Textbook - Coordinate Geometry 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) 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) 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 $E(-3, -5)$.

(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.