Plotting a Point in the Plane Questions in English

(N/A) To locate the points in the Cartesian plane,we follow these steps:
$1$. Draw the $x$-axis and $y$-axis on a graph paper,intersecting at the origin $O(0, 0)$.
$2$. Choose a suitable scale,for example,$1 \text{ cm} = 1 \text{ unit}$.
$3$. For each point $(x, y)$,start from the origin $O$. Move $x$ units along the $x$-axis (right if $x > 0$,left if $x < 0$) and then move $y$ units parallel to the $y$-axis (up if $y > 0$,down if $y < 0$).
$4$. The positions of the given points are plotted as follows:
- $(5, 0)$: On the $x$-axis,$5$ units to the right of the origin.
- $(0, 5)$: On the $y$-axis,$5$ units above the origin.
- $(2, 5)$: $2$ units right and $5$ units up.
- $(5, 2)$: $5$ units right and $2$ units up.
- $(-3, 5)$: $3$ units left and $5$ units up.
- $(-3, -5)$: $3$ units left and $5$ units down.
- $(5, -3)$: $5$ units right and $3$ units down.
- $(6, 1)$: $6$ units right and $1$ unit up.

(N/A) The pairs of numbers given in the table can be represented by the points $(-3, 7)$,$(0, -3.5)$,$(-1, -3)$,$(4, 4)$,and $(2, -3)$.
To plot these points:
$1$. For $(-3, 7)$,start at the origin,move $3$ units to the left along the $X$-axis,and then $7$ units upwards parallel to the $Y$-axis.
$2$. For $(0, -3.5)$,start at the origin,move $0$ units along the $X$-axis,and then $3.5$ units downwards along the $Y$-axis.
$3$. For $(-1, -3)$,start at the origin,move $1$ unit to the left along the $X$-axis,and then $3$ units downwards parallel to the $Y$-axis.
$4$. For $(4, 4)$,start at the origin,move $4$ units to the right along the $X$-axis,and then $4$ units upwards parallel to the $Y$-axis.
$5$. For $(2, -3)$,start at the origin,move $2$ units to the right along the $X$-axis,and then $3$ units downwards parallel to the $Y$-axis.
The locations of these points are shown in the figure.

(N/A) The given points are $A(-2, 8), B(-1, 7), C(0, -1.25), D(1, 3)$ and $E(3, -1)$. To plot these points:
$(i)$ We draw $X'OX$ and $YOY'$ as axes.
$(ii)$ We choose suitable units of distance on the axes.
$(iii)$ To plot $A(-2, 8)$,we start from the origin $O$,move $2$ units to the left on the $x$-axis and $8$ units upwards on the $y$-axis.
$(iv)$ To plot $B(-1, 7)$,we start from the origin $O$,move $1$ unit to the left on the $x$-axis and $7$ units upwards on the $y$-axis.
$(v)$ To plot $C(0, -1.25)$,we start from the origin $O$,move $0$ units on the $x$-axis and $1.25$ units downwards on the $y$-axis.
$(vi)$ To plot $D(1, 3)$,we start from the origin $O$,move $1$ unit to the right on the $x$-axis and $3$ units upwards on the $y$-axis.
$(vii)$ To plot $E(3, -1)$,we start from the origin $O$,move $3$ units to the right on the $x$-axis and $1$ unit downwards on the $y$-axis.

(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) 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.

(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 $(0, -5)$.

(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) 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) 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) $(i)$ 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 $3 - 3 = 0$.

(C) In the Cartesian coordinate system,the $x$-axis is the horizontal line where the vertical distance from the origin is zero.
Therefore,for any point lying on the $x$-axis,the $y$-coordinate (ordinate) must be $0$.
Thus,any point on the $x$-axis is represented in the form $(x, 0)$,where $x$ is any real number.

(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) 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 $(3,5)$.

(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 $(4, 3)$.

(N/A) To find the coordinates of any point $(x, y)$ in the Cartesian plane,we determine its perpendicular distance from the $y$-axis (which gives the $x$-coordinate) and its perpendicular distance from the $x$-axis (which gives the $y$-coordinate).

Point	Coordinates $(x, y)$
$A$	$(3, 6)$
$B$	$(-4, 0)$
$C$	$(2, -5)$
$D$	$(0, 7)$
$E$	$(-3, 4)$
$F$	$(6, 0)$
$G$	$(-5, -4)$
$H$	$(0, -6)$
$O$	$(0, 0)$

(N/A) To find the coordinates of a point $(x, y)$,we determine its perpendicular distance from the $y$-axis (the $x$-coordinate) and its perpendicular distance from the $x$-axis (the $y$-coordinate).

