# Coordinate Geometry - Class 9 - Mathematics

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## Examples - Coordinate Geometry | NCERT | Mathematics | Class 9

See Fig. 3.11 and complete the following statements:

(i) The abscissa and the ordinate of the point $\mathrm{B}$ are $\_~\_~\_$and $_~\_~\_$* *respectively. Hence, the coordinates of $\mathrm{B}$ are $(\_~\_, \_~\_)$

(ii) The $x$-coordinate and the $y$-coordinate of the point $\mathrm{M}$ are $\_~ \_~ \_$and $\_~ \_~ \_$, respectively. Hence, the coordinates of $\mathrm{M}$ are $(\_~\_, \_~\_)$

(iii) The $x$-coordinate and the $y$-coordinate of the point $\mathrm{L}$ are $\_~ \_~ \_$* *and $\_~ \_~ \_$ respectively. Hence, the coordinates of $\mathrm{L}$ are $(\_~\_, \_~\_)$.

(iv) The $x$-coordinate and the $y$-coordinate of the point $\mathrm{S}$ are $\_~ \_~ \_$ and $\_~ \_~ \_$ respectively. Hence, the coordinates of $S$ are $(\_~\_, \_~\_)$.

(i) The abscissa and the ordinate of the point $\mathrm{B}$ are $4$and $3$* *respectively. Hence, the coordinates of $\mathrm{B}$ are $(4, 3)$

(ii) The $x$-coordinate and the $y$-coordinate of the point $\mathrm{M}$ are $-3$and $4$, respectively. Hence, the coordinates of $\mathrm{M}$ are $(-3, 4)$

(iii) The $x$-coordinate and the $y$-coordinate of the point $\mathrm{L}$ are $-5$* *and $-4$ respectively. Hence, the coordinates of $\mathrm{L}$ are $(-5, -4)$.

(iv) The $x$-coordinate and the $y$-coordinate of the point $\mathrm{S}$ are $3$ and $-4$ respectively. Hence, the coordinates of $S$ are $(3, -4)$.

Write the coordinates of the points marked on the axes in Fig. 3.12.

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Sign up now## Extra Questions - Coordinate Geometry | NCERT | Mathematics | Class 9

The equation of the tangent drawn to the circle $x^{2}+y^{2}=4$ at the point whose polar coordinates are given by $\left(2, \frac{\pi}{3}\right)$ is

A $x+\sqrt{3} y=2$

B $\sqrt{3} x+y=2$

C $x+\sqrt{3} y=4$

D $\sqrt{3} x+y=4$

The correct answer is **Option C** $x+\sqrt{3} y=4$.

The point on the circle is given in polar coordinates as $(r, \theta) = \left(2, \frac{\pi}{3}\right)$. To convert this to Cartesian coordinates, we use:

- $x = r \cos(\theta)$,
- $y = r \sin(\theta)$.

Thus, $$ (x, y) = \left( 2 \cos \left( \frac{\pi}{3} \right), 2 \sin \left( \frac{\pi}{3} \right) \right) = (1, \sqrt{3}). $$

Given that the equation of the circle is $x^2 + y^2 = 4$, the tangent to the circle at $(x_1, y_1)$ is found using the formula: $$xx_1 + yy_1 = r^2$$ where $r^2 = 4$ for this circle. Plugging in $(x_1, y_1) = (1, \sqrt{3})$: $$ x(1) + y(\sqrt{3}) = 4 \quad \Rightarrow \quad x + \sqrt{3}y = 4 $$

Hence, the equation of the tangent line to the circle at the given point is **$x + \sqrt{3} y = 4$**.

The equation of the straight line passing through the point $(a, b, c)$ and parallel to the $z$-axis is [MP PET 1995; Pb. CET 2000]:

(A) $\frac{x-a}{1} = \frac{y-b}{1} = \frac{z-c}{0}$ (B) $\frac{x-a}{0} = \frac{y-b}{1} = \frac{z-c}{1}$ (C) $\frac{x-a}{1} = \frac{y-b}{0} = \frac{z-c}{0}$ (D) $\frac{x-a}{0} = \frac{y-b}{0} = \frac{z-c}{1}`

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Sign up now## Exercise 3.1 - Coordinate Geometry | NCERT | Mathematics | Class 9

How will you describe the position of a table lamp on your study table to another person?

You can describe the position of the table lamp on your study table using Cartesian Coordinate system. Make one table edge as X-axis, and another perpendicular edge as the Y-axis.

Let's say, starting from the corner where the two edges meet and moving rightward covers the X-axis, whereas moving forward covers the Y-axis.

Now, measure how far along the X-axis (horizontally from left to right of the table), and how far up the Y-axis (vertically from front to back of the table) the table lamp is. Assuming table lamp is 30 cm from the left along the X-axis and 20 cm from the front along the Y-axis.

So, the lamp's position would be described as (30,20), which signifies 30 cm along the X-axis and 20 cm along the Y-axis.

**(Street Plan)** : A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction.

All the other streets of the city run parallel to these roads and are $200 \mathrm{~m}$ apart. There are 5 streets in each direction. Using $1 \mathrm{~cm}=200 \mathrm{~m}$, draw a model of the city on your notebook. Represent the roads/streets by single lines.

There are many cross- streets in your model. A particular cross-street is made by two streets, one running in the North - South direction and another in the East - West direction. Each cross street is referred to in the following manner : If the $2^{\text {nd }}$ street running in the North - South direction and $5^{\text {th }}$ in the East - West direction meet at some crossing, then we will call this cross-street $(2,5)$. Using this convention, find:

(i) how many cross - streets can be referred to as $(4,3)$.

(ii) how many cross - streets can be referred to as $(3,4)$.

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## Exercise 3.2 - Coordinate Geometry | NCERT | Mathematics | Class 9

Write the answer of each of the following questions:

(i) What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane?

(ii) What is the name of each part of the plane formed by these two lines?

(iii) Write the name of the point where these two lines intersect.

(i) The name of the horizontal line in the Cartesian plane is the **x-axis**, and the name of the vertical line is the **y-axis**.

(ii) Each part of the plane formed by the intersection of these two lines is called a **quadrant**. There are four quadrants in the Cartesian plane.

(iii) The point where these two lines intersect is known as the **origin**.

See Fig.3.14, and write the following:

(i) The coordinates of $\mathrm{B}$.

(ii) The coordinates of $\mathrm{C}$.

(iii) The point identified by the coordinates $(-3,-5)$.

(iv) The point identified by the coordinates $(2,-4)$.

(v) The abscissa of the point $\mathrm{D}$.

(vi) The ordinate of the point $\mathrm{H}$.

(vii) The coordinates of the point $\mathrm{L}$.

(viii) The coordinates of the point $\mathrm{M}$.