Cartesian System Questions in English

(N/A) Consider that the lamp is placed on the table. Choose two adjacent edges,$DC$ and $AD$.
Draw perpendiculars from the position of the lamp to the edges $DC$ and $AD$ and measure the lengths of these perpendiculars.
Let the length of these perpendiculars be $30\, cm$ and $20\, cm$ respectively.
Now,the position of the lamp from the left edge $(AD)$ is $20\, cm$ and from the lower edge $(DC)$ is $30\, cm$.
This can be written as $(20, 30)$,where $20$ represents the perpendicular distance of the lamp from edge $AD$ and $30$ represents the perpendicular distance of the lamp from edge $DC$.

(A) Based on the convention provided,the first number represents the street running in the North-South direction,and the second number represents the street running in the East-West direction.
$(i)$ The cross-street $(4, 3)$ is formed by the intersection of the $4^{th}$ North-South street and the $3^{rd}$ East-West street. Since there is only one such intersection point,there is only $1$ cross-street that can be referred to as $(4, 3).$
$(ii)$ The cross-street $(3, 4)$ is formed by the intersection of the $3^{rd}$ North-South street and the $4^{th}$ East-West street. Similarly,there is only $1$ cross-street that can be referred to as $(3, 4).$

(N/A) $(i)$ Since the distance of the point $B$ from the $y$-axis is $4$ units,the $x$-coordinate or abscissa of the point $B$ is $4$. The distance of the point $B$ from the $x$-axis is $3$ units; therefore,the $y$-coordinate,i.e.,the ordinate,of the point $B$ is $3$. Hence,the coordinates of the point $B$ are $(4, 3)$.
As in $(i)$ above:
$(ii)$ The $x$-coordinate and the $y$-coordinate of the point $M$ are $-3$ and $4$,respectively. Hence,the coordinates of the point $M$ are $(-3, 4)$.
$(iii)$ The $x$-coordinate and the $y$-coordinate of the point $L$ are $-5$ and $-4$,respectively. Hence,the coordinates of the point $L$ are $(-5, -4)$.
$(iv)$ The $x$-coordinate and the $y$-coordinate of the point $S$ are $3$ and $-4$,respectively. Hence,the coordinates of the point $S$ are $(3, -4)$.

(N/A) You can see that:
$(i)$ The point $A$ is at a distance of $+4$ units from the $y$-axis and at a distance of $0$ from the $x$-axis. Therefore,the $x$-coordinate of $A$ is $4$ and the $y$-coordinate is $0$. Hence,the coordinates of $A$ are $(4, 0)$.
(ii) The point $B$ is on the $y$-axis at a distance of $3$ units from the origin. Therefore,the $x$-coordinate is $0$ and the $y$-coordinate is $3$. Hence,the coordinates of $B$ are $(0, 3)$.
(iii) The point $C$ is on the $x$-axis at a distance of $5$ units to the left of the origin. Therefore,the $x$-coordinate is $-5$ and the $y$-coordinate is $0$. Hence,the coordinates of $C$ are $(-5, 0)$.
(iv) The point $D$ is on the $y$-axis at a distance of $4$ units below the origin. Therefore,the $x$-coordinate is $0$ and the $y$-coordinate is $-4$. Hence,the coordinates of $D$ are $(0, -4)$.
$(v)$ The point $E$ is on the $x$-axis at a distance of $\frac{2}{3}$ units from the origin. Therefore,the $x$-coordinate is $\frac{2}{3}$ and the $y$-coordinate is $0$. Hence,the coordinates of $E$ are $(\frac{2}{3}, 0)$.

(N/A) $(i)$ The name of the horizontal line is the $x$-axis and the name of the vertical line is the $y$-axis.
$(ii)$ Each part of the plane formed by these two lines is called a quadrant.
$(iii)$ The point where these two lines intersect is called the origin.

(N/A) $(i)$ The $x$-coordinate and the $y$-coordinate of point $B$ are $-5$ and $2$ respectively. Therefore,the coordinates of point $B$ are $(-5, 2)$.
$(ii)$ The $x$-coordinate and the $y$-coordinate of point $C$ are $5$ and $-5$ respectively. Therefore,the coordinates of point $C$ are $(5, -5)$.
$(iii)$ The point whose $x$-coordinate and $y$-coordinate are $-3$ and $-5$ respectively is point $E$.
$(iv)$ The point whose $x$-coordinate and $y$-coordinate are $2$ and $-4$ respectively is point $G$.
$(v)$ The $x$-coordinate of point $D$ is $6$. Therefore,the abscissa of point $D$ is $6$.
$(vi)$ The $y$-coordinate of point $H$ is $-3$. Therefore,the ordinate of point $H$ is $-3$.
$(vii)$ The $x$-coordinate and the $y$-coordinate of point $L$ are $0$ and $5$ respectively. Therefore,the coordinates of point $L$ are $(0, 5)$.
$(viii)$ The $x$-coordinate and the $y$-coordinate of point $M$ are $-3$ and $0$ respectively. Therefore,the coordinates of point $M$ are $(-3, 0)$.

