2} |21| = 10.5$ square units.

(B) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Given vertices are $(x_1, y_1) = (-5, -1)$,$(x_2, y_2) = (3, -5)$,and $(x_3, y_3) = (5, 2)$.
Substituting these values into the formula:
Area $= \frac{1}{2} |(-5)(-5 - 2) + 3(2 - (-1)) + 5(-1 - (-5))|$
Area $= \frac{1}{2} |(-5)(-7) + 3(3) + 5(4)|$
Area $= \frac{1}{2} |35 + 9 + 20|$
Area $= \frac{1}{2} |64| = 32$ square units.

(B) For points to be collinear,the area of the triangle formed by them must be equal to $0$.
The formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$.
Substituting the given points $(7, -2), (5, 1),$ and $(3, k)$ into the formula:
$\frac{1}{2} |7(1 - k) + 5(k - (-2)) + 3(-2 - 1)| = 0$
$\frac{1}{2} |7 - 7k + 5(k + 2) + 3(-3)| = 0$
$\frac{1}{2} |7 - 7k + 5k + 10 - 9| = 0$
$|8 - 2k| = 0$
$8 - 2k = 0$
$2k = 8$
$k = 4$

(C) For points to be collinear,the area of the triangle formed by them must be $0$.
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$.
Substituting the given points $(8,1), (k,-4),$ and $(2,-5)$ into the formula:
$\frac{1}{2} |8(-4 - (-5)) + k(-5 - 1) + 2(1 - (-4))| = 0$
$\frac{1}{2} |8(-4 + 5) + k(-6) + 2(1 + 4)| = 0$
$\frac{1}{2} |8(1) - 6k + 2(5)| = 0$
$|8 - 6k + 10| = 0$
$18 - 6k = 0$
$6k = 18$
$k = 3$

(1:4) Let the vertices of the triangle be $A(0,-1), B(2,1), C(0,3)$.
Let $D, E, F$ be the mid-points of the sides $AB, AC, BC$ respectively. The coordinates of $D, E,$ and $F$ are calculated using the mid-point formula:
$D = \left(\frac{0+2}{2}, \frac{-1+1}{2}\right) = (1,0)$
$E = \left(\frac{0+0}{2}, \frac{-1+3}{2}\right) = (0,1)$
$F = \left(\frac{2+0}{2}, \frac{1+3}{2}\right) = (1,2)$
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$.
Area of $\Delta DEF = \frac{1}{2} |1(1-2) + 0(2-0) + 1(0-1)| = \frac{1}{2} |-1 + 0 - 1| = \frac{1}{2} |-2| = 1$ square unit.
Area of $\Delta ABC = \frac{1}{2} |0(1-3) + 2(3-(-1)) + 0(-1-1)| = \frac{1}{2} |0 + 2(4) + 0| = \frac{1}{2} |8| = 4$ square units.
Therefore,the required ratio is $1:4$.

(B) Let the vertices of the quadrilateral be $A(-4,-2), B(-3,-5), C(3,-2),$ and $D(2,3)$. Join $AC$ to form two triangles $\triangle ABC$ and $\triangle ACD$.
Area of a triangle $= \frac{1}{2} |x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})|$
Area of $\triangle ABC = \frac{1}{2} |(-4){(-5)-(-2)} + (-3){(-2)-(-2)} + 3{(-2)-(-5)}|$
$= \frac{1}{2} |(-4)(-3) + (-3)(0) + 3(3)| = \frac{1}{2} |12 + 0 + 9| = \frac{21}{2} = 10.5$ square units.
Area of $\triangle ACD = \frac{1}{2} |(-4){(-2)-(3)} + 3{(3)-(-2)} + 2{(-2)-(-2)}|$
$= \frac{1}{2} |(-4)(-5) + 3(5) + 2(0)| = \frac{1}{2} |20 + 15 + 0| = \frac{35}{2} = 17.5$ square units.
Area of quadrilateral $ABCD = \text{Area of } \triangle ABC + \text{Area of } \triangle ACD$
$= 10.5 + 17.5 = 28$ square units.

(C) Let the vertices of the triangle be $A (4, -6), B (3, -2),$ and $C (5, 2)$.
Let $D$ be the mid-point of side $BC$ of $\triangle ABC$. Therefore,$AD$ is the median of $\triangle ABC$.
Coordinates of point $D = \left(\frac{3+5}{2}, \frac{-2+2}{2}\right) = (4, 0)$.
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area of $\triangle ABD = \frac{1}{2} |4(-2 - 0) + 3(0 - (-6)) + 4(-6 - (-2))|$
$= \frac{1}{2} |4(-2) + 3(6) + 4(-4)|$
$= \frac{1}{2} |-8 + 18 - 16| = \frac{1}{2} |-6| = 3$ square units.
Area of $\triangle ADC = \frac{1}{2} |4(0 - 2) + 4(2 - (-6)) + 5(-6 - 0)|$
$= \frac{1}{2} |4(-2) + 4(8) + 5(-6)|$
$= \frac{1}{2} |-8 + 32 - 30| = \frac{1}{2} |-6| = 3$ square units.
Since the area of $\triangle ABD = \text{Area of } \triangle ADC = 3$ square units,the median $AD$ divides $\triangle ABC$ into two triangles of equal areas.

(A) Let the line $2x + y - 4 = 0$ divide the line segment joining $A(2, -2)$ and $B(3, 7)$ in the ratio $k:1$.
Using the section formula,the coordinates of the point of division are $\left(\frac{3k + 2}{k + 1}, \frac{7k - 2}{k + 1}\right)$.
Since this point lies on the line $2x + y - 4 = 0$,we substitute these coordinates into the equation:
$2\left(\frac{3k + 2}{k + 1}\right) + \left(\frac{7k - 2}{k + 1}\right) - 4 = 0$
Multiplying by $(k + 1)$,we get:
$2(3k + 2) + (7k - 2) - 4(k + 1) = 0$
$6k + 4 + 7k - 2 - 4k - 4 = 0$
$9k - 2 = 0$
$9k = 2$
$k = \frac{2}{9}$
Thus,the required ratio is $2:9$.

(N/A) If the given points are collinear,then the area of the triangle formed by these points must be $0$.
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the given points $(x, y), (1, 2)$ and $(7, 0)$ into the formula:
$0 = \frac{1}{2} |x(2 - 0) + 1(0 - y) + 7(y - 2)|$
$0 = \frac{1}{2} |2x - y + 7y - 14|$
$0 = \frac{1}{2} |2x + 6y - 14|$
Multiplying by $2$ on both sides:
$2x + 6y - 14 = 0$
Dividing the entire equation by $2$:
$x + 3y - 7 = 0$
Thus,the required relation between $x$ and $y$ is $x + 3y - 7 = 0$.

(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 $(1, 4)$.