Samacheer Kalvi Class 9 Maths Chapter 5 Coordinate Geometry Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 9th class students by providing Solutions for Samacheer Kalvi Class 9 Maths chapter 5 Coordinate Geometry. Here students can easily find all the solutions for Coordinate Geometry Exercise 5.1, 5.2, 5.3, 5.4, 5.5 and 5.6. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 5 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.
Coordinate Geometry Exercise 5.2 Solutions :
(1) Find the distance between the following pairs of points.
(i) (1, 2) and (4, 3)
(ii) (3,4) and (– 7, 2)
(ii) (a, b) and (c, b)
(iv) (3,– 9) and (–2, 3)
Solution :
(i) (1, 2) and (4, 3)
Distance = √(4-1)² + (3-2)²
= √3² + 1²
= √10
(ii) (3,4) and (– 7, 2)
Distance = √(-7-3)² + (2-4)²
= √100 + 4
= √104
(iii) (a, b) and (c, b)
Distance = √(c-a)² + (b-b)²
= √(c-a)²
= c – a
(iv) (3,– 9) and (–2, 3)
Distance = √(-2-3)² + (3+9)²
= √(-5)² + (12)²
= √169
= 13
(2) Determine whether the given set of points in each case are collinear or not.
(i) (7,–2), (5,1), (3,4)
(ii) (a,–2), (a,3), (a,0)
Solution :
(i) Let, A = (7, 2), B = (5, 1), C = (3,4)
AB = √(5-7)² + (1+2)² = √4+9 = √ 13
BC = √(3-5)² + (4-1)² = +9 = √13
AC = √(7-3)² + (-2-4)² = √ 16+36 = √52 = √13
So, AB + BC = AC So, the points are collinear.
(ii) Let, A = (a, -2), B = (a,3), C = (a, 0)
AB = √(a-a)² + (3+2)² = √25 = 5
BC = √(a-a)² + (0-3)² = √9 = 3
CA = √(a-a)² + (-2-0)² = √4 = 2
AB + AC = 5 = BC
Therefore, the points are collinear.
(3) Show that the following points taken in order form an isosceles triangle.
(i) A (5,4), B(2,0), C (–2,3)
(ii) A (6,–4), B (–2, –4), C (2,10)
Solution :
(i) AB = √(2 – 5)² + (0+4)² = √25 = 5
BC = √(-2 – 2)² + (3- 0)² = √25 = 5
CA = √(5 + 2)² + (4 – 3)² = √50 = 5√2
Since, AB + BC > AC
Thus, ABC is an isosceles triangle.
(ii) AB = √(-2 – 6)² + (-4 + 4)² = √64 = 8
BC = √(2 + 2)² + (10 + 4)² = √212 = 2√53
AC = √(6 – 2)² + (-4 – 10)² = √212 = 2√53
Since. AB + BC > AC
Hence, ABC is an isosceles triangle.
(4) Show that the following points taken in order form an equilateral triangle in each case.
(i) A(2, 2), B(–2, –2), C(-2√3, 2√3)
(ii) A(√3,2), B(0,1), C(0,3)
Solution :
(i) AB = √( – 2 – 2)² + (-2 – 2)² = √32 = 4√2
BC = √( – 2√3 + 2)² + (2√3 + 2)² = √32 = 4√2
=CA = √(2 + 2√3)² + (2 – 2√3)² = √32 = 4√2
Since, AB = BC = CA
So, ABC is an equilateral triangle.
(ii) AB = √(0 – √3)² + (1 – 2)² = √4 = 2
BC = √(0 – 0)² + (3 – 1)² = √4 = 2
AC = √(√3 – 0)² + (2 – 3)² = √4 = 2
Similarly, AB = BC = AC
Therefore, ABC is an equilateral triangle.
(5) Show that the following points taken in order form the vertices of a parallelogram.
Solution :
(i) AB = √( – 6 + 3)² + (-7 – 1)² = √9 + 64 = √73
BC = √(3 + 6)² + (-9 – 7)² = √81 + 4 = √85
CD = √(6 – 3)² + (-1 + 9)²= √9 + 64 = √73
AD = √(-3 + 6)² + (1 + 1)²= √81 + 4 =√85
Since, AB = CD and BC = AD
Thus, ABCD is parallelogram
(ii) AB = √(5 + 7)² + (10 + 3)² = √144 + 169 = √313
BC = √(15 – 5)² + (8 – 10)² = √100 + 4 = √104
CD = √(3 – 15)² + (-5 – 8)² = √144 + 169 = √313
AD = √( – 7 – 3)² + ( – 3 + 5)² = √100 + 4 = √104
Since, AB = CD and BC = AD
Hence, ABCD is parallelogram.
