Mix Examples - Coordinate Geometry Questions in English

(N/A) $(i)$ The abscissa of a point is its $x$-coordinate. Points with abscissa $0$ lie on the $y$-axis. From the figure,the points on the $y$-axis are $A(0, 3)$,$L(0, -4)$,and the origin $O(0, 0)$.
$(ii)$ The ordinate of a point is its $y$-coordinate. Points with ordinate $0$ lie on the $x$-axis. From the figure,the points on the $x$-axis are $I(-2, 0)$,$G(5, 0)$,and the origin $O(0, 0)$.
$(iii)$ The abscissa of a point is its $x$-coordinate. We need to find points where the $x$-coordinate is $-5$. From the figure,these points are $D(-5, 1)$ and $H(-5, -3)$.

(N/A) The equation of the line passing through $A(1, -1)$ and $B(4, 5)$ is given by the slope-intercept form.
Slope $m = \frac{5 - (-1)}{4 - 1} = \frac{6}{3} = 2$.
Using point-slope form: $y - (-1) = 2(x - 1) \implies y + 1 = 2x - 2 \implies y = 2x - 3$.
$(i)$ To find a point between $A$ and $B$,we can choose an $x$-coordinate between $1$ and $4$. Let $x = 2$. Then $y = 2(2) - 3 = 1$. Thus,$M(2, 1)$ is a point on the line segment $AB$.
$(ii)$ To find a point outside the line segment $AB$,we can choose an $x$-coordinate greater than $4$ or less than $1$. Let $x = 5$. Then $y = 2(5) - 3 = 7$. Thus,$N(5, 7)$ is a point on the line which lies outside the line segment $AB$.

(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 any point $(x, y)$,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). Given the scale $1 \,cm = 5$ units,we count the grid units accordingly.

Point	Coordinates $(x, y)$
$P$	$(10, 15)$
$Q$	$(-15, 0)$
$R$	$(20, -15)$
$S$	$(0, 25)$
$T$	$(-20, 20)$
$U$	$(-20, -25)$
$V$	$(0, -15)$
$W$	$(25, 0)$

(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 find the coordinates of any point $(x, y)$,we determine its perpendicular distance from the $Y$-axis (which gives the $x$-coordinate) and from the $X$-axis (which gives the $y$-coordinate). Given the scale $1 \, cm = 10$ units,we count the grid units accordingly.

Point	Coordinates
$P$	$(-50, 40)$
$Q$	$(80, -60)$
$R$	$(0, 40)$
$S$	$(-70, 0)$
$T$	$(-70, -70)$
$U$	$(0, -40)$
$V$	$(80, 0)$
$W$	$(30, 30)$

(N/A)

Coordinates of the point	Position of the point
$A(5, 0)$	On the $x$-axis
$B(3, -2)$	In the $4^{th}$ quadrant
$C(-2, 5)$	In the $2^{nd}$ quadrant
$D(0, 4)$	On the $y$-axis
$E(-3, -4)$	In the $3^{rd}$ quadrant
$F(4, 3)$	In the $1^{st}$ quadrant
$G(0, -4)$	On the $y$-axis
$H(-5, 0)$	On the $x$-axis

(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

(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 $(0,0)$.

(B) The coordinates of any point on the $x$-axis are of the form $(x, 0)$,where $x$ is any real number.
Since the given point is $(5, -5)$,its $y$-coordinate is $-5$,which is not equal to $0$.
Therefore,the point $(5, -5)$ does not lie on the $x$-axis.
Thus,the statement is False.

(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 $3$.

(C) The line segment joins points $P(5,3)$ and $Q(5,-8)$.
Since the $x$-coordinates of both points are equal to $5$,the line is a vertical line represented by the equation $x = 5$.
$A$ vertical line $x = 5$ is parallel to the $y$-axis and intersects the $x$-axis at the point where $y = 0$.
Therefore,the line intersects the $x$-axis at $(5,0)$.

(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.

(D) The points $A(3, -8)$ and $B(3, 5)$ have the same $x$-coordinate,which is $3$.
This means the line joining these two points is a vertical line represented by the equation $x = 3$.
$A$ vertical line $x = 3$ is parallel to the $y$-axis and intersects the $x$-axis at a point where the $y$-coordinate is $0$.
Therefore,the intersection point is $(3, 0)$.

(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: $|8 - (-3)| = |8 + 3| = 11$.

(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.

(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.

(B) Given $a=5$ and $b=7$.
Substituting these values into the coordinates $(a-b, b-a)$:
$x$-coordinate $= a - b = 5 - 7 = -2$.
$y$-coordinate $= b - a = 7 - 5 = 2$.
The point is $(-2, 2)$.
Since the $x$-coordinate is negative and the $y$-coordinate is positive,the point $(-2, 2)$ lies in the second 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 $7$.

(D) The distance of a point $(x, y)$ from the origin $(0, 0)$ is given by the formula $d = \sqrt{x^2 + y^2}$.
Substituting the coordinates $(0, -6)$ into the formula:
$d = \sqrt{0^2 + (-6)^2}$
$d = \sqrt{0 + 36}$
$d = \sqrt{36}$
$d = 6$
Thus,the distance is $6$ units.

