7 CO-ORDINATE GEOMETRY CHAPTER

CHAPTER
7
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CO-ORDINATE GEOMETRY
Points to Remember :
1. The abscissa and ordinate of a given point are the distances of the point from y-axis and x-axis respectively.
2. The coordinates of a point on the x-axis are of the form (x, 0) and of a point on the y-axis are of the form
(0, y).
3. Distance Formula : The distance between two points P(x 1, y 1) and Q(x 2, y 2) is given by
PQ  ( x2  x1 )2  ( y2  y1 ) 2
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4. Distance of a point P(x, y) from the origin O(0, 0) is given by OP  x 2  y 2 .
5.
6.
7.
8.
9.
Three points P, Q and R are collinear, then PQ + QR = PR or PQ + PR = QR or PR + RQ =PQ.
A ABC is an isosceles triangle, if AB = BC or AB = AC or BC = AC.
A ABC is an equilateral triangle, if AB = BC = AC.
A ABC is a right-angle triangle, if AB2 + BC2 = AC2 or AB2 + AC2 = BC2 or BC2 + AC2 = AB2.
Section Formula : The co-ordinates of the point which divides the join of points P(x1, y1) and Q(x2, y2)
B
mx  nx1 my2  ny1 
internally in the ratio m : n are  2
.
,
m  n 
 mn
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10. Mid-point Formula : The co-ordinates of the mid-point of the line segment joining the points P(x1, y1)
and Q(x2, y2) are  x1  x2 , y1  y2  .
2 
 2
11. The coordinates of the centroid of a triangle formed by the points P(x1, y1), Q(x2, y2) and R(x3, y3) are
 x1  x2  x3 y1  y2  y3 
,


3
3


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12. The area of the triangle formed by the points P(x1, y1), Q(x2, y2) and R(x3, y3) is given by
1
.| x1 ( y2  y3 )  x2 ( y3  y1 )  x3 ( y1  y2 ) |
2
1
| x1 y2  x2 y3  x3 y1  x2 y1  x3 y2  x1 y3 |
2
13. The points P(x1, y1), Q(x2, y2) and R(x3, y3) are collinear if area of PQR = 0 i.e. x1 (y2 – y3) + x2(y3 – y1) +
x3(y1 – y2) = 0.

14. The area of a quadrilateral formed by the points P(x1, y1), Q(x2, y2), R(x3, y3) and S(x4, y4), taken in order is
given by :
1
| x1 y2  x2 y3  x3 y4  x4 y1  x2 y1  x3 y2  x4 y3  x1 y4 |
2
MATHEMATICS–X
CO-ORDINATE GEOMETRY
117
Some Properties of different type of quadrilaterals :
[To be used for proving a given quadrilateral to be a parallelogram, or a rectangle, or a rhombus (but not a
square), or a square]
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15. The quadrilateral ABCD is a parallelogram if and only if (a) AB = DC, AD = BC or (b) mid-points of BD
and AC are the same.
16. The quadrilateral ABCD is a rectangle if and only if (a) AB = CD, AD = BC and AC2 = AB2 + BC2 or (b) AB
= CD, AD = BC, AC = BD or (c) the mid-points of AC and BD are the same and AC = BD.
17. The quadrilateral ABCD is a rhombus (but not a square) if and only if (a) AB = BC= CD = DA and AC 
BD or (b) the mid points of AC and BD are the same and AB = AD, but AC  BD.
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18. The quadrilateral ABCD is a square if and only if (a) AB = BC = CD = DA and AC = BD or (b) the midpoints of AC and BD are the same and AC = BD, AB = AD.
ILLUSTRATIVE EXAMPLES
Example 1. Find the distance between the points :
(i) A (0, 0), B (–5, 12)
(iii) P (cos , –sin ), Q( – cos , sin )
(iv) P (a + b, a – b), Q (a – b, a + b)
Solution.
(i) AB  ( 5  0) 2  (12  0) 2
 52  122  25  144
 169  13 units
(ii)
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AB  ( 7  5) 2  [ 3  ( 8)]2
(ii) A (5, –8), B(–7, –3)
B
[NCERT]
( using distance formula)
 ( 12) 2  (5) 2  144  25  169  13 units
(iii) PQ  (  cos   cos ) 2  [sin   (  sin  )]2
 ( 2 cos  ) 2  (2 sin ) 2  4 cos 2   4 sin 2 
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 4(cos 2   sin 2 )  4  1  2 units
(iv) PQ  ( a  b  a  b) 2  ( a  b  a  b ) 2
 ( 2b) 2  (2b) 2  4b 2  4b 2  8b 2  2 2b units
Example 2. Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.
Solution. Let A(5, –2), B(6, 4) and C(7, –2) be the vertices of a ABC.