(6) Verify that the following points taken in order form the vertices of a rhombus.
(i) A (3, –2), B (7, 6), C (–1, 2), D (– 5, – 6)
(ii) A (1, 1), B (2, 1), C (2, 2), D (1, 2)
Solution :
(ii) AB = √(2 – 1)² + (1 – 1)² = √1 = 1
BC = √(2 – 2)² + (2 – 1)² = √1 = 1
CD = √(1 – 2)² + (2 – 2)² = √1 = 1
AD = √(1 – 1)² + (1 – 2)² = √1 = 1
Since AB = BC = CD = AD
Therefore, ABCD is a Rhombus
(7) A (–1, 1), B (1, 3) and C (3, a) are points and if AB = BC, then find ‘a’.
Solution :
AB = √(1+1)² + (3-1)²
= √4 + 4
= √8
BC = √(3-1)² + (a-3)²
= √4 + a² – 6a + 9
So, AB = BC
= 8 = 13 + a² – 6a
= a2 – 6a + 5 = 0
= a2 – 5a – 1a + 5 = 0
= a (a – 5) – 1 (a – 5) = 0
= (a – 5) (a – 1) = 0
Therefore, a = 1 or 5
(8) The abscissa of a point A is equal to its ordinate, and its distance from the point B(1, 3) is 10 units, What are the coordinates of A?
Solution :
Let, P the point
So PA = √(x-3)² + (y-4)²
PB = √(x- (-5)² + (y-6)²
PA = PB
= √(x-3)² + (y-4)² = √x – (-5)² + (y-6)²
= x² – 6x + 9 + y² – 8y + 16 = x² + 10x + 25 + y² – 12y + 36
= – 6x – 10x – 8y + 12y + 25 – 25 = 36
= – 16x + 4y = 36
= 16x – 4y + 36 = 0
= 4x – y + 9 = 0
= y = 4x + 9
(10) Let A (2, 3) and B (2, – 4) be two points. if P lies on the x-axis, such that AP = 3/7 AB, find the coordinates of P
Solution :
AD = 3/7 AB
AP = √(x-2)² + (0-3)², AB = √(-4-3)² + (2-2)²
So, √(x-2)² + 9 = 3/7 √(-7)²
= 7 √x² – 4x + 13 = 3×7
= x² – 4x + 13 = 9
= x² – 4x – 4 = 0
= (x – 2) (x – 2) = 0
x = 2, 2
∴ The co-ordinates of P (x, 0) = P (2, 0)
(11) Show that the point (11, 2) is the centre of the circle passing through the points (1, 2), (3, -4) and (5, -6)
Solution :
Let o be the centre,
OA = √(1 – 11)² + (-2 – 2)² = √100 = 10
OB = √(3 – 11)² + (-4 – 2)² = √64 + 36 = √100 = 10
OC = √(5 – 11)² + (-6 – 2)² = (36 + 64 = √100 = 10
Since, OA = OB = OC
Hence, O is the centre of Circle.
(12) The radius of a circle with centre at origin is 30 units. Write the coordinates of the points where the circle intersects the axes. Find the distance between any two such points.
Solution :
Radius of the circle = 30
Equation of circle
√x² + y² = 30 [Passing through origin]
√(x – 0)2 + (0 – 0)² = 30
= x² = 900
= x = ± 30
Again, √(0 – 0)² + (y – 0)² = 30
= y² = 900
= y = ± 30
The distance between two points,
√(0-30)² + (30-0)²
= √900 + 900
= 30√2
Therefore, the distance between any such two points will be 30√2
Coordinate Geometry Exercise 5.3 Solutions :
(1) Find the mid-points of the line segment joining the points
(i) (-2, 3) and (-6, -5)
(ii) (8, -2) and (-8, 0)
(iii) (a, b) and (a + 2b, 2a – b)
(iv) (1/2, -3/7) and (3/2, -11/7)
Solution :
(i) Mid point,
= (-2 + (-6)/2, 3 + (-5)/2)
= (-4, -1)
(ii) Mid point,
= (8 – 8/2, -2 + 0/2)
= (0, -1)
(iii) Mid point,
= (a + a + 2b/2, b + 2a-b/2)
= (a + b, a)
(iv) Mid point,
= (4/2)/, (-3-11)/2
= (4/4, (-14/14)
= (1, -1)
(2) The centre of a circle is (−4, 2). If one end of the diameter of the circle is (−3,7), then find the other end.
