Mix Examples - Coordinate Geometry Questions in English

(A) Let the coordinates of a point be $(x, y)$.
According to the problem,the abscissa is equal to the ordinate,so $x = y$.
If $x = y > 0$,the point $(x, x)$ lies in the $I$ quadrant.
If $x = y < 0$,the point $(x, x)$ lies in the $III$ quadrant.
Since the origin $(0, 0)$ is excluded,the points lie in the $I$ and $III$ quadrants.

(B) In the Cartesian plane,the sign convention for quadrants is as follows:
- First quadrant: $(+, +)$
- Second quadrant: $(-, +)$
- Third quadrant: $(-, -)$
- Fourth quadrant: $(+, -)$
For the point $(-3, 5)$,the abscissa ($x$-coordinate) is $-3$ (negative) and the ordinate ($y$-coordinate) is $5$ (positive).
Since the point has a negative $x$-coordinate and a positive $y$-coordinate,it lies in the second quadrant.

(C) In a Cartesian plane,the coordinate system is divided into four quadrants.
In the first quadrant,both $x$ and $y$ coordinates are positive $(+,+)$.
In the second quadrant,the $x$-coordinate (abscissa) is negative and the $y$-coordinate (ordinate) is positive $(-,+)$.
In the third quadrant,both coordinates are negative $(-,-)$.
In the fourth quadrant,the $x$-coordinate is positive and the $y$-coordinate is negative $(+,-)$.
Therefore,for a point in the second quadrant,the signs are $(-,+)$.

(D) In a Cartesian coordinate system,a point is of the form $(x, y)$.
If the $x$-coordinate is $0$,the point lies on the $y$-axis.
If the $y$-coordinate is $0$,the point lies on the $x$-axis.
For the point $(0, -7)$,the $x$-coordinate is $0$,therefore,it lies on the $y$-axis.

(A) In a Cartesian coordinate system,a point $(x, y)$ lies on the $x$-axis if its $y$-coordinate is $0$.
Since the given point is $(-10, 0)$,the $y$-coordinate is $0$,which means it lies on the $x$-axis.
Because the $x$-coordinate is $-10$ (which is less than $0$),the point lies on the negative direction of the $x$-axis.

(B) In a Cartesian coordinate system,any point on the $x$-axis is represented as $(x, 0)$,where $x$ is any real number.
The first coordinate of a point $(x, y)$ is called the abscissa.
Therefore,for any point on the $x$-axis,the abscissa can be any real number.

(C) In a Cartesian coordinate system,any point on the $x$-axis is represented as $(x, 0)$.
Here,the first coordinate $x$ is called the abscissa,and the second coordinate $0$ is called the ordinate.
Therefore,the ordinate of all points on the $x$-axis is $0$.

(D) In a Cartesian coordinate system,the two axes (the $x$-axis and the $y$-axis) intersect at a specific point. This point of intersection is known as the origin,and its coordinates are $(0, 0)$.

(A) In a Cartesian plane,the coordinate system is divided into four quadrants based on the signs of the $x$ and $y$ coordinates:
$1$. $I$ quadrant: Both $x$ and $y$ are positive $(+, +)$.
$2$. $II$ quadrant: $x$ is negative and $y$ is positive $(-, +)$.
$3$. $III$ quadrant: Both $x$ and $y$ are negative $(-, -)$.
$4$. $IV$ quadrant: $x$ is positive and $y$ is negative $(+, -)$.
Therefore,a point whose both coordinates are negative will lie in the $III$ quadrant.

(B) In the Cartesian plane,a point $(x, y)$ lies in the $IV$ quadrant if $x > 0$ and $y < 0$. $A$ point $(x, y)$ lies in the $III$ quadrant if $x < 0$ and $y < 0$.
For the given points:
$1$. $(1, -1)$: $x > 0, y < 0$,so it lies in the $IV$ quadrant.
$2$. $(2, -2)$: $x > 0, y < 0$,so it lies in the $IV$ quadrant.
$3$. $(4, -5)$: $x > 0, y < 0$,so it lies in the $IV$ quadrant.
$4$. $(-3, -4)$: $x < 0, y < 0$,so it lies in the $III$ quadrant.
Since the points lie in different quadrants ($IV$ and $III$),they do not lie in the same quadrant.