[NCERT]
AB  (6  5) 2  (4  2) 2  12  6 2  1  36  37
BC  (7  6) 2  ( 2  4) 2  12  ( 6) 2  1  36  37
AC  (7  5) 2  ( 2  2) 2  2 2  0 2  4  2
Since, AB = BC = 37 .
We observe that A, B, C are the vertices of an isosceles triangle.
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MATHEMATICS–X
Example 3. Find value(s) of y for which the distance between the points P (2, –3) and Q (10, y) is 10 units.
[NCERT]
Solution. Let P (2, –3) and Q (10, y) be the two given points such that PQ = 10.

PQ  (10  2) 2  ( y  3) 2  10

PQ 2  64  y 2  6 y  9  100
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(given)
 y2 + 6y – 27 = 0 y2 + 9y – 3y – 27 = 0
 y (y + 9) – 3 (y + 9) = 0  (y + 9) (y – 3) = 0
 y = – 9 or y = 3 Ans.
Example 4. Show that (1, –2), (3, 0), (1, 2) and (–1, 0) are the vertices of a square.
Solution. Let A(1, –2), B(3, 0), C(1, 2) and D(–1, 0) be the given points.
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AB  (3  1) 2  (0  2) 2  2 2  2 2  4  4  8  2 2

BC  (1  3) 2  (2  0) 2  ( 2) 2  2 2  4  4  8  2 2
CD  ( 1  1) 2  (0  2) 2  ( 2) 2  2 2  4  4  8  2 2
2
2
B
2
2
DA  (1  1)  ( 2  0)  2  ( 2)  4  4  8  2 2
diagonal AC  (1  1) 2  (2  2) 2  0 2  4 2  16  4
and diagonal BD  ( 1  3) 2  (0  0) 2  ( 4) 2  16  4
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Now, we have AB = BC = CD = DA and diagonal AC = diagonal BD.
 The given points are the vertices of a square.
Example 5. Show that the points (1, –1), (5, 2) and (9, 5) are collinear.
Solution. Let A(1, –1), B(5, 2) and C(9, 5) be the given points.
[CBSE 2006(C)]
Then, we have, AB  (5  1) 2  (2  1) 2  16  9  25  5
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BC  (5  9) 2  (2  5) 2  16  9  25  5
AC= (1  9) 2  ( 1  5) 2  64  36  100  10
and,
clearly,
AC = AB + BC.
Hence, A, B, C are collinear points.
Example 6. Find the coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, –2) and
(2, –2). Also, find its circum-radius.
Solution.
We know circumcentre of a triangle is equidistant from the vertices of a triangle.
Let A(8, 6), B(8, –2) and C(2, –2) be the vertices of a given triangle and let P(x, y) be the circumcentre
of this triangle. Then,
PA = PB = PC

PA2 = PB2 = PC2
Now, PA2 = PB2
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CO-ORDINATE GEOMETRY
119
 (x– 8)2 + (y – 6)2 = (x – 8)2 + (y + 2)2
 x2 + y2 – 16x – 12 y + 100 = x2 + y2 – 16x + 4y + 68
 16y = 32  y = 2.
and PB2 = PC2
 (x – 8)2 + (y + 2)2 = (x – 2)2 + (y + 2)2
 x2 + y2 – 16x + 4y + 68 = x2+ y2 – 4x + 4y + 8
 12x = 60  x = 5
So, the coordinates of the circumcentre P are (5, 2).
Also, circum-radius = PA = PB = PC
 (5  8) 2  (2  6) 2  ( 3) 2  ( 4) 2
 9  16  25 = 5 units Ans.
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Example 7. Find the co-ordinates of the points of trisection of the line joining the points (4, –1) and (–2, –3).
[NCERT]
Solution.
Let C and D be the points of trisection of AB
1 : 2
A (4, –1)

C
B
D
AC : CB = 1 : 2
So, coordinates of C
 1 (2)  2(4) 1 (3)  2(1) 

,

1 2
1 2


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 2  8 3  2   5 

,
   2,

3   3 
 3
Since D is the mid-point of BC
[ Using section formula]
5


 2  (2)  3  (3) 
 Coordinates of D = mid point of BC  
,

2
2




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B (–2, –3)
[ mid-point formula]
 7 
  0, 
 3 
Hence, points of trisection are
 –5 
 –7 
C  2,  and D  0,
 . Ans
3


 3 
Example 8. Find the ratio in which the join of (–3, 10) and (6, –8) is divided by (–1, 6).
Solution.
Let the given points be A(–3, 10), B(6, –8) and C(–1, 6).
k
: 1
A (–3, 10)
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[NCERT]
C(–1, 6)
CO-ORDINATE GEOMETRY
B (6, –8)
MATHEMATICS–X
Let the point C(–1, 6) divides AB in the ratio k : 1.