Point	Coordinates
$A$	$(2, 5)$
$B$	$(-6, 0)$
$C$	$(-3, 4)$
$D$	$(-8, -8)$
$E$	$(3, -2)$
$F$	$(6, 0)$
$G$	$(0, -5)$
$H$	$(0, 7)$

(N/A) To determine the position of a point $(x, y)$ in the Cartesian plane:
- If $x > 0$ and $y > 0$,it lies in the $I$ quadrant.
- If $x < 0$ and $y > 0$,it lies in the $II$ quadrant.
- If $x < 0$ and $y < 0$,it lies in the $III$ quadrant.
- If $x > 0$ and $y < 0$,it lies in the $IV$ quadrant.
- If $x = 0$,it lies on the $y$-axis.
- If $y = 0$,it lies on the $x$-axis.

Points	Position of the point
$P(-3, -5)$	$III$ quadrant
$Q(0, 5)$	$y$-axis
$R(3, 4)$	$I$ quadrant
$S(4, 3)$	$I$ quadrant
$T(5, 0)$	$x$-axis
$U(-3, 5)$	$II$ quadrant
$V(5, -3)$	$IV$ quadrant
$W(0, -3)$	$y$-axis
$X(-3, 0)$	$x$-axis
$Y(3, -5)$	$IV$ quadrant
$Z(-5, 3)$	$II$ quadrant

(N/A) To determine the position of a point $(x, y)$ in the Cartesian plane:
- If $x > 0, y > 0$,it lies in the first quadrant.
- If $x < 0, y > 0$,it lies in the second quadrant.
- If $x < 0, y < 0$,it lies in the third quadrant.
- If $x > 0, y < 0$,it lies in the fourth quadrant.
- If $y = 0$,it lies on the $x$-axis.
- If $x = 0$,it lies on the $y$-axis.

Point	Position
$A(15, -10)$	Fourth quadrant
$B(-10, -20)$	Third quadrant
$C(25, 0)$	$x$-axis
$D(-15, 25)$	Second quadrant
$E(-5, 0)$	$x$-axis
$F(0, -15)$	$y$-axis
$G(25, 5)$	First quadrant
$H(0, 15)$	$y$-axis

(A) To determine the position of a point $(x, y)$ in the Cartesian plane:
$1$. If $x > 0$ and $y > 0$,the point lies in the first quadrant.
$2$. If $x < 0$ and $y > 0$,the point lies in the second quadrant.
$3$. If $x < 0$ and $y < 0$,the point lies in the third quadrant.
$4$. If $x > 0$ and $y < 0$,the point lies in the fourth quadrant.
$5$. If $y = 0$,the point lies on the $x$-axis.
$6$. If $x = 0$,the point lies on the $y$-axis.

Point	Position
$A(10, 40)$	First quadrant
$B(-30, 0)$	$x$-axis
$C(50, -40)$	Fourth quadrant
$D(0, -30)$	$y$-axis
$E(-20, 50)$	Second quadrant
$F(0, 60)$	$y$-axis
$G(70, 0)$	$x$-axis
$H(-40, -50)$	Third quadrant

(D) In the Cartesian plane,the coordinates of a point are represented as $(x, y)$.
For the first quadrant,both $x > 0$ and $y > 0$.
For the second quadrant,$x < 0$ and $y > 0$.
For the third quadrant,both $x < 0$ and $y < 0$.
For the fourth quadrant,$x > 0$ and $y < 0$.
Given the point $(2.8, 4.9)$,we observe that $x = 2.8$ (which is $> 0$) and $y = 4.9$ (which is $> 0$).
Since both coordinates are positive,the point $(2.8, 4.9)$ lies in the first quadrant.

(C) The distance of a point $(x, y)$ from the origin $(0, 0)$ is given by the formula $\sqrt{x^2 + y^2}$.
Here,the point is $(7, 0)$,so $x = 7$ and $y = 0$.
Substituting these values into the formula:
Distance $= \sqrt{7^2 + 0^2} = \sqrt{49 + 0} = \sqrt{49} = 7$.
Therefore,the distance of the point $(7, 0)$ from the origin is $7$.

(A) The $x$-axis is the horizontal line where the $y$-coordinate is always $0$.
Since the point lies on the right side of the $y$-axis,the $x$-coordinate is positive.
Given the distance from the origin is $5$ units,the $x$-coordinate is $5$.
Therefore,the coordinates of the point are $(5, 0)$.