(C) In a Cartesian coordinate system,a point is represented as $(x, y)$.
If the $y$-coordinate (ordinate) of a point is $0$,the point is of the form $(x, 0)$.
Any point with a $y$-coordinate of $0$ lies on the $x$-axis,as it has no vertical displacement from the origin.

(D) In the Cartesian plane,the quadrants are defined as follows:
$I$ quadrant: $(+, +)$
$II$ quadrant: $(-, +)$
$III$ quadrant: $(-, -)$
$IV$ quadrant: $(+, -)$
For the point $(-5, 2)$,the $x$-coordinate is negative and the $y$-coordinate is positive,so it lies in the $II$ quadrant.
For the point $(2, -5)$,the $x$-coordinate is positive and the $y$-coordinate is negative,so it lies in the $IV$ quadrant.
Therefore,the points $(-5, 2)$ and $(2, -5)$ lie in the $II$ and $IV$ quadrants,respectively.

(A) The perpendicular distance of a point $P(x, y)$ from the $x$-axis is given by $|y|$.
Given that the perpendicular distance is $5$ units,we have $|y| = 5$,which implies $y = 5$ or $y = -5$.
The foot of the perpendicular from point $P$ to the $x$-axis is the point $(x, 0)$.
Since the foot of the perpendicular lies on the negative direction of the $x$-axis,the $x$-coordinate must be negative (i.e.,$x < 0$).
However,the question asks for the possible values of the $y$-coordinate of point $P$.
Since $y$ can be either $5$ or $-5$,the point $P$ can be $(x, 5)$ or $(x, -5)$ where $x < 0$.

(B) The coordinates of the vertices are $O(0,0), A(3,0), B(3,4)$,and $C(0,4)$.
Calculating the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$OA = \sqrt{(3-0)^2 + (0-0)^2} = \sqrt{3^2} = 3$ units.
$AB = \sqrt{(3-3)^2 + (4-0)^2} = \sqrt{4^2} = 4$ units.
$BC = \sqrt{(0-3)^2 + (4-4)^2} = \sqrt{(-3)^2} = 3$ units.
$CO = \sqrt{(0-0)^2 + (0-4)^2} = \sqrt{(-4)^2} = 4$ units.
Since opposite sides are equal ($OA = BC = 3$ units and $AB = CO = 4$ units) and the adjacent sides are perpendicular (as they lie along the axes),the figure $OABC$ is a rectangle.

(C) In a Cartesian coordinate system,the fourth quadrant is defined by points $(x, y)$ where the $x$-coordinate is positive $(x > 0)$ and the $y$-coordinate is negative $(y < 0)$.
Let us examine the given points:
$P(-1, 1)$: $x$ is negative,$y$ is positive (Quadrant $II$).
$Q(3, -4)$: $x$ is positive,$y$ is negative (Quadrant $IV$).
$R(1, -1)$: $x$ is positive,$y$ is negative (Quadrant $IV$).
$S(-2, -3)$: $x$ is negative,$y$ is negative (Quadrant $III$).
$T(-4, 4)$: $x$ is negative,$y$ is positive (Quadrant $II$).
Therefore,the points located in the fourth quadrant are $Q$ and $R$.

(A) The abscissa of a point $(x, y)$ is its $x$-coordinate.
For point $P(-2, 3)$,the abscissa is $-2$.
For point $Q(-3, 5)$,the abscissa is $-3$.
Therefore,(abscissa of $P$) $-$ (abscissa of $Q$) $= -2 - (-3)$.
$= -2 + 3 = 1$.