1 
(3)(1)  6( k )
k 1
6
and
6k  3
k 1
–k – 1 = 6k – 3
7k = 2
2
k
7
1 
(10)(1)  ( 8)( k )
k 1
8k  10
k 1
6k + 6 = – 8k + 10
14k = 4
2
k
7
[ Section formula]
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6




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 The required ratio is 2 :1 or 2 : 7 Ans.
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Example 9. If A(5, –1), B(–3, –2) and C(–1, 8) are the vertices of ABC, find the length of medians through
A and the coordinates of the centroid.
[CBSE 2006(C)]
Solution. Let AD be the median through the vertex A of ABC. Then, D is the mid-point of BC. So, the
coordinates of D are  3  1 , 2  8  i.e. (–2, 3).
2 
 2
A(5, –1)
G
B
C(–1, 8)
B( –3, –2)

D (–2, 3)
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AD  (5  2) 2  ( 1  3) 2  49  16  65 units
Let G be the centroid of ABC. Then the coordinate of G are
 5  (3)  (1) 1  (2)  8 
1 5
,

 i.e.  3 , 3  Ans.
3
3




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Example 10. Find the area of the triangle whose vertices are (–5, –1), (3, –5), (5, 2).
Solution. Let A(–5, –1), B(3, –5), C(5, 2) be the given points.
[NCERT]
1
Area of ABC =   | x1 ( y2  y3 )  x2 ( y3  y1 )  x3 ( y1  y2 ) |
2
1
 | 5(5  2)  3(2  (5))  5(1  (5)) |
2
1
1
 | 35  21  20 |  | 76 |  38
2
2
 ar (ABC) = 38 sq. units Ans.
Example 11. Find the area of the quadrilateral whose vertices taken in order are (–4, –2), (–3, –5), (3, –2),
(2, 3).
[NCERT]
Solution. Let A(–4, –2), B(–3, –5), C(3, –2) and D (2, 3) be the given points.
Now, area of quad. ABCD = ar (ABD) + ar (BCD)
For ar (ABD) : A(–4, –2), B(–3, –5), D(2, 3)
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CO-ORDINATE GEOMETRY
121
ar (ΔABD) 
1
| x1 ( y2  y3 )  x2 ( y3  y1 )  x3 ( y1  y2 ) |
2

1
| 4(5  3)  (3)(3  (2))  2(2  (5)) |
2

1
| (4)(8)  (3)(5)  2(3) |
2

1
1
23
| 32  15  6 |  | 23 | 
2
2
2
D(2, 3)
B(–3, –5)
23
 ar (ΔABD) = sq. units.
2
For ar (BCD) : B(–3, –5), C(3, –2), D(2, 3)
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1
ar (BCD) = | x1 ( y2  y3 )  x2 ( y3  y1 )  x3 ( y1  y2 ) |
2

1
| 3(2  3)  3(3  (5))  2(5  (2)) |
2

1
| (3)(5)  3(8)  2(3) |
2

1
1
33
|15  24  6 | | 33 |
2
2
2

B
33
sq. units
2
area of quad. ABCD = ar (ABD) + ar (BCD)
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 ar (BCD) 
23 33
 sq. units
=
2 2
=
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C(3, –2)
A(–4, –2)
56
sq. units = 28 sq. units Ans.
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Example 12. Prove that the points (2, –2), (–3, 8) and (–1, 4) are collinear.
Solution. Let  be the area of the triangle formed by given three points A(2, –2), B(–3, 8) and C(–1, 4).
We have,
1
| x1 ( y2  y3 )  x2 ( y3  y1 )  x3 ( y1  y2 ) |
2


