9th Standard Syllabus & Materials
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NEW9th Standard
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NEW9th Standard
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TN 9th Tamil உயிருக்கு வேர்-இலக்கணம் - பகுபத உறுப்பிலக்கணம் Important Questions And Answers Study Material - QB365 Set A
NEW9th Standard
TN 9th Tamil அமுதென்று பேர் - இலக்கணம்-எழுத்து -அளபெடை Important Questions And Answers Study Material - QB365 Set A
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Published on: 30/07/2019
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 following pairs of points. (1, 2) and (4, 3)
2.
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)
3.
In which quadrant does the following points lie?(2, 5)
4.
In which quadrant does the following points lie? (3,–8)
5.
Show that the following points taken in order form the vertices of a parallelogram.
A(–3, 1), B(–6, –7), C (3, –9) and D(6, –1)
6.
Determine whether the given set of points in each case are collinear or not (a, –2), (a, 3), (a, 0)
7.
Let A(2, 3) and B(2, –4) be two points. If P lies on the x-axis, such that AP = \(\frac{3}{7}\) AB, find the coordinates of P.
8.
Calculate the distance between the points A (7, 3) and B which lies on the x-axis whose abscissa is 11.
9.
Find the distance between the points (–4, 3), (2, –3).
10.
A(–1, 1), B(1, 3) and C(3, a) respectively and if AB = BC, then find ‘a’.
11.
If the distance between the points (5,–2), (1, a), is 5 units, find the values of a.
12.
Plot the following points (0, –3), (0, 4), (0, –1), (0, 5) in the Cartesian plane. Where do they lie?
13.
The distance between the points (a, 0) and (0, b) is____________
a unit
b unit
\(\sqrt{a^2+{b^2}}\ unit\)
\(\sqrt{a^2-{b^2}}\ unit\)
14.
A point on the y-axis is ________________
(1, 1)
(6,0)
(0,6)
(-1, -1)
15.
The point whose abscissa is 5 and lies on the x-axis is__________
(-5, 0)
(5,5)
(0,5)
(5,0)
16.
The point whose ordinate is 4 and which lies on the y-axis is ______.
( 4, 0 )
(0, 4)
(1, 4)
(4, 2)
17.
On plotting the points O(0, 0), A(3, – 4), B(3, 4) and C(0, 4) and joining OA, AB, BC and CO, which of the following figure is obtained?
Square
Rectangle
Trapezium
Rhombus
18.
The point M lies in the IV quadrant. The coordinates of M is _______
(a,b)
(–a, b)
(a, –b)
(–a, –b)
19.
Point (–10, 0) lies ___________
on the negative direction of x-axis
on the negative direction of y-axis
in the III quadrant
in the IV quadrant
20.
Point (–3, 5) lie in the ________ quadrant
I
II
III
IV
21.
The axis intersect at a point called _____________
22.
The abscissa of every point on y-axis is______________
23.
The ordinate of every point on the x-axis is________________
24.
The abscissa of the origin is ____________
1.
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
2.
(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
3.
The x-coordinate is positive and y – coordinate is positive. So Point (2,5) lies in the I quadrant
4.
The x- coordinate is positive and y – coordinate is negative. So, Point(3,–8) lies in the IV quadrant.
5.
Distance =\(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
AB =\(\sqrt { (-6+3)^{ 2 }+(-7-1)^{ 2 } } \)
= \(\sqrt { (-3)^{ 2 }+(-8)^{ 2 } } =\sqrt { 9+64 } =\sqrt { 73 } \)
BC = \(\sqrt { (3+6)^{ 2 }+(-9+7)^{ 2 } } \)
= \(\sqrt { { 9 }^{ 2 }+(-2)^{ 2 } } =\sqrt { 81+4 } =\sqrt { 85 } \)
CD = \(\sqrt { (6-3)^{ 2 }+(-1+9)^{ 2 } } \)
= \(\sqrt { (3)^{ 2 }+({ 8 })^{ 2 } } =\sqrt { 9+64 } =\sqrt { 73 } \)
AD = \(\sqrt { (6+3)^{ 2 }+(-1-1)^{ 2 } } \)
= \(\sqrt { (9)^{ 2 }+(-2)^{ 2 } } =\sqrt { 81+4 } =\sqrt { 85 } \)
AB = CD =\(\sqrt { 73 } \) and BC = AD = \(\sqrt { 85 } \)(Opposite sides sre equal)
∴ ABCD is a parallelogram.
6.
Distance AB = \(\sqrt { (a-a)^{ 2 }+(3+2)^{ 2 } } \)
= \(\sqrt { 0+{ 5 }^{ 2 } } =\sqrt { 25 } =5\)
BC = \(\sqrt { (a-a)^{ 2 }+(0-3)^{ 2 } } \)
= \(\sqrt { 0+9 } =\sqrt { 9 } =3\)
AC = \(\sqrt { (a-a)^{ 2 }+(0+2)^{ 2 } } \)
= \(\sqrt { 0+2^{ 2 } } =\sqrt { 4 } =2\)
AC + BC = AB ⇒ 2 + 3 = 5
∴ The given three points are collinear
7.