(A) We know that if a point lies on the $x$-axis,its ordinate (the $y$-coordinate) must be $0$.
Given points are $P(5,1)$,$Q(8,0)$,$R(0,4)$,$S(0,5)$,and $O(0,0)$.
Checking the $y$-coordinates of each point:
- For $P(5,1)$,$y = 1 \neq 0$.
- For $Q(8,0)$,$y = 0$.
- For $R(0,4)$,$y = 4 \neq 0$.
- For $S(0,5)$,$y = 5 \neq 0$.
- For $O(0,0)$,$y = 0$.
Thus,the points $Q(8,0)$ and $O(0,0)$ lie on the $x$-axis.

(B) In a Cartesian coordinate system,the abscissa is the $x$-coordinate of a point. The $x$-coordinate is positive to the right of the $y$-axis. This region corresponds to the $I$ quadrant (where both $x$ and $y$ are positive) and the $IV$ quadrant (where $x$ is positive and $y$ is negative). Therefore,the abscissa is positive in the $I$ and $IV$ quadrants.

(C) In the Cartesian plane,the signs of coordinates $(x, y)$ in different quadrants are as follows:
$I$ quadrant: $(+, +)$ (both signs are same)
$II$ quadrant: $(-, +)$ (signs are different)
$III$ quadrant: $(-, -)$ (both signs are same)
$IV$ quadrant: $(+, -)$ (signs are different)
Therefore,the points whose abscissa ($x$-coordinate) and ordinate ($y$-coordinate) have different signs lie in the $II$ and $IV$ quadrants.

(D) To find the coordinates of point $P$,we observe its position on the Cartesian plane.
$1$. The $x$-coordinate is the perpendicular distance of the point from the $y$-axis. Looking at the figure,point $P$ is at a distance of $2$ units to the left of the $y$-axis,so the $x$-coordinate is $-2$.
$2$. The $y$-coordinate is the perpendicular distance of the point from the $x$-axis. Looking at the figure,point $P$ is at a distance of $4$ units above the $x$-axis,so the $y$-coordinate is $4$.
$3$. Therefore,the coordinates of point $P$ are $(-2, 4)$.

(A) To identify the point with coordinates $(-5, 3)$,we look for the point that is $-5$ units along the $x$-axis and $3$ units along the $y$-axis.
Looking at the Cartesian plane:
$1$. Start at the origin $(0, 0)$.
$2$. Move $5$ units to the left along the negative $x$-axis to reach $x = -5$.
$3$. From there,move $3$ units upwards along the positive $y$-axis to reach $y = 3$.
The point located at these coordinates is $L$.

(B) In a Cartesian coordinate system,any point on the $y$-axis has an abscissa (x-coordinate) of $0$.
Given that the ordinate (y-coordinate) is $4$.
Therefore,the coordinates of the point are $(0,4)$.

(C) We know that any point lying on the $x$-axis must have its $y$-coordinate (ordinate) equal to $0$.
Let us examine the given points:
$P(0,3)$: The $y$-coordinate is $3 \neq 0$,so it does not lie on the $x$-axis.
$Q(1,0)$: The $y$-coordinate is $0$,so it lies on the $x$-axis.
$R(0,-1)$: The $y$-coordinate is $-1 \neq 0$,so it does not lie on the $x$-axis.
$S(-5,0)$: The $y$-coordinate is $0$,so it lies on the $x$-axis.
$T(1,2)$: The $y$-coordinate is $2 \neq 0$,so it does not lie on the $x$-axis.
Therefore,the points that do not lie on the $x$-axis are $P, R,$ and $T$.

(D) point on the $y$-axis always has its abscissa (x-coordinate) equal to $0$.
The negative direction of the $y$-axis represents values less than $0$.
Since the point is at a distance of $5$ units in the negative direction,its ordinate (y-coordinate) is $-5$.
Therefore,the coordinates of the point are $(0, -5)$.

(A) The perpendicular distance of any point $(x, y)$ from the $y$-axis is given by the absolute value of its $x$-coordinate (abscissa).
For the point $P(3, 4)$,the $x$-coordinate is $3$.
Therefore,the perpendicular distance from the $y$-axis is $|3| = 3$ units.

