Coordinate Geometry - Class 9 Mathematics - Chapter 3 - Notes, NCERT Solutions & Extra Questions
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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.
(i) Point A is at $(4, 0)$
(ii) Point B is at $(0, 3)$
(iii) Point C is at $(-5, 0)$
(iv) Point D is at $(0, -4)$
(v) Point E is at $(\frac{2}{3}, 0)$
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Ask Chatterbot AIExtra 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}`
The correct option is (D): $$ \frac{x-a}{0} = \frac{y-b}{0} = \frac{z-c}{1} $$
To find the equation of the line passing through the point $(a, b, c)$ and parallel to the z-axis, we start by considering the general form of a line's equation in 3D: $$ \frac{x-a}{l} = \frac{y-b}{m} = \frac{z-c}{n} $$ where $(l, m, n)$ is the direction vector of the line.
Since the line is parallel to the $z$-axis, its direction vector is $(0, 0, 1)$. This implies that any point on this line has the form $(a, b, z_1)$, with $z_1$ varying along the line while $x$ and $y$ remain constant at $a$ and $b$ respectively.
Substituting $l=0$, $m=0$ and $n=1$ into the general equation, we get: $$ \frac{x-a}{0} = \frac{y-b}{0} = \frac{z-c}{1} $$
This line equation correctly restricts $x$ and $y$ to be constant (as when you divide by zero, the numerator must also be zero for the equation to hold), and allows $z$ to vary freely, reflecting parallelism to the $z$-axis.
The perpendicular from the origin to the line $y = mx + c$ meets it at the point $(-1,2)$. Find the values of $m$ and $c.
Solution
To find the values of $m$ and $c$, we start by determining the equation of the perpendicular dropped from the origin $(0,0)$ to the point $(-1,2)$ on the line.
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Equation of the Perpendicular Line: The slope of a line connecting points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $$ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} $$ Applying this to points $(0,0)$ and $(-1,2)$: $$ \text{slope} = \frac{2 - 0}{-1 - 0} = \frac{2}{-1} = -2 $$ Thus, the equation of the line is $y = mx + c$ becomes: $$ y = -2x $$ Which simplifies to: $$ y + 2x = 0 \quad \text{(i)} $$
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Equation of the Given Line (which is perpendicular to line (i)): If two lines are perpendicular, their slopes multiply to $-1$. Therefore, the slope of the line perpendicular to one with slope $-2$ is: $$ m_{\perp} = \frac{1}{2} $$ We then write the equation of the line passing through $(-1,2)$ using point-slope form: $$ y - 2 = \frac{1}{2}(x + 1) $$ Expanding and rearranging gives: $$ y - 2 = \frac{1}{2}x + \frac{1}{2} \Rightarrow 2y - 4 = x + 1 \Rightarrow 2y = x + 5 \Rightarrow y = \frac{x}{2} + \frac{5}{2} $$
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Comparison with Line Equation $y = mx + c$: Comparing $y = \frac{x}{2} + \frac{5}{2}$ with $y = mx + c$, we find that: $m = \frac{1}{2}$ and $c = \frac{5}{2}$.
Thus, the values of $m$ and $c$ for the line $y = mx + c$ given that it is perpendicular to the origin at point $(-1,2)$ are $m = \frac{1}{2}$ and $c = \frac{5}{2}$.
Which of the following represents the direction of point G with respect to the point J?
South East
North West
South West
South
North East
The correct option is A: South East.
To determine the direction of point G with respect to point J, you can refer to the provided map or image. It shows the relative positions of points G and J. From the image, it is clear that point G is located towards the bottom-right of point J, which corresponds to the South East direction.
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Ask Chatterbot AIExercise 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 $(3,4)$.
(i) Since a cross-street is determined uniquely by the intersection of one street running in the North-South direction and one in the East-West direction, a particular cross-street referred to by a pair of numbers, such as $(4,3)$, indicates a unique intersection. Therefore, there can only be 1 cross-street that can be referred to as $(4,3)$, which means the $4^{\text{th}}$ street in the North-South direction intersects with the $3^{\text{rd}}$ street in the East-West direction at one unique point.
(ii) Similarly, the notation $(3,4)$ signifies a unique intersection point between the $3^{\text{rd}}$ street in the North-South direction and the $4^{\text{th}}$ street in the East-West direction. So, there is 1 cross-street that can be referred to as $(3,4)$.
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Ask Chatterbot AIExercise 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 $\mathrm{M}$.
(i) The coordinates are $(-5, 2)$
(ii) The coordinates are $(5, -5)$
(iii) The point is $\mathrm{E}$
(iv) The point is $\mathrm{G}$
(v) The abscissa is $6$
(vi) The ordinate of $\mathrm{H}$ is $-3$
(vii) The coordinates of $\mathrm{L}$ are $(0, 5)$
(viii) The coordinates of $\mathrm{M}$ are $(-3, 0)$
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