9th Standard Syllabus & Materials
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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Maths Test1.
Find the coordinates of the point which divides the line segment joining A(−5,11) and B(4,−7) in the ratio 7:2.
2.
Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) in such a way that AP = \(\frac{2}{5}\)AB
3.
The line segment joining A(6, 3) and B(−1, −4) is doubled in length by adding half of AB to each end. Find the coordinates of the new end points.
4.
Using section formula, show that the points A(7, −5), B(9, −3) and C(13, 1) are collinear.
5.
A line segment AB is increased along its length by 25% by producing it to C on the side of B. If A and B have the coordinates (−2,−3) and (2,1) respectively, then find the coordinates of C.
6.
If the centroid of a triangle is at (4, −2) and two of its vertices are (3, −2) and (5, 2) then find the third vertex of the triangle.
7.
The vertices of a triangle are (1, 2), (h, −3) and (−4, k). If the centroid of the triangle is at the point (5, −1) then find the value of \(\sqrt { { (h+k) }^{ 2 }+{ (h+3k) }^{ 2 } } \)
8.
Orthocentre and centroid of a triangle are A(−3, 5) and B(3,3) respectively. If C is the circumcentre and AC is the diameter of this cicle, then find the radius of the circle.
9.
ABC is a triangle whose vertices are A(3, 4), B(−2, −1) and C(5, 3) . If G is the centroid and BDCG is a parallelogram then find the coordinates of the vertex D.
10.
If \(\left( \frac { 3 }{ 2 } ,5 \right) ,\left( 7,\frac { -9 }{ 2 } \right) \)and\((\frac{13}{2},\frac{-13}{2})\) are mid-points of the sides of a triangle, then find the centroid of the triangle.
1.
x1 y1
Here (-5,11)
x2 y2
(4,-7)
m n
7 : 2
\(=P(x,y)\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) \)
\(=\left( \frac { 7\times 4+2\times -5 }{ 7+2 } ,\frac { 7\times -7+2\times 11) }{ 7+2 } \right) \)
\(=\left( \frac { 28-10 }{ 9 } ,\frac { -49+22 }{ 9 } \right) \)
\(=\left( \frac { 18 }{ 9 } ,\frac { -27 }{ 9 } \right) =(2,-3)\)
2.
x1 y1 x2 y2
A(1, 2) B (6, 7)
\(=P(x,y)\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) \)
\(=\left( \frac { 2\times 6+3\times 1 }{ 2+3 } ,\frac { 2\times 7+3\times 2) }{ 2+3 } \right) \)
\(=\left( \frac { 12-3 }{ 5 } ,\frac { 14+6 }{ 5 } \right) =\left( \frac { 15 }{ 5 } ,\frac { 20 }{ 5 } \right) =(3,4)\)
3.
\(\bar { AB } =\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (-1-6) }^{ 2 }+{ (-4-3) }^{ 2 } } \)
\(=\sqrt { { (-7) }^{ 2 }+{ (-7) }^{ 2 } } =\sqrt { 49+49 } =\sqrt { (49)2 } \)
\(\frac { 1 }{ 2 } AB=\frac { 7\sqrt { 2 } }{ 2 } =\frac { 7\sqrt { 2 } }{ \sqrt { 2 } \times \sqrt { 2 } } =\frac { 7 }{ \sqrt { 2 } } \)
\(AB=\left( \frac { 6+(-1) }{ 2 } ,\frac { 3+(-4) }{ 2 } \right) =\left( \frac { 5 }{ 2 } ,\frac { -1 }{ 2 } \right) \)
Mid point of \(AB=\left( \frac { 6+(-1) }{ 2 } ,\frac { 3+(-4) }{ 2 } \right) =\left( \frac { 5 }{ 2 } ,\frac { -1 }{ 2 } \right) \)
SinceC is at the distance \(\frac{1}{2}\) AB from B
Midpoint of MC = B
\(\left( \frac { \frac { 5 }{ 2 } +{ x }_{ 3 } }{ 2 } ,\frac { \frac { -1 }{ 2 } +{ y }_{ 3 } }{ 2 } \right) =(-1,-4)\)
\(\left( \frac { \frac { 5 }{ 2 } +{ x }_{ 3 } }{ 2 } \right) =-1\)
\(\frac { 5 }{ 2 } +{ x }_{ 3 }=-2\)
\({ x }_{ 3 }=-2-\frac { 5 }{ 2 } \)
\(=\frac { -4-5 }{ 2 } \)
\(=\frac { -9 }{ 2 } \)
\(\therefore C({ x }_{ 3 },{ y }_{ 3 })=\left( \frac { -9 }{ 2 } ,\frac { -15 }{ 2 } \right) \)
\(\left( \frac { -\frac { 1 }{ 2 } +{ y }_{ 3 } }{ 2 } \right) =-4\)
|\(-\frac { 1 }{ 2 } +{ y }_{ 3 }=-8\)
\({ y }_{ 3 }=-8+\frac { 1 }{ 2 } \)
\(=\frac { -16+1 }{ 2 } \)
\(=\frac { -15 }{ 2 } \)
Similarly by Mid point of DM = A(6, 3)
\(\left( \frac { { x }_{ 4 }+\frac { 5 }{ 2 } }{ 2 } ,\frac { { y }_{ 4 }+\left( \frac { -1 }{ 2 } \right) }{ 2 } \right) =(6,3)\)
\(\frac { { x }_{ 4 }+\frac { 5 }{ 2 } }{ 2 } =6\)
\({ x }_{ 4 }+\frac { 5 }{ 2 } =12\)
\({ x }_{ 4 }=12-\frac { 5 }{ 2 } =\frac { 24-5 }{ 2 } =\frac { 19 }{ 2 } \)
\(\frac { { y }_{ 4 }+\left( \frac { -1 }{ 2 } \right) }{ 2 } =3\)
\({ y }_{ 4 }+\left( \frac { -1 }{ 2 } \right) =6\)
\({ y }_{ 4 }=6+\frac { 1 }{ 2 } =\frac { 12+102 }{ 2 } =\frac { 13 }{ 2 } \)
\(\therefore\) The other end D (x4, y4)\(=\left( \frac { 19 }{ 2 } ,\frac { 13 }{ 2 } \right) \)
4.