(A) $(i)$ True. $A$ point on the $y$-axis has an $x$-coordinate of $0$. Since the given point is $(0, -2)$,it lies on the $y$-axis.
$(ii)$ False. The perpendicular distance of a point $(x, y)$ from the $x$-axis is given by the absolute value of its $y$-coordinate (the ordinate). For the point $(4, 3)$,the ordinate is $3$. Therefore,the perpendicular distance from the $x$-axis is $3$,not $4$.

(N/A) $(i)$ The point $(3,0)$ has an abscissa (x-coordinate) of $3$ and an ordinate (y-coordinate) of $0$. If the ordinate of a point is $0$,the point lies on the x-axis,not in any quadrant.
Hence,the given statement is False.
$(ii)$ The point $(1,-1)$ has a positive x-coordinate and a negative y-coordinate,so it lies in the $IV$ quadrant. The point $(-1,1)$ has a negative x-coordinate and a positive y-coordinate,so it lies in the $II$ quadrant.
Since they lie in different quadrants,the given statement is False.

(A) $(i)$ False. We know that in the coordinates of a point,the abscissa ($x$-coordinate) is written first,followed by the ordinate ($y$-coordinate). Therefore,the coordinates of the point are $(1, -\frac{1}{2})$,not $(-\frac{1}{2}, 1)$.
$(ii)$ False. Any point that lies on the $y$-axis must have an abscissa of $0$ and is of the form $(0, y)$. Since the point is at a distance of $2$ units from the $x$-axis,its coordinates must be $(0, 2)$ or $(0, -2)$,not $(2, 0)$.

(TRUE) In the Cartesian plane,the $II$ quadrant is defined by points where the $x$-coordinate (abscissa) is negative and the $y$-coordinate (ordinate) is positive.
For the point $(-1, 7)$,the $x$-coordinate is $-1$ (which is $< 0$) and the $y$-coordinate is $7$ (which is $> 0$).
Since the signs are $(-, +)$,the point $(-1, 7)$ lies in the $II$ quadrant.
Therefore,the statement is True.

(N/A) $1$. The point $P(-6, 2)$ is located in the second quadrant where the $x$-coordinate is $-6$ and the $y$-coordinate is $2$.
$2$. $A$ perpendicular $PM$ is drawn from $P$ to the $x$-axis. Since $M$ lies on the $x$-axis,its $y$-coordinate is $0$. The $x$-coordinate remains the same as $P$,which is $-6$. Thus,the coordinates of $M$ are $(-6, 0)$.
$3$. $A$ perpendicular $PN$ is drawn from $P$ to the $y$-axis. Since $N$ lies on the $y$-axis,its $x$-coordinate is $0$. The $y$-coordinate remains the same as $P$,which is $2$. Thus,the coordinates of $N$ are $(0, 2)$.

(N/A) $(i)$ To find the coordinates of a point,we look at its position relative to the $X$-axis and $Y$-axis.
- For point $B$,the $x$-coordinate is $-5$ and the $y$-coordinate is $2$. So,$B = (-5, 2)$.
- For point $C$,the $x$-coordinate is $-2$ and the $y$-coordinate is $-3$. So,$C = (-2, -3)$.
- For point $E$,the $x$-coordinate is $3$ and the $y$-coordinate is $-1$. So,$E = (3, -1)$.
$(ii)$ The coordinates $(0, -2)$ mean $x = 0$ and $y = -2$. Looking at the graph,this point lies on the $Y$-axis at $-2$,which is point $F$.
$(iii)$ The abscissa is the $x$-coordinate of a point. For point $H$,the coordinates are $(1, 4)$. Therefore,the abscissa is $1$.
$(iv)$ The ordinate is the $y$-coordinate of a point. For point $D$,the coordinates are $(4, 0)$. Therefore,the ordinate is $0$.