Solution :
Diameter = (-3, 7), centre = (-4, 2)
Let another point (x, y)
Mid point (-3 + x/2, 7 + y/2)
So, -3 + x/2= -4, 7 + y/2 = 2
= x = -5, = y = -3
Therefore, the other end will be (-5, -3)
(3) If the mid-point (x, y) of the line joining (3, 4) and (p, 7) lies on 2x + 2y +1 = 0, then what will be the value of p?
Solution :
Any point (x, y)
So, (3 + P/2, 4 + 7/2) = (x, y)
So, 3 + P/2 = x, 4 + 7/2 = y
= x = 3 + P/2 = 11/2 = y
Hence, 2x + 2y + 1 = 0
= 2× 3 + P/2 + 2× 11/2 + 1 = 0
= 3 + P + 11 + 1 = 0
= P = -15
Therefore, the value of p is – 15.
(5) O(0, 0) is the centre of a circle whose one chord is AB, where the points A and B are (8,6) and (10,0) respectively. OD is the perpendicular from the centre to the chord AB. Find the coordinates of the mid-point of OD.
Solution :
Mid point = (x_1 + x_2/2, y_4 + y_2/2)
Mid point of AB = (8 + 10/2, 6 + 0/2)
= (9, 3)
So, 0 + 9/2, 0 + 3/2
= 9/2, 3/2 is the co-ordination of the mid point of OD
(7) The points A (−3, 6), B (0, 7) and C(1, 9) are the mid-points of the sides DE, EF and FD of a triangle DEF. Show that the quadrilateral ABCD is a parallelogram.
Solution :
Mid point of DE A(-3, 6) = x_1 + x_2/2, y_1 + y_2/2
So, x_1 + x_2/2 = -3
= x_1 + x_2 = – 6
y_1 + y_2/2 = 6
= y_1 + y_2 = 12
Mid point of EF B (0, 7), = (x_2 + x_3/2, y_2 + y_3/2)
x_2 + x_3/2 = 0
= x_2 + x_3 = 0
y_2 + y_3/2 = 7
= y_2 + y_3 = 14
Mid point of AC = (-3-3/2, 6-2/2) = (-3, 2)
Mid point of BD = -6+0/2, -3+7/2 = (-3, 2)
Since mid point of AC and BD are similar
Therefore, ABCD obviously a parallelogram.
Coordinate Geometry Exercise 5.4 Solutions :
(1) Find the coordinates of the point which divides the line segment joining the points A (4, -3) and B (9, 7) in the ratio 3 : 2
Solution :
P(x, y) = (3(9) + 2 (4)/3 + 2, 3(7) +2 (-3)/3 + 2)
= (27 + 8/5, 21 – 6/5)
= (7, 3)
(2) In what ratio does the point P (2, -5) divide the line segment joining A (-3, 5) and B (4, -9)
Solution :
P (2, -5) = m × 4 + n(-3)/m + n, m(-9) + n (5)/m + n
= 38 m = 35 n
4m – 3n/m + n = 2
= 4m – 3n = 2m + 2n
= 2m = 5n
= m : n = 5 : 2
(3) 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 = 2/5 AB
Solution :
P (x, y) = (2 × 6 + 3 × 1/3 + 2, 2 × 7 + 3 × 2/3 + 2)
= (15/5, 20/5)
= (3, 4)
Therefore, the coordinates of a point P is (3, 4)
(4) Find the coordinates of the points of trisection of the line segment joining the points A (-5, 6) and B (4, -3)
Solution :
P (x, y) (1 × 4 + 2 × (-5)/1 + 2, 1 × (-3) + 2 × 6/1 + 2)
= (-6/3, 9/3)
= (-2, 3)
And Q (x, y) = (2 × 4 + 1 × (-5)/2 + 1, 2 × (-3) + 1 × 6/2 + 1)
= (3/3, 0/3)
= (1, 0)
Therefore, the coordinates of the points of trisection (1, 0)
(5) 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.
