19. Chapter 19. Coordinate Geometry NIOS Class 10 Mathematics Textbook Solutions
Chapter 19. Coordinate Geometry NIOS Class 10 Mathematics Textbook Solutions for Exam Preparation
Chapter 19. CO-ORDINATE GEOMETRY
CHECK YOUR PROGRESS 19.1
1. Write down x and y co-ordinates for each of the following points :
(a) (3, 3) (b) (–6, 5) (c) (–1, –3) (d) (4, –2)
Solution: (a) x co-ordinate is 3 .
y co-ordinate is 3 .
(b) x co-ordinate is –6 .
y co-ordinate is 5.
(c) x co-ordinate is –1
y co-ordinate is –3.
(d) x co-ordinate is 4.
y co-ordinate is –2 .
2. Write down distances of each of the following points from the y and x axis respectively.
(a) A(2, 4) (b) B(–2, 4) (c) C(–2, –4) (d) D(2, –4)
Solution: (a) The distance of the point A from the y-axis is 4 units to the right of origin and from the x-axis is 2 units above the origin.
(b) The distance of the point B from the y-axis is 4 units to the left of the origin and from the x-axis is 2 unit above the origin.
(c) The distance of the point C from the y-axis is 4 units to the left of the origin and from the x-axis is also 2 units below the origin.
(d) The distance of the point D from the y-axis is 4 units to the right of the origin and from the x-axis is 2 units below the origin.
3. Group each of the following points quadrantwise :
A(–3, 2) , B (2, 3) , C(7, -6) , D(l, 1) , E(–9, –9) , F (–6, 1) , G (–4, –5) , H(11, –3) , P(3, 12) , Q(–13, 6)
Solution: A(– 3 , 2) lies in the II quadrant .
B(2, 3) lies in the I quadrant .
C(7, –6) lies in the III quadrant .
D(1, 1) lies in the I quadrant .
E(– 9,– 9) lies in the III quadrant .
F(–6, 1) lies in the II quadrant .
G(–4, –5) lies in the III quadrant .
H(11, –3) lies in the IV quadrant .
P(3, 12) lies in the I quadrant .
Q(–13, 6) lies in the II quadrant .
CHECK YOUR PROGRESS 19.2
1. Find the distance between each of the following pair of points:
(a) (3, 2) and (11, 8) (b) (–1, 0) and (0, 3) (c) (3, –4) and (8, 5) (d) (2, –11) and (–9, –3)
Solution: (a) (3, 2) and (11, 8)
Let, the points are P(3, 2) and Q(11, 8) .
Here, 
By using distance formula, we have
units
(b) (–1, 0) and (0, 3)
Let, the points are P(– 1, 0) and Q(0 , 3) .
Here, 
By using distance formula, we have
units
(c) (3, –4) and (8, 5)
Let, the points are P(3, 2) and Q(11, 8) .
Here, 
By using distance formula, we have
units
(d) (2, –11) and (–9, –3)
Let, the points are P(2, – 11) and Q(– 9,– 3) .
Here, 
By using distance formula, we have
units
2. Find the radius of the circle whose centre is at (2, 0) and which passes through the point (7, –12).
Solution: Let O(2, 0) and P(7,–12) be the given points of the line segment OP.
Here, 
By using distance formula, we have
units
3. Show that the points (–5, 6), (–1, 2) and (2, –1) are collinear.
Solution: Let, the three points are A(–5, 6), B(–1, 2) and C(2, –1) .
By using distance formula, we have
units 
units
units
Therefore, the points (–5, 6), (–1, 2) and (2, –1) are collinear.
CHECK YOUR PROGRESS 19.3
1. Find the co-ordinates of the point which divides internally the line segment joining the points:
(a) (1, –2) and (4, 7) in the ratio 1 : 2 (b) (3, –2) and (–4, 5) in the ratio 1 : 1
Solution: (a) Let,(x,y) be the co-ordinates of the point .
Here, 
Using section formula, we have
Therefore, (2,1) be the co-ordinates of the point .
(b) (3, –2) and (–4, 5) in the ratio 1 : 1 .
Let,(x,y) be the co-ordinates of the point .
Here, 
Using section formula, we have
Therefore, 
be the co-ordinates of the point .
2. Find the mid-point of the line joining:
(a) (0, 0) and (8, –5) (b) (–7, 0) and (0, 10)
Solution: (a) (0, 0) and (8, –5)
Let,(x,y) be the mid-point of the line.
Here, 
Using mid-Point Formula , we have
and 
Therefore, the mid-point of the line is 
.
(b) (–7, 0) and (0, 10)
Let,(x,y) be the mid-point of the line.
Here, 
Using mid-Point Formula , we have
and 
Therefore, the mid-point of the line is 
.
3. Find the centroid of the triangle whose vertices are (5, –1), (–3, –2) and (–1, 8).
Solution: Let, A(5, –1), B(– 3,– 2) and C(– 1, 3) be the vertices of a triangle and also G(x, y) be the centroid of the triangle .
Here, 
and 
Therefore, the centroid of the triangle is 
.
TERMINAL EXERCISE
1. In Fig. 19.15, AB = AC. Find x.
Solution: Given, the points are A(0,2) , B(0,5) and C(x,2)
A/Q, 
[Squaring both side]
Therefore, the value of x is 0 .
2. The length of the line segment joining two points (2, 3) and (4, x) is 
units. Find x.
Solution: Given, the two points are (2, 3) and (4, x) .
Using distance formula, we have
A/Q , 
[squaring both side]
and 
3. Find the lengths of the sides of the triangle whose vertices are A(3, 4), B(2, –1) and C(4, –6).
Solution: Given, the vertices of the triangle are A(3, 4), B(2, –1) and C(4, –6).
Using distance formula, We have 
units
units
CA 
units
4. Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle.
Solution: Let, the vertices of the triangle are A(2, – 2), B(– 2, 1) and C(5, 2).
Using distance formula, We have 
Thus, the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle.
5. Find the co-ordinates of a point which divides the join of (2, –1) and (–3, 4) in the ratio of 2 : 3 internally.
Solution: Let,(x,y) be the co-ordinates of the point .
Here, 
Using section formula, we have
Therefore, (0,1) be the co-ordinates of the point .
6. Find the centre of a circle, if the end points of a diameter are P(–5, 7) and Q(3, –11).
Solution: Let, O(x,y) be the centre of a circle .
Here, 
Since, O be the mid-point of the segment PQ (diameter) .
Using mid-Point Formula , we have
and 
Therefore, the point of the centre of a circle is 
.
7. Find the centroid of the triangle whose vertices are P(–2, 4), Q(7, –3) and R(4, 5).
Solution: Given, P( –2,4), Q(7,– 3) and R(4, 5) be the vertices of a triangle .
Let, G(x, y) be the centroid of the triangle .
Here, 
and 
Therefore, the centroid of the triangle is 
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