(N/A) To find the coordinates of a point $(x, y)$,we determine its distance from the $Y$-axis (x-coordinate) and the $X$-axis (y-coordinate).
- For point $P$: The x-coordinate is $1$ and the y-coordinate is $1$. So,$P = (1, 1)$.
- For point $Q$: The x-coordinate is $-3$ and the y-coordinate is $0$ (since it lies on the $X$-axis). So,$Q = (-3, 0)$.
- For point $R$: The x-coordinate is $-2$ and the y-coordinate is $-3$. So,$R = (-2, -3)$.
- For point $S$: The x-coordinate is $2$ and the y-coordinate is $1$. So,$S = (2, 1)$.
- For point $T$: The x-coordinate is $4$ and the y-coordinate is $-2$. So,$T = (4, -2)$.
- For point $O$: This is the origin,where both axes intersect. So,$O = (0, 0)$.

(N/A) $1$. Draw the coordinate axes $X'OX$ and $Y'OY$ on a graph paper.
$2$. Plot the points $P(-3, 2)$,$Q(-7, -3)$,$R(6, -3)$,and $S(2, 2)$ by locating their respective $x$ and $y$ coordinates.
$3$. Join the points in the order $P$ to $Q$,$Q$ to $R$,$R$ to $S$,and $S$ to $P$.
$4$. Observe the figure formed. The side $PS$ is parallel to the side $QR$ because both lie on horizontal lines ($y=2$ and $y=-3$ respectively),but they are of different lengths ($PS = 5$ units,$QR = 13$ units).
$5$. Since one pair of opposite sides is parallel and the other is not,the figure obtained is a trapezium.

(N/A) To plot the points $(x, y)$ on a Cartesian plane,follow these steps:
$1$. Draw the coordinate axes $X'OX$ and $Y'OY$.
$2$. For each point $(x, y)$,start from the origin $(0, 0)$.
$3$. Move $x$ units along the $X$-axis (right if $x > 0$,left if $x < 0$).
$4$. From that position,move $y$ units parallel to the $Y$-axis (up if $y > 0$,down if $y < 0$).
The points are plotted as follows:
- $(2, 4)$: Move $2$ units right and $4$ units up.
- $(4, 2)$: Move $4$ units right and $2$ units up.
- $(-3, 0)$: Move $3$ units left and $0$ units up/down (lies on the $X$-axis).
- $(-2, 5)$: Move $2$ units left and $5$ units up.
- $(3, -3)$: Move $3$ units right and $3$ units down.
- $(0, 0)$: The origin.

(N/A) $1$. Plot the points $A(1,3)$,$B(-1,-1)$,and $C(-2,-3)$ on the Cartesian plane.
$2$. Observe that all three points $A$,$B$,and $C$ lie on the same straight line.
$3$. Since the points lie on the same straight line,they are collinear.

(N/A) $1$. Plot the points $A(1, 1)$,$B(2, -3)$,and $C(-1, -2)$ on a Cartesian plane.
$2$. Observe the positions of these points on the graph.
$3$. $A$ set of points is said to be collinear if they all lie on the same straight line.
$4$. By connecting these points,we can see that they do not form a single straight line.
$5$. Therefore,the given points are not collinear.

(N/A) $1$. Plot the points $A(0,0)$,$B(2,2)$,and $C(5,5)$ on a Cartesian plane.
$2$. Observe that all three points lie on the same straight line passing through the origin.
$3$. Since the points $A, B,$ and $C$ lie on a single straight line,they are collinear.

(N/A) $(i)$ In the point $(-3, 5)$,the abscissa ($x$-coordinate) is $-3$ (negative) and the ordinate ($y$-coordinate) is $5$ (positive). Points with a negative $x$ and a positive $y$ lie in the second quadrant.
$(ii)$ In the point $(-5, -3)$,the abscissa ($x$-coordinate) is $-5$ (negative) and the ordinate ($y$-coordinate) is $-3$ (negative). Points with both negative $x$ and $y$ coordinates lie in the third quadrant.

(A) $(i)$ For the point $(-5, 3)$,the abscissa ($x$-coordinate) is $-5$ (negative) and the ordinate ($y$-coordinate) is $3$ (positive). Points with a negative $x$ and a positive $y$ lie in the second quadrant.
$(ii)$ For the point $(3, 5)$,the abscissa ($x$-coordinate) is $3$ (positive) and the ordinate ($y$-coordinate) is $5$ (positive). Points with both positive $x$ and $y$ coordinates lie in the first quadrant.

