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Published on: 21/09/2019
Coordinate Geometry
Download Tamil Nadu 9th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Maths Test1.
Find the distance between the two given points A (3, 4) and B (3, 8)
2.
Find the distance between the two given points A (8, 3) and B (- 4, 3)
3.
Plot the points A( -1, 0), B( 3, 0), C(3, 4), and O(-1, 4) on a graph sheet. Join them to form a rectangle. Draw the mirror image of the diagram in clockwise direction:
(i) about x-axis.
(ii) about y-axis.
What is your observation on the coordinates of the mirror image?
4.
Find the distance between the following pairs of points. (3,– 9) and (–2, 3)
5.
Find the distance between the following pairs of points. (3,4) and (– 7, 2)
6.
Find the distance between the following pairs of points. (1, 2) and (4, 3)
7.
Plot the following points in the coordinate plane. Join them in order. What type of geometrical shape is formed?
(–3, 3) (2, 3) (–6, –1) (5, –1)
8.
Plot the following points in the coordinate plane. Join them in order. What type of geometrical shape is formed?
(0, 0) (–4, 0) (–4, –4) (0, –4)
9.
Plot the following points in the coordinate plane and join them. What is your conclusion about the resulting figure?
(0, –4) (0, –2) (0, 4) (0, 5)
10.
Write down the abscissa and ordinate of the following.
(i) P
(ii) Q
(iii) R
(iv) S
11.
Plot the following points in the coordinate system and identify the quadrants
P(–7, 6), Q(7, –2), R(–6, –7), S(3, 5) and T(3, 9)
12.
In which quadrant does the following points lie? (–7, 3)
13.
In which quadrant does the following points lie? (2, 5)
14.
In which quadrant does the following points lie? (–1,–3)
15.
In which quadrant does the following points lie? (3,–8)
1.
AB =\(\sqrt{(3-3)^2+(8-4)^2}=\sqrt{0^2+4^2}=\sqrt{16}=4 \ units\)
2.
A (8, 3) and B (- 4, 3)
Distance =\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(-4-8)^2+(3-3)^2}\)
\(\sqrt{(-12)+0}=\sqrt{144}=12 \ units\)
3.
Take a graph sheet and plot all the points and join the points to form a rectangle.
Take a mirror and see the image of the rectangle formed.
4.
Distance between the two points (3,- 9) and (-2, 3)
= \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+({ y }_{ 2 }-{ y }_{ 1 })^{ 2 } } \)
= \(\sqrt { (-2-3)^{ 2 }+(3+9)^{ 2 } } =\sqrt { (-5)^{ 2 }+(12)^{ 2 } } \)
=\(\sqrt { 25+144 } =\sqrt { 169 } \) = 13 units
5.
Distance between the points (3,4) and (- 7, 2)
= \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+({ y }_{ 2 }-{ y }_{ 1 })^{ 2 } } \)
= \(\sqrt { (-7-3)^{ 2 }+(2-4)^{ 2 } } =\sqrt { (-10)^{ 2 }+(-2)^{ 2 } } \)
= \(\sqrt { 100+4 } =\sqrt { 104 } =\sqrt { { 2 }^{ 2 }\times 2\times 13 } \)
= \(2\sqrt { 26 } \) units
6.
Distance between the points (1, 2) and (4, 3)
= \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-{ y }_{ 1 })^{ 2 } } \)
= \(\sqrt { (4-1)^{ 2 }+(3-2)^{ 2 } } =\sqrt { { 3 }^{ 2 }+{ 1 }^{ 2 } } \)
= \(\sqrt { 9+1 } =\sqrt { 10 } \) = units
7.
The shape of the geometrical figure Trapezium.
8.
The geometrical shape of the figure is square
9.
The line is on the y - axis
10.
(i) P is (-4, 4) [-4 is abscissa and 4 is ordinate]
(ii) Q is (3, 3) [3 is abscissa and 3 is ordinate]
(iii) R is (4, -2) [4 is abscissa and -2 is ordinate]
(iv) S is (-5, -3) [-5 is abscissa and -3 is ordinate]
11.
(i) P (-7, 6) lies in the II quadrant because the x-coordinate is negative and y-coordinate is positive
(ii) Q (7, -2) lies in the IV quadrant because the x-coordinate is positive and y-coordinate is negative
(iii) R (-6, -7) lies in the III quadrant because the x-coordinate is negative and y-coordinate is negative
(iv) S (3, 5) lies in the I quadrant because the x-coordinate is positive and y-coordinate is also positive
(v) T (3, 9) lies in the I quadrant because the x-coordinate is positive and y-coordinate is also positive
12.
The x-coordinate is negative and y – coordinate is positive. So, Point(–7,3) lies in the II quadrant
13.
The x-coordinate is positive and y – coordinate is positive. So Point (2,5) lies in the I quadrant
14.
The x-coordinate is negative and y – coordinate is negative. So, Point (–1,–3) lies in the III quadrant
15.
The x- coordinate is positive and y – coordinate is negative. So, Point(3,–8) lies in the IV quadrant.
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Tamilnadu Stateboard 9th Standard Subjects
Tamilnadu Stateboard Standards