1
| 2(8  4)  (3)(4  (2))  (1)(2  8) |
2

1
| 2(4)  (3)(6)  (1)(10) |
2

1
0
| 8  18  10 |   0
2
2
Hence, the given points are collinear.
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MATHEMATICS–X
PRACTICE EXERCISE
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Question based on distance formula :
1. Find the distance between the points:
2.
(i) A (–3, –2) and B (–6, –7)
(ii) P(a, 0) and Q(0, b)
(iii) A (–m, –n) and B (m, n)
(iv) R( 3  1, 1) and S (0, 3)
(v) M (3  3, 3  3) and N (0, 0)
(vi) A (a sin , – b cos ) and B (–a cos , b sin )
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Find the value of p if the distance between the points (3, p) and (4, 1) is 10 units.
3. Find the value of y if the distance between the points (2, –3) and (10, y) be 10 units.
4. For what value of x, the distance between P(x, 7) and Q(–2, 3) is 4 5 units.
5. If the points (3, 2) and (2, –3) are equidistant from (a, b) show that a + 5b = 0.
6. Find the coordinates of the points on the x-axis which are at a distance of 5 units from the point (5, 4).
7. Find the coordinates of the points on the y-axis which are at a distance of 13 units from the point (12, 9).
8. Find the point on x-axis which is equidistant from the points (–4, 6) and (5, 9).
B
9. Find the point on the y-axis which is equidistant from the points (3, 2) and (–5, –2).
10. A point P is at a distance of 13 units from the point (5, 4). Find the coordinates of P, if its ordinate is
thrice of its abscissa.
11. Show that the following points are the vertices of a right triangle.
(i) A(–2, 3), B(8, 3) and C(6, 7)
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(ii) A(8, 4), B(5, 7) and C(–1, 1)
12. Show that the following points are the vertices of isosceles triangle.
(i) A(10, –18), B(–5, 2) and C(3, 6)
(ii) P(–2, –2), Q(–1, 2) and R(3, 1)
13. Show that the following points are the vertices of isosceles right triangle :
(i) A(6, 4), B(3, 0) and C(–1, 3)
(ii) P(0, 0), Q(5, 5) and R(–5, 5)
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[CBSE 2005]
14. Show that the following points are the vertices of an equilateral triangle :
(i) A(1, 1), B(–1, –1) and C(– 3 , 3 )
(ii) P(2, 4), Q(2, 6) and R( 2  3, 5)
15. Show that the following points are vertices of a scalene triangle :
(i) P(1, 0), Q(3, 5) and R(–7, 3)
(ii) A(–1, 2), B(–5, 9) and C(3, 0)
16. Show that the following points are the vertices of a parallelogram :
(i) A(–2, –1), B(1, 0), C(4, 3) and D(1, 2)
(ii) P(1, –2), Q(3, 6), R(5, 10) and S(3, 2)
17. Show that the following points are the vertices of a rectangle :
(i) A(2, –2), B (8, 4), C(5, 7) and D(–1, 1)
(ii) P(0, –1), Q(–2, 3), R(6, 7) and S (8, 3)
18. Show that the following points are the vertices of a square :
(i) A(4, 3), B(6, 4), C(5, 6) and D(3, 5)
(ii) P(0, –1), Q(2, 1), R(0, 3) and S(–2, 1)
19. Show that the following points are the vertices of a rhombus :
(i) A (2, –1), B(3, 4), C(–2, 3) and D(–3, –2)
(ii) P(1, –2), Q(2, 3), R(–3, 2) and S(–4, –3)
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CO-ORDINATE GEOMETRY
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20. Show that the given points are collinear :
(i) A (1, 1), B(3, –3) and C(–2, 7)
(ii) P(–1, 3), Q(5, 2) and R(–13, 5)
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21. Show that the points A(2, –1), B(3, 4), C(–2, 3) and D(–3, –2) forms a rhombus but not a square. Find the
area of the rhombus also.
22. Find the value of x such that PQ = QR where the coordinates of P, Q and R are (6, –1), (1, 3) and (x, 8)
respectively.
23. If the points (5, 1) and (–1, 5) are equidistant from the points (x, y), show that 3x = 2y.
[CBSE 2005]
24. If P(x, y) is equidistant from A(a + b, b – a) and B(a – b, a + b), show that bx = ay.
25. The centre of a circle is (3k + 1, 2k – 1). If the circle passes through the point (–1, –3) and the length of
its diameter be 20 units, find the value of k.
Question based on section formula :
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26. Find the coordinates of the point which divides the line segment joining the points (4, –7) and (–5, 6)
internally in the ratio 7 : 2.
27. Find the coordinate of point P which divides the join of A(6, 5) and B(9, 2) in the ratio 1 : 2.
28. A and B are the points (1, 2) and (2, 3). Find the coordinates of point C on the line segment AB such that
3AC = 4 BC.
B
29. Find the points of trisection of the line segment joining the points :
(i) (3, –2) and (–3, –4)
(ii) (1, –2) and (–3, 4)
30. Three consecutive vertices of a parallelogram are (–2, –1), (1, 0) and (4, 3). Find the coordinate of the
fourth vertex.
31. Find the ratio in which the line-segment joining the points (6, 4) and (1, –7) is divided internally by x-axis.
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32. Find the ratio in which the line segment joining the points (7, 3) and (–4, 5) is divided internally by y-axis.
33. Find the ratio in which the point (11, 15) divides the line segment joining the points (15, 5) and (9, 20).
34. Determine the ratio in which the line 3x + y – 9 = 0 divides the segment joining the points (1, 3) and (2, 7).
35. In what ratio the point (–3, k) divides the line segment joining the points (–5, –4) and (–2, 3). Hence, find
the value of k.
36. Find the ratio in which the line segment joining (–2, –3) and (5, 6) is divided by (i) x-axis (ii) y-axis. Also,
find the coordinates of the point of division in each case.
37. If the coordinates of the mid-points of the sides of a triangle are (1, 2), (0, –1) and (2, –1). Find the
coordinates of its vertices.
38. The coordinates of the middle points D, E, F of the sides BC, CA and AB respectively of a ABC are
(–3, 2), (5, –7) and (11, 7) respectively, find the coordinates of the vertices A, B and C.
39. The line segment joining the points (–6, 8) and (8, –6) is divided into four equal parts. Find the coordinates of the point of section.
40. The line segment joining the points (3, –4) and (1, 2) is trisected at the ponts P and Q. If the coordinates
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5 
of P and Q are (p, –2) and  , q  respectively, find the values of p and q.
3 
[CBSE 2005]
41. The line joining the points (2, 1) and (5, –8) is trisected at the points P and Q. If P lies on the line 2x – y
+ k = 0, find the value of k.
[CBSE 2005]
42. If the points (10, 5), (8, 4) and (6, 6) are the mid-points of the sides of a triangle, find its vertices.
[CBSE 2006]
124
CO-ORDINATE GEOMETRY
MATHEMATICS–X
43. Find the centre of a circle, the end points of whose one diameter are (–3, –1) and (5, 8).
44. Find the lengths of medians of a ABC having the vertices A(5, 1), B(1, 5) and C(–3, –1).
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45. Find the coordinates of the point of intersection of medians of ABC whose vertices are A(–7, 5), B(–
1, –3) and C(5, 7).
46. Find the centroid of a triangle whose vertices are :
(i) (–2, 3), (2, –1), (4, 0)
(ii) (4, –8), (–9, 7), (18, 13)
47. Two vertices of a triangle are (1, 2) and (3, 5) and its centroid is at the origin. Find the coordinates of the
third vertex.
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5 1
48. A (3, 2) and B(–2, 1) are two vertices of ABC whose centroid G has the coordinates  ,   . Find the
 3 3
coordinates of the third vertex C of the triangle.
[CBSE 2004]
49. If the coordinates of the mid-points of the sides of a triangle are (1, 1), (2, –3) and (3, 4). Find its centroid.
50. If the coordinates of the mid-points of the sides of a triangle are (4, –3), (4, 5) and (–2, 3). Find the
coordinates of its centroid.
Questions based on Area of a Triangle
51. Find the area of the triangle whose vertices are :
B
(i) A (1, –1), B(–4, 6) and C(–3, –5)
(ii) P (4, 2), Q(4, 5) and R (–2, 2)
(iii) A (1, 2), B(–2, 3) and C(–3, –4)
(iv) P (5, 2), Q(4, 7) and R (7, –4)
52. Using the concept of area of the triangle, prove that the following points are collinear :
(i) (1, 4), (3, –2) and (–3, 16)
(ii) (2, 5), (4, 6) and (8, 8)
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53. (i) For what value of k, the points (k, –1), (5, 7) and (8, 11) are collinear?
(ii) For what value of k are the points (k, 2 – 2k), (–k + 1, 2k) and (–4 – k, 6 – 2k) lie on a straight line?
54. The vertices of some triangles are given below alongwith their areas. Find the value of a.
Vertices
(i)
(2, 3), (6, –2), (–2, a)
(ii)
(3, 8), (4, a), (5, –2)
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Area
6
8
55. Find the area of the quadrilateral, the coordinates of whose vertices are :
(i) (–3, 2), (5, 4), (7, –6) and (–5, –4)
(ii) (–4, –2), (–3, –5), (3, –2) and (2, 3)
(iii) (–5, 7), (–4, –5), (–1, –6) and (4, 5)
(iv) (6, 9), (7,4), (4, 2) and (3, 7)
56. A (–7, 5), B(5, –7) and C(15, 9) are vertices of a triangle ABC. If D, E and F are the mid-points of sides AB,
BC and AC respectively. Show that : area of ABC = 4 (area of DEF)
57. If the points (a, 0), (0, b) and (1, 1) are collinear, show that
1 1
 1.
a b
58. If the area of the quadrilateral whose angular points, taken in order, are (1, 2), (–5, 6), (7, –4), (p, –2) be
zero, find the value of p.
59. Triangle ABC has vertices A(4, –6), B(3, –2) and C(5, 2). Show that the median AD divides the triangle
ABC into two triangles of equal areas.
60. Show that the area of ABC, where A(p, p – 2), B(p + 2, p + 2) and C(p + 3, p) does not depend on the
value of p.
MATHEMATICS–X
CO-ORDINATE GEOMETRY
125
Some More Problems
61. ABCD is a square with the opposite angular points A(3, 4) and C(1, –1). Find the coordinates of B and D.
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62. Find the coordinates of the circumcentre of a triangle whose vertices are A(5, 1), B(11, 1) and C(11, 9).
63. An equilateral triangle has two vertices at the points (0, 0) and (3, 3) . Find the coordinates of the third
vertex.
64. Find the coordinates of the points equidistant from three given points A(5, 1), B(–3, –7) and C(7, –1).
[CBSE 2006]
65. Find the coordinates of a point whose distance from (3, 5) is 5 units and that from (0, 1) is 10 units.
66. A circle passes through the points A(3, 1), B(1, –3) and C(6, –8). Find the coordinates of the centre of the
circle.
67. If a is the distance between the poins (2, 3) and (1, –1), b is the distance between (0, 2) and (13, 14) and
c is the distance between (–1, 0) and (2, –5). Prove that no triangle can be constructed with sides
a, b and c.
68. The coordinates of a vertex of a triangle are (2, 5) and the coordinates of the mid-points of the sides
passing through this vertex are (8, 0) and (9, 3). Find the coordinates of the remaining vertices.
69. If the coordinates of two points A and B are (3, 4) and (5, –2) respectively. Find the coordinates of any
point P, if PA = PB and area of PAB = 10 square units.
70. The area of a triangle is 5 square units. Two of its vertices are (2, 1) and (3, –2). The third vertex lies on
y = x + 3. Find the third vertex.
B
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HINTS TO SELECTED QUESTIONS
10. Let the co-ordinates of P be (x, 3x). Then, (x – 5)2 + (3x – 4)2 = ( 13)2 .
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30. Since diagonals of a parallelogram bisect each other, O is the mid-point of AC and BD.
2  4 1  3 
Now co-ordinates of O are 
,
  (1, 1)
2 
 2
D(x, y)
A(–2, –1)
 x 1 y  0 
,
Also, 
  (1, 1).
2 
 2
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C(4, 3)
O
B(1, 0)
3x + y – 9 = 0
34. Let the required ratio be k : 1.
11  2  k
3 1  7  k
Using section formula, a 
,b
k 1
k 1
(1, 3)
k : 1
(a, b)
(2, 7)
Now, point (a, b) lies on the line 3x + y – 9 = 0, so it will
statisfy its equation. Find the value of k now.
37. Let co-ordinates of vertices be A(x1, y1), B(x2, y2) and C(x3, y3). Since, D is the mid point of AB, using midpoint theorem, we have
x1  x2
y  y2
 1 and 1
2
2
2