(N/A) $(i)$ Clearly,the distance of point $P$ from the $y$-axis is $3$ units and its distance from the $x$-axis is $2$ units. Since $P$ lies in the first quadrant,its coordinates are $(3, 2)$.
Point $R$ lies on the $x$-axis,and its distance from the $y$-axis is $3$ units and from the $x$-axis is $0$ units. So,its coordinates are $(3, 0)$.
Clearly,point $Q$ lies in the fourth quadrant. The distance of $Q$ from the $y$-axis is $3$ units and from the $x$-axis is $1$ unit. So,the coordinates of $Q$ are $(3, -1)$.
$(ii)$ From the given figure,we observe that the points $L$ and $M$ lie on the same vertical line $x = 3$. Therefore,the abscissa (the $x$-coordinate) of both points $L$ and $M$ is $3$.
Hence,the difference between the abscissa of the points $L$ and $M$ is $3 - 3 = 0$.

(N/A) $(i) \, (-3, 5)$ lies in the $II$ quadrant because $x < 0$ and $y > 0$.
$(ii) \, (4, -1)$ lies in the $IV$ quadrant because $x > 0$ and $y < 0$.
$(iii) \, (2, 0)$ lies on the $x$-axis because the $y$-coordinate is $0$.
$(iv) \, (2, 2)$ lies in the $I$ quadrant because $x > 0$ and $y > 0$.
$(v) \, (-3, -6)$ lies in the $III$ quadrant because $x < 0$ and $y < 0$.

(C, D, E, G) We know that if a point lies on the $y$-axis,its abscissa ($x$-coordinate) must be $0$. Therefore,the coordinates of such a point are of the form $(0, y)$.
Checking the given points:
$A(1,1)$: $x=1 \neq 0$
$B(1,0)$: $x=1 \neq 0$
$C(0,1)$: $x=0$ (Lies on $y$-axis)
$D(0,0)$: $x=0$ (Lies on $y$-axis)
$E(0,-1)$: $x=0$ (Lies on $y$-axis)
$F(-1,0)$: $x=-1 \neq 0$
$G(0,5)$: $x=0$ (Lies on $y$-axis)
$H(-7,0)$: $x=-7 \neq 0$
$I(3,3)$: $x=3 \neq 0$
Thus,the points that lie on the $y$-axis are $C(0,1), D(0,0), E(0,-1),$ and $G(0,5)$.

(N/A) To plot the points, we first identify the coordinates as ordered pairs $(x, y)$:
$1$. $(1.25, -0.5)$
$2$. $(0.25, 1)$
$3$. $(1.5, 1.5)$
$4$. $(-1.75, -0.25)$
Steps to plot:
- Draw the Cartesian plane with $X'OX$ and $Y'OY$ as the coordinate axes.
- Set the scale as $1 \ cm = 0.25 \ \text{units}$ on both axes.
- Locate each point by moving $x$ units along the $X$-axis and $y$ units along the $Y$-axis.
- The plotted points are shown in the graph below.

(A) point on the $x$-axis has a $y$-coordinate of $0$. Since its distance from the $y$-axis is $7$ units,its $x$-coordinate is $7$. Thus,the coordinates are $(7, 0)$.
$A$ point on the $y$-axis has an $x$-coordinate of $0$. The distance from the $x$-axis is given as $-7$ units. In coordinate geometry,the distance from the $x$-axis represents the $y$-coordinate. Therefore,the $y$-coordinate is $-7$. Thus,the coordinates are $(0, -7)$.

(N/A) $(i)$ The point that lies on both the $x$ and $y$ axes is the origin,which has coordinates $(0,0)$.
$(ii)$ $A$ point on the $y$-axis has an abscissa of $0$. Given the ordinate is $-4$,the coordinates are $(0,-4)$.
$(iii)$ $A$ point on the $x$-axis has an ordinate of $0$. Given the abscissa is $5$,the coordinates are $(5,0)$.

