Mix Examples - Coordinate Geometry Questions in English

(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) $1$. 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) $1$. 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.

(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 $(0, -7)$.

(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 $(0,-3).$

(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 $(x, y)$,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 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

(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.

(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) 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 $(3, 0)$.

(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.

(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.

(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.