(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 point $P$ are $(3, 2)$ and point $Q$ are $(3, -5)$.
Since the $x$-coordinate for both points is $3$,the line joining these points is a vertical line represented by the equation $x = 3$.
$A$ vertical line parallel to the $y$-axis intersects the $x$-axis at the point $(3, 0)$.
Therefore,the line intersects the $x$-axis.

(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 $(0, 0)$.

(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 $(5, 0)$.

(C) In a Cartesian coordinate system,the origin is the point where the $x$-axis and $y$-axis intersect. By definition,the coordinates of the origin are $(0, 0)$.

(B) In the Cartesian coordinate system,the horizontal line is known as the $x$-axis and the vertical line passing through the origin $(0, 0)$ is known as the $y$-axis. Therefore,the correct answer is the $y$-axis.

(C) In a Cartesian plane,the axes $XOX'$ and $YOY'$ divide the plane into four quadrants.
- The first quadrant is formed by the region between the positive $X$-axis and positive $Y$-axis.
- The second quadrant is formed by the region between the negative $X$-axis $(OX')$ and positive $Y$-axis $(OY)$.
- The third quadrant is formed by the region between the negative $X$-axis $(OX')$ and negative $Y$-axis $(OY')$.
- The fourth quadrant is formed by the region between the positive $X$-axis $(OX)$ and negative $Y$-axis $(OY')$.
Therefore,the interior of $\angle X'OY'$ corresponds to the third quadrant.

(B) In a Cartesian coordinate system,the quadrants are defined as follows:
$I$ quadrant: $(+, +)$
$II$ quadrant: $(-, +)$
$III$ quadrant: $(-, -)$
$IV$ quadrant: $(+, -)$
Given the point $\left(-\frac{7}{2}, \frac{5}{2}\right)$,the $x$-coordinate is negative and the $y$-coordinate is positive.
Therefore,the point lies in the $II$ quadrant.

(C) In a Cartesian plane,the $X$-axis and $Y$-axis divide the plane into four quadrants. The $X^{\prime}OX$ line represents the $X$-axis,where $OX$ is the positive $X$-axis and $OX^{\prime}$ is the negative $X$-axis. Similarly,$YOY^{\prime}$ represents the $Y$-axis,where $OY$ is the positive $Y$-axis and $OY^{\prime}$ is the negative $Y$-axis. The third quadrant is bounded by the negative $X$-axis $(OX^{\prime})$ and the negative $Y$-axis $(OY^{\prime})$. Therefore,the interior of $\angle X^{\prime}OY^{\prime}$ represents the third quadrant.

(A) In a Cartesian coordinate system,any point on the $y$-axis has a horizontal distance of $0$ from the origin. Therefore,the $x$-coordinate (also known as the abscissa) of any point on the $y$-axis is always $0$. The coordinates of a point on the $y$-axis are represented as $(0, y)$.

(B) In the Cartesian plane,a point $(x, y)$ has coordinates with opposite signs if one coordinate is positive and the other is negative.
Case $1$: $x > 0$ and $y < 0$. This corresponds to the Fourth quadrant.
Case $2$: $x < 0$ and $y > 0$. This corresponds to the Second quadrant.
Therefore,a point with coordinates of opposite signs lies in either the Second quadrant or the Fourth quadrant.

(A) The coordinates of the two points are $(8, 8)$ and $(-8, 8)$.
Since the $y$-coordinate is the same for both points $(y = 8)$,the line segment joining these points is a horizontal line.
$A$ horizontal line is always parallel to the $x$-axis.
Therefore,the line is parallel to the $x$-axis.

(D) In a Cartesian coordinate system,the $X$-axis and $Y$-axis divide the plane into four quadrants.
- The first quadrant is bounded by ray $OX$ and ray $OY$.
- The second quadrant is bounded by ray $OX'$ and ray $OY$.
- The third quadrant is bounded by ray $OX'$ and ray $OY'$.
- The fourth quadrant is bounded by ray $OX$ and ray $OY'$.
Therefore,the region bounded by ray $OY'$ and ray $OX$ is the fourth quadrant.

(D) In a Cartesian coordinate system,the coordinate axes consist of the $x$-axis and the $y$-axis.
These two axes intersect each other at the origin $(0, 0)$.
By definition,the $x$-axis and the $y$-axis are perpendicular to each other.
Therefore,the angle formed by the intersection of the coordinate axes is $90^{\circ}$.

(B) In a Cartesian coordinate system,the ray $OX$ represents the positive $x$-axis,where $x > 0$. The ray $OX^{\prime}$ represents the negative $x$-axis,where $x < 0$. Since the point $(a, 0)$ lies on the ray $OX^{\prime}$,the $x$-coordinate $a$ must be a negative number.

(A) point is said to lie on the $x$-axis if its $y$-coordinate is $0$.
Given the point $\left(7 \frac{3}{4}, 0\right)$,the $y$-coordinate is $0$.
Therefore,the point lies on the $x$-axis.

(B) In a 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 negative $x$-coordinate and a positive $y$-coordinate,it lies in the second quadrant.

(C) Given values are $a=5, b=3, c=-8, d=-5$.
We need to find the coordinates of the point $(a+c, b+d)$.
First,calculate the $x$-coordinate: $a+c = 5 + (-8) = -3$.
Next,calculate the $y$-coordinate: $b+d = 3 + (-5) = -2$.
The point is $(-3, -2)$.
Since both the $x$-coordinate and $y$-coordinate are negative,the point lies in the third quadrant.