(N/A) For the point $(-2, 4)$,the abscissa ($x$-coordinate) is negative and the ordinate ($y$-coordinate) is positive. Since $(-, +)$ belongs to the $2nd$ quadrant,$(-2, 4)$ lies in the $2nd$ quadrant.
For the point $(3, -1)$,the abscissa is positive and the ordinate is negative. Since $(+, -)$ belongs to the $4th$ quadrant,$(3, -1)$ lies in the $4th$ quadrant.
For the point $(-1, 0)$,the ordinate is $0$. Any point with an ordinate of $0$ lies on the $x$-axis. Since the abscissa is negative,$(-1, 0)$ lies on the negative $x$-axis.
For the point $(1, 2)$,both the abscissa and the ordinate are positive. Since $(+, +)$ belongs to the $1st$ quadrant,$(1, 2)$ lies in the $1st$ quadrant.
For the point $(-3, -5)$,both the abscissa and the ordinate are negative. Since $(-, -)$ belongs to the $3rd$ quadrant,$(-3, -5)$ lies in the $3rd$ quadrant.
These points are plotted in the Cartesian plane as shown in the figure: $A(-2, 4)$,$B(3, -1)$,$C(-1, 0)$,$D(1, 2)$,and $E(-3, -5)$.

(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 $x < 0$.

(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 $Q$) $= -2 - (-3)$.
$= -2 + 3 = 1$.

(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.

(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) $(i)$ 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.

(C) In a Cartesian coordinate system,the position of any point is represented by an ordered pair $(x, y)$.
For any point lying on the $y$-axis,the horizontal distance from the origin (the $x$-coordinate) is always $0$.
Therefore,any point on the $y$-axis is of the form $(0, y)$,where $y$ represents the vertical distance from the origin.

(D) The points of the form $(a, a)$ have the same $x$ and $y$ coordinates.
Since the $x$-coordinate is equal to the $y$-coordinate,the point satisfies the equation $y=x$.
Therefore,the point of the form $(a, a)$ always lies on the line $y=x$.

(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 $(5,0)$.

(B) The Cartesian plane is divided into four quadrants based on the signs of the coordinates $(x, y)$:
$1$. First Quadrant: Both $x$ and $y$ are positive $(+, +)$.
$2$. Second Quadrant: $x$ is negative and $y$ is positive $(-, +)$.
$3$. Third Quadrant: Both $x$ and $y$ are negative $(-, -)$.
$4$. Fourth Quadrant: $x$ is positive and $y$ is negative $(+, -)$.
For the point $(-3, -5)$,the $x$-coordinate is $-3$ (negative) and the $y$-coordinate is $-5$ (negative). Since both coordinates are negative,the point lies in the third quadrant. Therefore,the statement is False.

(A) The statement is True.
In a Cartesian coordinate system,the $x$-axis and $y$-axis are two perpendicular lines that intersect at a point called the origin.
The coordinates of the origin are defined as $(0,0)$.

(A) point is on the $y$-axis if its $x$-coordinate is $0$.
In the given point $(0, 7)$,the $x$-coordinate is $0$.
Therefore,the point $(0, 7)$ lies on the $y$-axis.
Thus,the statement is True.

(TRUE) The statement is True.
The Cartesian plane is formed by two perpendicular lines,the $x$-axis and the $y$-axis,which intersect at the origin $(0, 0)$.
These two axes divide the plane into four regions,which are known as quadrants.

(B) If two points $(x_1, y_1)$ and $(x_2, y_2)$ are the same,then their corresponding coordinates must be equal,i.e.,$x_1 = x_2$ and $y_1 = y_2$.
Given the points are $(x+2, 7)$ and $(2x-1, 7)$.
Since the $y$-coordinates are already equal $(7 = 7)$,we equate the $x$-coordinates:
$x + 2 = 2x - 1$
Subtract $x$ from both sides:
$2 = x - 1$
Add $1$ to both sides:
$x = 3$
Therefore,the value of $x$ is $3$.

(D) Given $a=5, b=7, c=6,$ and $d=10.$
Substitute these values into the point $(a-b, c-d)$:
$x = a - b = 5 - 7 = -2$
$y = c - d = 6 - 10 = -4$
The point is $(-2, -4).$
Since both the $x$-coordinate and $y$-coordinate are negative,the point lies in the third quadrant.