x1 y1 x2 y2
A(7, -5) B(9, -3)
\(\bar { AB } =\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } \)
\(=\sqrt { { (9-7) }^{ 2 }+{ (-3-(-5)) }^{ 2 } } =\sqrt { { 2 }^{ 2 }+{ (-3+5) }^{ 2 } } =\sqrt { 4+4 } \)
\(=\sqrt { 8 } =\sqrt { 4\times 2 } =2\sqrt { 2 } \)
\(\bar { BC } =\sqrt { { (13-7 })^{ 2 }+{ (1-(-5)) }^{ 2 } } =\sqrt { { 4 }^{ 2 }+{ 4 }^{ 2 } } =\sqrt { 16+16 } \)
\(=\sqrt { 32 } =\sqrt { 16\times 2 } =4\sqrt { 2 } \)
\(\bar { AC } =\sqrt { { (13-7) }^{ 2 }+{ (1-(-5)) }^{ 2 } } \)
\(=\sqrt { { 6 }^{ 2 }+{ 6 }^{ 2 } } =\sqrt { 36+36 } \)
\(=\sqrt { { 6 }^{ 2 }+{ 6 }^{ 2 } } =\sqrt { 36+36 } \)
\(2\sqrt { 2 } +4\sqrt { 2 } =6\sqrt { 2 } \)
∴ AB + BC = AC, Here B is the common Point
∴ A, B, C are collinear
5.
x1 y1 x2 y2
A(-2, -3) B(2, 1)
m : n = 3 : 1
The point P divides AB in the ratio 3 : 1
\(P(x,y)=\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) \)
\(=\left( \frac { 3\times 2+1\times -2 }{ 3+1 } ,\frac { 3\times 1+1\times -3) }{ 3+1 } \right) \)
\(=\left( \frac { 6-2 }{ 4 } ,\frac { 3-3 }{ 4 } \right) =\left( \frac { 4 }{ 4 } ,\frac { 0 }{ 4 } \right) =(1,0)\)
P is at 25% distance from B on its left and C is at 25% distance from B on its right
\(\therefore\) B is the mid point of PC
Mid point of \(\bar { PC } =\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
\((2,1)=\left( \frac { 1+{ x }_{ 2 } }{ 2 } ,\frac { 0+{ y }_{ 2 } }{ 2 } \right) \)
\(\frac { 1+{ x }_{ 2 } }{ 2 } =2\quad \frac { { y }_{ 2 } }{ 2 } =1\)
1 + x2 = 4 y2 = 2
⇒ x2 = 3, y2 = 2
∴ C(x2, y2) = (3, 2) is the solutions.
6.
Centroid G (x,y) = (4, -2)
two vertices (x1, y1) = (3, -2)
(x2, y2) = (5, 2), (x3, y3) =?
\(\frac { 8+{ x }_{ 3 } }{ 3 } =4\ \ \frac { { y }_{ 3 } }{ 3 } =-2\)
8 + x3 = 12 y3 = -6
x3 = 4
∴ The third vertex (x3, y3) = (4, -6)
7.