Solution :
AB = √(-1-6)² + (-4-3)² = √49×2
1/2 AB = 7 × √2/2 = 7/√2
Mid point of AB = (6 + (-1)/2, 3-4/2)
= (5/2, -1/2)
Mid point CM
(5/2 + x_3)/2 ; (- 1/2 + y-3)/2 = (-1, -4)
= (5/2 + x_3)/2 = -1
= 5/2 + x_3 = -2
= x_3 = -2 – 5/2
= x_3 = -9/2
(- ½ + y_3)/2 = -4
= – ½ + y_3 = -8
= y_3 = – 8 + 1/2
= -15/2
Similarly, Mid point of DM
(x_4 + 5/2)/2, (y_4 – 1/2)/2 = 6, 3
(x_4 + 5/2)/2 = 6
= x_4 = 12 – 5/2
= x = 19/2
(y_4 – 1/2)/2 = 3
= y_3 = 6+ 1/2
= y_3 = 13/2
Coordinate Geometry Exercise 5.5 Solutions :
(1) Find the centroid of the triangle whose vertices are
(i) (2,−4), (−3,-7) and (7,2)
(ii) (−5,−5), (1,-4) and (−4,−2)
Solution :
Centroid G,
= (-5+1-4/3, -5-4-2/3)
= (-8/3, -11/3)
Thus, the centroid of the triangle is (-8/3, -11/3).
(2) 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.
Solution :
Centroid (4, -2)
Again, 3 + 5 + x/3, -2 + 2 + y/3 = 4, -2
= 8 + x/3 = 4
= x = 4
Y = 6
Thus, The third vertex is (4, -6)
(3) Find the length of median through A of a triangle whose vertices are A(−1,3), B(1,−1) and C(5,1)
Solution :
Let, D be the mid point of opposite side of A.
So, mid point of DC,
= (1+5/2, -1+1/2)
= (3, 0)
So, Length of AD,
= √(3- (-1))² + (0-3)²
= √4² + 3²
= √25
= 5
Therefore, the length of median is 5
(4) 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 √(h+k)² + (h+3k)²
Solution :
Centroid = (1 + h – 4/3, 2-3 + k/3)
Again, given centroid = (5, -1)
-3 + h/3 = 5
= h = 18
-1 + K/3 = -1
= k = -2
So, √(h + k)² + (h + 3k)²
= √(18-2)² + (18 – 6)²
= √256 + 144 = √400
= 20
Hence, the value of √(h + k)² + (h + 3k)² is 20.
(5) 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 circle, then find the radius of the circle.
Solution :
G divides the line in 1 : 2 ratio
So, P (x, y) = (3, 3)
= (1.x + 2 × (.3)/2 + 1, 1 × y + 2×5/3)
= x – 6/3 = 3
= x = 15
y + 10/3 = 3
= y = -1
So, x = 15, y = 1
AC = √(5 – (-3))² + (-1 -5)²
= √324 + 36
= √360
So, radius = 1/2 AC
= 1/2 √360
= 3√10
Therefore, the radius of the circle is 3√10
(6) 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
Solution :
Centroid = (3-2 + 5/3, 4 – 1 + 3/3)
= (2, 2)
Mid point of DE,
= (x + 2/2, y + 2/2)
= (-2 + 5/2, 4 + 3/2)
= x + 2/2 = 3/2
= x = 1
∴ y + 2/2 = 2/2
= y = 2-2 = 0
The co-ordinates of the vertex (1, 0)
(7) If (3/2, 5), (7, -9/2) and (13/2, -13/2) are mid points of the sides of a triangle, then find the centroid of the triangle
Solution :
Centroid of the triangle,
= ((3/2 + 7/1 + 13/2)/3, (5/1 – 9/2 – 13/2)/3)
= (3+ 14 + 13/2)/3 (10 – 9 – 13/2)/3
= (30/6, 12/6)
= (5, -2)
Thus, the centroid of the triangle is (5, -2)
Coordinate Geometry Exercise 5.6 Solutions :
(1) If the y-coordinate of a point is zero, then the point always lies ___
(1) in the I quadrant
(2) in the II quadrant
(3) on x-axis
(4) on y-axis
Solution :
Correct option – (3)
If the y-coordinate of a point is zero, then the point always lies On x-axis.
(2) The points (–5, 2) and (2, –5) lie in the ____
(1) Same quadrant
(2) II and III quadrant respectively
(3) II and IV quadrant respectively
(4) IV and II quadrant respectively
Solution :
Correct option – (3)
The points (–5, 2) and (2, –5) lie in the II and IV quadrant respectively.
(3) 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
(1) Square
(2) Rectangle
(3) Trapezium
(4) Rhombus
Solution :
Correct option – (3)
Trapezium will be obtain.