126
x1 + x2 = 2
...(1), and y1 + y2 = 4
...(4)
CO-ORDINATE GEOMETRY
MATHEMATICS–X
Similarly, we will get
x2 + x3 = 0
...(2) and, y2 + y3 = – 2
...(5)
x1 + x3 = 4
...(3) and, y1 + y3 = – 2
...(6)
A(x1, y1)
adding (1), (2), (3) we get
2(x1 + x2 + x3) = 6  x1 + x2 + x3 = 3
B(x2, y2)
...(7)
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F(2, –1)
D(1, 2)
C(x3, y3)
E(0, –1)
Substracting (1), (2) and (3) from eqn. (7) respectively, we get
x3 = 1, x1 = 3 and x2 = – 1
Similarly using eqn. (4), (5), (6) we will get the values of y1, y2 and y3.
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53. (ii) The points A(k, 2 – 2k), B(–k + 1, 2k) and C(–4 – k, 6 – 2 k) lie on a straight line if ar (ABC) = 0
1
 | k (2k  6  2k )  (k  1)(6  2k  2  2k )  (4  k )(2  2k  2 k ) | 0
2
 | k (4k  6)  (k  1)(4)  (4  k )(2  4k ) |  0
1
.
2
61. Let B(x, y) be the unknown vertex.
 2 k 2  k  1  0  k  1 or
Now,
AB = BC  AB2 = BC2
B
 (x – 3)2 + (y – 4)2 = (x – 1)2 + (y + 1)2
 4x + 10y – 23 = 0
 x
23  10 y
4
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In right ABC, AB2 + BC2 = AC2
...(1)