Given points are A(2, 3) and B(2, -4)
The point P lines on the x-axis.
∴ The point P is (x, 0)
AP = \(\frac { 3 }{ 7 } \) AB
\(\frac { AP }{ AB } \) = \(\frac { 3 }{ 7 } \)
\(\frac { AP }{ PB } \) = \(\frac { 3 }{ 7 } \) ........(1)
Distance =\(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
AP = \(\sqrt { (x-2)^{ 2 }+(0-3)^{ 2 } } \)
= \(\sqrt { { x }^{ 2 }-4x+4+9 } \)
= \(\sqrt { { x }^{ 2 }-4x+13 } \)
BP =\(\sqrt { (x-2)^{ 2 }+(0+4)^{ 2 } } \)
= \(\sqrt { x^{ 2 }-4x+4+16 } \)
= \(\sqrt { x^{ 2 }-4x+20 } \)
From (1) we get
\(\frac { AP }{ PB } \) = \(\frac { 3 }{4 } \)
\(\frac { \sqrt { { x }^{ 2 }-4x+13 } }{ \sqrt { { x }^{ 2 }-4x+20 } } =\frac { 3 }{ 4 } \) (Squaring on both sides)
\(\frac { { x }^{ 2 }-4x+13 }{ { x }^{ 2 }-4x+20 } =\frac { 9 }{ 16 } \)
16x2- 64x + 208 = 9x2 - 36x + 180
7x2 - 28x + 28 = 0
x2- 4x + 4 = 0
(x-2)2= 0
x-2 =0
x = 2
∴ The point P is (2, 0)
8.
Since B is on the x-axis, the y-coordinate of B is 0.
So, the coordinates of the point B is (11, 0)
By the distance formula the distance between the points A (7, 3), B (11, 0) is
\(d=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(AB=\sqrt { \left( 11-7 \right) ^{ 2 }+\left( 0-3 \right) ^{ 2 } } \)
\(=\sqrt { \left( 4 \right) ^{ 2 }+\left( -3 \right) ^{ 2 } } =\ \sqrt { 16+9 } =\sqrt { 25 } =5\)
9.
The distance between the points (-4, 3), (2, -3) is
\(d=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 2+4 \right) ^{ 2 }+\left( -3-3 \right) ^{ 2 } } \)
\(=\sqrt { \left( { 6 }^{ 2 }+\left( -6 \right) ^{ 2 } \right) } =\sqrt { \left( 36+36 \right) } \)
\(=\sqrt { 36\times 2 } \)
\(=6\sqrt { 2 } \)
10.
(-1,1), (1, 3) and (3, a)
Distance =\(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+({ y }_{ 2 }-y_{ 1 })^{ 2 } } \)
AB =\(\sqrt { (1+1)^{ 2 }+(3-1)^{ 2 } } \)
= \(\sqrt { (2)^{ 2 }+(2)^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } \)
BC =\(\sqrt { (3-1)^{ 2 }+(a-3)^{ 2 } } \)
= \(\sqrt { (2)^{ 2 }+(a-3)^{ 2 } } =\sqrt { 4+(a-3)^{ 2 } } \)
But given AB = BC ⇒\(\sqrt { 8 } =\sqrt { 4+(a-3)^{ 2 } } \)
a-3 = 2 (or) a - 3 = -2
a = 2+3 (or) a = 3-2
a = 5 (or) a = 1
∴ 4+(a - 3)2 = 8
(a - 3)2 = 8 - 4
(a - 3)2 = 4
a - 3 = \(\sqrt { 4 } =\pm 2\)
∴ The value of a = 5 or a = 1
11.
The two given points are (5,-2), (1, a) and d = 5.
By distance formula
\(d=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 1-5 \right) ^{ 2 }+\left( a+2 \right) ^{ 2 } } =5\)
\(=\sqrt { 16+\left( a+2 \right) ^{ 2 } } =5\)
16 + (a+2)2 = 25 (By squaring on both the sides)
(a+2)2 = 25–16
(a+2)2 = 9
(a+2) = \(\pm \) 3 (By taking the square root on both side)
a = –2 \(\pm \) 3
a = –2 + 3 (or) a = –2 –3
a = 1 or –5.
12.
All points lie on y-axis.
13.
(c)
\(\sqrt{a^2+{b^2}}\ unit\)
14.
(c)
(0,6)
15.
(d)
(5,0)
16.
(b)
(0, 4)
17.
(c)
Trapezium
18.
(c)
(a, –b)
19.
(a)
on the negative direction of x-axis
20.
(b)
II
21.
( )
Origin
22.
( )
0
23.
( )
0
24.
( )
0
9th Standard Syllabus & Materials
9th Standard
TN 9th Tamil உள்ளத்தின் சீர் - திருக்குறள் Important Questions And Answers Study Material - QB365 Set A
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TN 9th Tamil உயிருக்கு வேர் - தண்ணீர் Important Questions And Answers Study Material - QB365 Set A
Tamilnadu Stateboard 9th Standard Subjects
Tamilnadu Stateboard Standards