Vertices of a triangle
(x1 ,y1) = (1,2)
(x2,y2) = (h,-3)
(x3, y3) = (-4, k)
Centroid G (x,y) = (5, -1)
\(G=\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) =(5,-1)\)
\(\left( \frac { 1+h+(-4) }{ 3 } ,\frac { 2+(-3)+k }{ 3 } \right) =(5,-1)\)
\(\Rightarrow \frac { -3+h }{ 3 } =5\)
-3 + h = 15
h = 18
\(\frac { 2-3+k }{ 3 } =-1\)
\(\therefore \sqrt { { (h+k) }^{ 2 }+{ (h+3k) }^{ 2 } } \)
\(=\sqrt { (18+(-2))^{ 2 }+{ (18+3(-2)) }^{ 2 } } \)
\(=\sqrt { { 16 }^{ 2 }+{ 12 }^{ 2 } } =\sqrt { 256+144 } =\sqrt { 400 } =20\)
8.
H = A (-3, 5)
G = B (3, 3)
S = Circumcentre (C)
G divides Hand S in the ratio 2 :1
m : n = 1 : 2
A(x2, y1) = (-3, 5)
C(x2, y2) = (x2, y2)
P (x, y) = (3, 3)
\(\therefore \ P(x,y)=\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) \)
\((3,3)=\left( \frac { { 1x }^{ 2 }+2(3) }{ 2+1 } ,\frac { 1({ y }_{ 2 })+2(5) }{ 2+1 } \right) \)
\((3,3)=\left( \frac { { x }_{ 2 }-6 }{ 3 } ,\frac { { y }_{ 2 }+10 }{ 3 } \right) \)
\(\frac { { x }_{ 2 }-6 }{ 3 } =3\)
x2 - 6 = 9
x2 = 9 + 6
x = 15
\(\frac { { y }_{ 2 }+10 }{ 3 } =3\)
y2+10 = 9
y2 = 9-10
y2 = -1
\(\bar { AC } =\sqrt { { (5-(-3)) }^{ 2 }+{ (-1-5) }^{ 2 } } \)
\(=\sqrt { { 8 }^{ 2 }+{ (-6) }^{ 2 } } =\sqrt { 324+36 } \)
\(=\sqrt { 360 } =\sqrt { 36\times 10 } =6\sqrt { 10 } \)
radius = \(\frac{1}{2}\)AC
\(=\frac { 1 }{ 2 } \times 6\sqrt { 10 } =3\sqrt { 10 } \)units.
Radius \(=\frac{3}{2} \sqrt{10}\) or \(3 \sqrt{\frac{10}{4}}\)
\(=3 \sqrt{\frac{5}{2}}\) units
\(\because \) AC is a diametre of circle
9.
Centroid G =\(\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) \)
\(\therefore G(x,y)=\left( \frac { 3+(-2)+5 }{ 3 } ,\frac { 4+(-1)+3 }{ 3 } \right) \)
\(=\left( \frac { 8-2 }{ 3 } ,\frac { 7-1 }{ 3 } \right) =\left( \frac { 6 }{ 3 } ,\frac { 6 }{ 3 } \right) =(2,2)\)
In a parallelogram diagonals bisect each other
∴ Mid point of DG = Mid point of BC
\(\left( \frac { x+2 }{ 2 } ,\frac { y+2 }{ 2 } \right) =\left( \frac { -2+5 }{ 2 } ,\frac { -1+3 }{ 2 } \right) \)
\(\frac { x+2 }{ 2 } =\frac { 3 }{ 2 } \)
x+2 = 3
x = 3 - 2 = 1
\(\frac { y+2 }{ 2 } =\frac { 2 }{ 2 } \)
y = 2 - 2 = 0
y + 2 = 2
\(\therefore\) The co-ordinates of the vertex D (x, y) = (1, 0)
10.
"The centroid of the triangle obtained by joining the mid points of the sides of a triangle is the same as the centroid of the original triangle."
\(\therefore\) The mid points of the sides of the triangle are given as
x1y1 , x2y, x3y3
\(\left( \frac { 3 }{ 2 } ,5 \right) \left( 7,\frac { -9 }{ 2 } \right) \left( \frac { 13 }{ 2 } ,\frac { -13 }{ 2 } \right) \)
\(\therefore\)Centroid \(=\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) \)
\(=\left( \frac { \frac { 3 }{ 2 } +\frac { 7 }{ 1 } +\frac { 13 }{ 2 } }{ 3 } ,\frac { 5+\left( \frac { -9 }{ 2 } \right) +\frac { (-13) }{ 2 } }{ 3 } \right) =\left( \frac { \frac { 3+14+13 }{ 2 } }{ 3 } ,\frac { \frac { 10-9-13 }{ 2 } }{ 3 } \right) \)
\(=\left( \frac { \frac { 30 }{ 2 } }{ 3 } ,\frac { \frac { -12 }{ 2 } }{ 3 } \right) =\left( \frac { 15 }{ 3 } ,\frac { -6 }{ 3 } \right) =(5,-2)\)
9th Standard Syllabus & Materials
9th Standard
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NEW9th Standard
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NEW9th Standard
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Tamilnadu Stateboard 9th Standard Subjects
Tamilnadu Stateboard Standards