(x – 3)2 + (y – 4)2 + (x – 1)2 + (y + 1)2 = (3 – 1)2 + (4 + 1)2

x2 + y2 – 4x – 3y – 1 = 0
Putting value of x from (1) in (2), we get
4y2 – 12y + 5 = 0
y
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1
5
or,
2
2
Putting values of y in (1), we get x 
D
A(3, 4)
C(1, –1)
B(x, y)
...(2)
9
1
and x  
2
2
9 1
 1 5
 Required vertices of square are  ,  and   , 
2 2
 2 2
62. Circumcentre is equidistant from the vertices of the triangle.
A(5, 1)
Let the co-ordinates of circumcentre O be (x, y).

OA = OB = OC

OA2 = OB2 = OC2

(x – 5)2 + (y – 1)2 = (x – 11)2 + (y – 1)2 = (x – 11)2 + (y – 9)2
O(x, y)
B(11, 1)
C
B(11, 9)
Now, OA2 = OB2 and OB2 = OC2 will give two linear equations in x and y. Solve them together to get the
coordinates of O.
MATHEMATICS–X
CO-ORDINATE GEOMETRY
127
63. Let the third vertex be C(x, y).
Then, AB = BC = AC

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AB2 = BC2 = AC2
C(x, y )
Now, AB2 = (3 – 0)2 + ( 3  0)2  12
3) 2 = x2 + y2 – 6x – 2 3y + 12
BC2 = (x – 3)2 + (y –
A(0, 0)
AC2 = x2 + y2

AB2 = BC2  x2 + y2 = 12
...(1)
and
BC2 = AC2  6x + 2
...(2)
3 y  12
Solve eqn. (1) and (2) together to get x and y.
69. Let the co-ordinates of P be (x, y).
B(3, 3)
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P(x, y)
PA = PB  PA2 = PB2
2
2
2
2
 (x – 3) + (y – 4) = (x – 5) + (y + 2)
 x – 3y – 1 = 0
Now, ar (PAB) = 10
...(1)
A(3, 4)