(A) To plot the points on a graph paper,follow these steps:
$1$. Draw two perpendicular lines,$X'OX$ (x-axis) and $Y'OY$ (y-axis),intersecting at the origin $O(0, 0)$.
$2$. Choose a scale: $0.5 \ cm = 1 \ unit$ on both axes.
$3$. Plot each point based on its coordinates $(x, y)$:
- For $A(1, 3)$,move $1 \ unit$ to the right of the y-axis and $3 \ units$ above the x-axis.
- For $B(-3, -1)$,move $3 \ units$ to the left of the y-axis and $1 \ unit$ below the x-axis.
- For $C(1, -4)$,move $1 \ unit$ to the right of the y-axis and $4 \ units$ below the x-axis.
- For $D(-2, 3)$,move $2 \ units$ to the left of the y-axis and $3 \ units$ above the x-axis.
- For $E(0, -8)$,stay on the y-axis and move $8 \ units$ below the x-axis.
- For $F(1, 0)$,move $1 \ unit$ to the right of the y-axis and stay on the x-axis.

(N/A) Let the three vertices of the rectangle be $A(3,2)$,$B(-4,2)$,and $C(-4,5)$ (see the figure).
We need to find the coordinates of the fourth vertex $D$ such that $ABCD$ forms a rectangle.
Since the opposite sides of a rectangle are equal and parallel to the axes,the abscissa (x-coordinate) of $D$ must be equal to the abscissa of $A$,which is $3$.
The ordinate (y-coordinate) of $D$ must be equal to the ordinate of $C$,which is $5$.
Therefore,the coordinates of the fourth vertex $D$ are $(3,5)$.

(C) $1$. Plot the points $A(5,3)$,$B(-2,3)$,and $D(5,-4)$ on the Cartesian plane.
$2$. Join the points $AB$ and $AD$. Since $ABCD$ is a square,all its sides are equal in length and each interior angle is $90^{\circ}$.
$3$. The side $AB$ is horizontal (parallel to the $x$-axis) with length $|5 - (-2)| = 7$ units. The side $AD$ is vertical (parallel to the $y$-axis) with length $|3 - (-4)| = 7$ units.
$4$. To complete the square,the vertex $C$ must have the same $x$-coordinate as $B$ (which is $-2$) and the same $y$-coordinate as $D$ (which is $-4$).
$5$. Therefore,the coordinates of the vertex $C$ are $(-2,-4)$.

(N/A) Given that the length and breadth of the rectangle are $5$ and $3$ units respectively.
One vertex is at the origin $(0,0)$.
The longer side (length $= 5$) lies on the $x$-axis.
Since one vertex lies in the third quadrant,the rectangle must extend towards the negative $x$-axis and negative $y$-axis.
Therefore,the vertices are:
$1$. The origin: $(0,0)$
$2$. Moving $5$ units along the negative $x$-axis: $(-5,0)$
$3$. Moving $3$ units down from $(-5,0)$ to reach the third quadrant: $(-5,-3)$
$4$. Moving $3$ units down from the origin along the $y$-axis: $(0,-3)$
Thus,the coordinates of the vertices are $(0,0), (-5,0), (-5,-3),$ and $(0,-3).$

(D) $1$. Plot the points $P(1, 0)$,$Q(4, 0)$,and $S(1, 3)$ on the Cartesian plane.
$2$. In a square $PQRS$,the side length $PQ$ is the distance between $(1, 0)$ and $(4, 0)$,which is $|4 - 1| = 3$ units.
$3$. Similarly,the side length $PS$ is the distance between $(1, 0)$ and $(1, 3)$,which is $|3 - 0| = 3$ units.
$4$. Since $PQ = PS = 3$ units and the sides are perpendicular (along the axes),the point $R$ must complete the square.
$5$. The $x$-coordinate of $R$ must match the $x$-coordinate of $Q$,which is $4$.
$6$. The $y$-coordinate of $R$ must match the $y$-coordinate of $S$,which is $3$.
$7$. Therefore,the coordinates of point $R$ are $(4, 3)$.