(A) Given $a=2, b=-3, c=-2, d=-5.$
First,calculate the coordinates of the point $(ab, cd)$:
$x = a \times b = 2 \times (-3) = -6$
$y = c \times d = (-2) \times (-5) = 10$
The point is $(-6, 10).$
In the Cartesian plane,a point with a negative $x$-coordinate and a positive $y$-coordinate lies in the second quadrant.
Therefore,the point $(-6, 10)$ is in the second quadrant.

(B) The coordinates of the two points are $(x_1, y_1) = (5, -3)$ and $(x_2, y_2) = (-5, -3)$.
Since the $y$-coordinates of both points are equal $(y_1 = y_2 = -3)$,the line segment joining these two points is a horizontal line.
$A$ horizontal line is always parallel to the $x$-axis.
Therefore,the line joining $(5, -3)$ and $(-5, -3)$ is parallel to the $x$-axis.

(A) Since the points $(2x - 3, y + 3)$ and $(x + 7, 2y - 2)$ are the same,their corresponding coordinates must be equal.
Equating the $x$-coordinates: $2x - 3 = x + 7$
$2x - x = 7 + 3$
$x = 10$
Equating the $y$-coordinates: $y + 3 = 2y - 2$
$3 + 2 = 2y - y$
$y = 5$
Thus,the values of $x$ and $y$ are $10$ and $5$ respectively.

(A) Given values are $a=5, b=7, c=3, d=-5.$
Substituting these values into the coordinates:
$x = \frac{a}{b} = \frac{5}{7}$
$y = \frac{c}{d} = \frac{3}{-5} = -0.6$
Since the $x$-coordinate is positive $(>0)$ and the $y$-coordinate is negative $( < 0)$,the point $(\frac{5}{7}, -0.6)$ lies in the fourth quadrant.

(B) For two points $(a, b)$ and $(b, a)$ to be the same point in the coordinate plane,their corresponding coordinates must be equal.
This implies $a = b$ and $b = a$.
Both conditions lead to the same requirement: $a = b$.
Looking at the given options,we check which pair satisfies $a = b$:
Option $A$: $2 \neq -2$
Option $B$: $-2 = -2$ (This satisfies the condition)
Option $C$: $-2 \neq 2$
Option $D$: $2 \neq \frac{1}{2}$
Therefore,the only possible case is $a = -2$ and $b = -2$.

(C) 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 $(5, -3)$ and $(5, 8)$.
Here,$x_1 = 5, y_1 = -3$ and $x_2 = 5, y_2 = 8$.
Substituting these values into the formula:
$d = \sqrt{(5 - 5)^2 + (8 - (-3))^2}$
$d = \sqrt{(0)^2 + (8 + 3)^2}$
$d = \sqrt{0 + (11)^2}$
$d = \sqrt{121} = 11$.
Since the $x$-coordinates are the same,the distance is simply the absolute difference between the $y$-coordinates: $|8 - (-3)| = |8 + 3| = 11$.

(A) Since both coordinates represent the same point,we equate the $x$-coordinates and $y$-coordinates separately.
$1$. Equating $x$-coordinates: $2a + 5 = a + 11$
$2a - a = 11 - 5$
$a = 6$
$2$. Equating $y$-coordinates: $3b + 2 = b + 14$
$3b - b = 14 - 2$
$2b = 12$
$b = 6$
Therefore,the values of $a$ and $b$ are $6$ and $6$ respectively.

(C) In the Cartesian coordinate system,the plane is divided into four quadrants by the $x$-axis and $y$-axis.
- In the $I$ quadrant,both coordinates are positive $(+, +)$.
- In the $II$ quadrant,the $x$-coordinate is negative and the $y$-coordinate is positive $(-, +)$.
- In the $III$ quadrant,both coordinates are negative $(-, -)$.
- In the $IV$ quadrant,the $x$-coordinate is positive and the $y$-coordinate is negative $(+, -)$.
Therefore,a point whose both coordinates are negative lies in the $III$ quadrant.

(A) The coordinates of the points are $X(3, 5)$ and $Y(-4, 5)$.
Since the $y$-coordinates of both points are equal $(y = 5)$,the line segment $XY$ is a horizontal line.
$A$ horizontal line is always parallel to the $x$-axis.
Therefore,the line joining $X$ and $Y$ is parallel to the $x$-axis.

(D) In the Cartesian plane,the quadrants are defined as follows:
$1$. First quadrant: $x > 0, y > 0$
$2$. Second quadrant: $x < 0, y > 0$
$3$. Third quadrant: $x < 0, y < 0$
$4$. Fourth quadrant: $x > 0, y < 0$
Since the given point has a positive $x$-coordinate and a negative $y$-coordinate,it lies in the fourth quadrant.

(A) The point where the $x$-axis and the $y$-axis intersect is known as the origin. Its coordinates are $(0, 0)$.