1
| x[4  (2)]  3(2  y )  5( y  4) |  10
2

| 6 x  2 y  26 |  20  6 x  2 y  26   20

6x + 2y – 46 = 0
or,
6x + 2y – 6 = 0
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B
...(2)
...(3)
B(5, – 2)
Solving (1) and (2); (1) and (3) we get the coordinates of P are (7, 2) or (1, 0).
70. Let co-ordinates of third vertex be (a, a + 3) as this vertex lies on the line y = x + 3. Now use area formula
and solve.
MULTIPLE CHOICE QUESTIONS
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Mark the correct alternative in each of the following :
1. The distance between the points (a, b) and (–b, a) is :
2.
3.
4.
5.
128
(a) 2 a 2  b 2
(b) a 2  b2
(c) 2( a 2  b 2 )
(d) none of these
If the distance between the points (1, 0) and (4, a) is 5, then value of a is :
(a) ± 4
(b) 4
(c) –4
(d) none of these
A line segment is of length 10 units. If the coordinates of its one end are (2, –3) and the abscissa of the
other end is 10, then its ordinate is :
(a) 9, 6
(b) 3, –9
(c) –3, 9
(d) 9, –6
The value of k for which the points (k, –1), (2, 1) and (4, 5) are collinear is:
(a) 0
(b) 1
(c) –1
(d) 2
The point of intersection of y-axis and the perpendicular bisector of the line segment joining the points
(3, 6) and (–3, 4) is :
(a) (0, 4)
(b) (0, –4)
(c) (0, 5)
(d) (0, –5)
CO-ORDINATE GEOMETRY
MATHEMATICS–X
6. The vertices of a triangle are (–2, 0), (2, 3) and (1, –3). The triangle is
(a) equilateral
(b) isosoceles
(c) right
(d) scalene
7. The coordinates of the point which divide the join of (–1, 7) and (4, –3) in the ratio 2 : 3 is :
(a) (1, 2)
(b) (1, 3)
(c) (–2, 3)
(d) none of these
8. The ratio in which the join of (1, –5) and (–4, 5) is divided by x-axis is :
(a) 1 : 1
(b) 1 : 2
(c) 2 : 3
(d) 3 : 5
9. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, then the value of x and y
is :
(a) x = 3, y = – 6
(b) x = – 6, y = 3
(c) x = 6, y = 3
(d) x = 6, y = –3
10. The area of a rhombus (in sq. units) whose vertices taken in order are (3, 0), (4, 5), (–1, 4) and (–2, –1) is:
(a) 12
(b) 24
(c) 48
(d) none of these
11. The vertices of a triangle are (3, –5), (–7, 4) and (10, –2). The coordinates of its centroid is :
(a) (–2, 1)
(b) (–2, –1)
(c) (2, –1)
(d) (1, 2)
12. If the centroid of a triangle formed by (7, x), (y, –6) and (9, 10) is at (6, 3), then (x, y) =
(a) (4, 5)
(b) (5, 4)
(c) (–5, –2)
(d) (5, 2)
13. The area of the triangle formed by (a, b + c), (b, c + a) and (c, a + b) is :
(a) abc
(b) a + b + c
(c) a2b2c2
(d) 0
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7 
 3 
 9
14. The value of k for which the points  2,  ,  3,
 and  k ,  are collinear is :
2
2

 