(4) If P (–1, 1), Q (3,–4), R ( 1, –1), S (–2, –3) and T ( –4, 4) are plotted on a graph paper, then the points in the fourth quadrant are
(1) P and T
(2) Q and R
(3) Only S
(4) P and Q
Solution :
Correct option – (2)
The points in the fourth quadrant are Q and R.
(5) The point whose ordinate is 4 and which lies on the y-axis is
(1) (4, 0)
(2) (0, 4)
(3) (1, 4)
(4) (4, 2)
Solution :
Correct option – (2)
The point whose ordinate is 4 and which lies on the y-axis is (0, 4)
(6) The distance between the two points (2, 3) and (1, 4) is ____
(1) 2
(2) √56
(3) √10
(4) √2
Solution :
Correct option – (4)
The distance between the two points (2, 3) and (1, 4) is √2
(7) If the points A (2, 0), B (-6, 0), C (3, a–3) lie on the x-axis then the value of a is
(1) 0
(2) 2
(3) 3
(4) –6
Solution :
Correct option – (3)
If the points A (2, 0), B (-6, 0), C (3, a–3) lie on the x-axis then the value of a is 3.
(8) If (x+2, 4) = (5, y–2), then the coordinates (x, y) are
(1) (7, 12)
(2) (6, 3)
(3) (3, 6)
(4) (2, 1)
Solution :
Correct option – (3)
So, the coordinates (x, y) are (3, 6).
(9) If Q_1, Q_2, Q_3, Q_4 are the quadrants in a Cartesian plane then Q_2∩Q_3 is __
(1) Q_1 ∪ Q_2
(2) Q_2 ∪ Q_3
(3) Null Set
(4) Negative x-axis
Solution :
Correct option – (3)
If Q_1, Q_2, Q_3, Q_4 are the quadrants in a Cartesian plane then Q_2∩Q_3 is Null Set.
(10) The distance between the point (5, –1) and the origin is ___
(1) √24
(2) √37
(3) √26
(4) √17
Solution :
Correct option – (3)
The distance between the point (5, –1) and the origin is √26.
(11) The coordinates of the point C dividing the line segment joining the points P (2, 4) and Q (5, 7) internally in the ratio 2:1 is
(1) (7/2, 11/2)
(2) (3, 5)
(3) (4, 4)
(4) (4, 6)
Solution :
Correct option – (4)
The ratio 2 : 1 is (4, 6)
(14) If the coordinates of one end of a diameter of a circle is (3, 4) and the coordinates of its centre is (−3, 2), then the coordinate of the other end of the diameter is
(1) (0, −3)
(2) (0, 9)
(3) (3, 0)
(4) (−9, 0)
Solution :
Correct option – (4)
The coordinate of the other end of the diameter is (−9, 0).
(15) The ratio in which the x-axis divides the line segment joining the points A (a_1, b_1) and B (a_2, b_2) is
(1) b_1 : b_2
(2) –b_1 : b_2
(3) a_1 : a_2
(4) –a_1 : a_2
Solution :
Correct option – (2)
The points A (a_1, b_1) and B (a_2, b_2) is –b_1 : b_2
(16) The ratio in which the x-axis divides the line segment joining the points (6, 4) and (1, −7) is
(1) 2 : 3
(2) 3 : 4
(3) 4 : 7
(4) 4 : 3
Solution :
Correct option – (3)
The ratio in which the x-axis divides the line segment joining the points (6, 4) and (1, −7) is 4 : 7.
(18) The mid-point of the line joining (−a, 2b) and (−3a, −4b) is
(1) (2a, 3b)
(2) (−2a, −b)
(3) (2a, b)
(4) (−2a, −3b)
Solution :
Correct option – (2)
The mid-point of the line joining (−a, 2b) and (−3a, −4b) is (−2a, −b).
(19) In what ratio does the y-axis divides the line joining the points (−5, 1) and (2, 3) internally
(1) 1 : 3
(2) 2 : 5
(3) 3 : 1
(4) 5 : 2
Solution :
Correct option – (4)
The required ratio will be 5 : 2
(20) If (1, −2), (3, 6), (x, 10) and (3, 2) are the vertices of the parallelogram taken in order, then the value of x is
(1) 6
(2) 5
(3) 4
(4) 3
Solution :
Correct option – (2)
Therefore, the value of x is 5.
Next Chapter Solution :