 2
5
5
(a) 4
(b) 5
(c)
(d)
2
2
15. If the area of the triangle formed by the points (x, 2x), (–2, 6) and (3, 1) is 5 sq. units, then value of x is :
(a)
2
or 2
3
(b)
3
or 2
5
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B
(c) 3
(d) 5
VERY SHORT ANSWER TYPE QUESTIONS (1 MARK QUESTIONS)
1.
2.
3.
4.
5.
6.
7.
8.
9.
What is the distance of the point (–6, 8) from origin?
What is the ordinate of any point on x-axis?
What is the distance of the point A(3, –4) from y-axis?
What is the distance between the points (1, –2) and (–3, 2)?
If the distance between the points (x, 0) and (5, 8) is 10 units, find the value(s) of x.
Find the mid-point of the line segment joining the points P(–3, 4) and Q (9, –6).
What are the coordinates of the centroid of triangle formed by points A(–2, 4), B(7, –3) and C(1, 5).
Find the coordinates of points which divides line joining (–4, 0) and (0, 6) in the ratio 1 : 3.
Point A(3, –4) lies on circle of radius 5 cm with centre (0, 0). Write the coordinates of the other end of the
diameter whose one end is A.
Find the third vertex of a triangle if two of its vertices are (3, –6) and (–5, 2) and its centroid is at the point
(2, 0).
What is the area of triangle formed by the points (–2, 0), (4, 0) and (2, 3).
Find the coordinates of fourth vertex of the rectangle formed by the points (0, 0), (3, 0) and (0, 5).
What is the area of ABC, if points A, B and C are collinear?
In what ratio is the line joining the points P(7, 7) and Q(–4, 4) is divided by (0, –1)?
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10.
11.
12.
13.
14.
15. C is point on the perpendicular bisector of AB. What is the relation between A, B, C?
MATHEMATICS–X
CO-ORDINATE GEOMETRY
129
PRACTICE TEST
M.M : 30
Time : 1 hour
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General Instructions :
Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.
1. Check whether (5, –2), (6, 4) and (7, 2) are the vertices of an isosceles triangle.
2. Show that the points (–3, 2), (1, –2) and (9, –10) are collinear.
3. Find the length of median AD of a triangle whose vertices are A(–1, 3), B(1, –1) and C(5, 1).
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4. What point on x-axis is equidistant from the points (–3, 4) and (7, 6)?
5. In what ratio does the line segment joining the points (6, 4) and (1, –7) is divided internally by the axis of
x?
6. Three vertices of a parallelogram, taken in order are (3, 1), (2, 2) and (–2, 1) respectively. Find the
coordinates of fourth vertex.
7. Find the coordinates of the point of trisection of the line segment AB whose end points are A (2, 1) and
B (5, –8).
8. The coordinates of the centroid of a triangle are (1, 3) and the two vertices are (8, 5) and (–7, 6). Find the
third vertex of the triangle.
B
9. If (3, 2), (4, 4) and (1, 3) are the mid-points of the sides of a triangle, find the coordinates of the vertices
of the triangle.
10. Find the area of the quadrilateral, the coordinates of whose vertices are (1, 2), (6, 2), (5, 3) and (3, 4).
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ANSWERS OF PRACTICE EXERCISE
1. (i)
34 units
(ii)
(iv) 2 2 units
2. p = 4 or –2
a 2  b 2 units
(v) 2 6 units
3. y = 3 or –9
(iii) 2 m 2  n 2 units
(vi)
a 2  b 2 (sin   cos )
4. x = 6 or –10
6. (8, 0) and (2, 0)
8. (3, 0)
9. (0, –2)
 7 21 
10. (2, 6) or  , 
5 5 
22. x = – 3 or 5
25. k = 2,
30. (1, 2)
 11 18 
28. C  , 
7 7
31. 4 : 7
10 
 8  
 1   5 
29. (i)  1,
 ,  1,
 (ii)   , 0  ,  , 2 
3 
 3  
 3   3 
32. 7 : 4
33. 2 : 1
34. 3 : 4
35. 2 : 1; k = 2/3
1 
36. (i) 1 : 2,  , 0  (ii) 2 : 5;
3 
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7. (0, 14) and (0, 4)
21. 24 sq. units.
27. P(7, 4)
37. (1, –4), (3, 2) and (–1, 2)
39.
130
 5 9 
 9 5 
 ,  ; (1, 1) ;  ,

2
2


2 2 
46
13
28 

26.  3,

9 

 3 
 0,

 7 
38. (19, –2), (3, 16), (–9, –12)
40. p 
CO-ORDINATE GEOMETRY
7
,q0
3
41. k = – 8 or – 13
MATHEMATICS–X
42. (8, 7), (12, 3), (4, 5)
 7
43.  1, 
 2
44.
45. (–1, 3)
4
46. (i)  ,
3
 2
49.  2, 
 3
 5
50.  2, 
 3
AD  37, BE  5, CF  2 13
47. (–4, –7)
48. (4, –4)
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2
 13 
 (ii)  , 4 
3
 3 
51. (i) 24 sq. units (ii) 9 sq. units (iii) 11 sq. units. (iv) 2 sq. units
53. (i) k = – 1 (ii) k  1 or
1
2
55. (i) 80 sq. units (ii) 28 sq. units (iii) 72 sq. units (iv) 17 sq. units
61.
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54. (i) a = 5 or 11 (ii) a = – 5 or 11
9 1
 1 5 
 ,  and  , 
2
2


 2 2
62. (8, 5)
64. (2, –4)
65. (6, 9)
66. (6, –3)
69. (7, 2) or (1, 0)
 7 13   3 3 
70.  ,  or  , 
 2 2   2 2
B
58. p = 3
63. (0, 2 3) or (3,  3)
68. (14, –5) and (16, 1)
ANSWERS OF MULTIPLE CHOICE QUESTIONS
1. (c)
6. (d)
11. (c)
2. (a)
7. (b)
12. (d)
3. (b)
8. (a)
13. (d)
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4. (b)
9. (c)
14. (b)
5. (c)
10. (b)
15. (a)
ANSWERS OF VERY SHORT ANSWER TYPE QUESTIONS
1. 10 units
2. 0
M
A
6. (3, –1)
7. (2, 2)
11. 9 sq. units 12. (3, 5)
3. 3 units
4. 4 2 units
5. –1 or 11
3

8.  3, 
2

9. (–3, 4)
10. (8, 4)
13. zero
14. 4 : 7
15. AC = BC
ANSWERS OF PRACTICE TEST
3. 5 units
7. (3, –2) and (4, –5)
10.
4. (3, 0)
8. (2, –2)
5. 4 : 7
6. (–1, 0)
9. (0, 1), (6, 3) and (2, 5)
11
sq.units
2
MATHEMATICS–X
CO-ORDINATE GEOMETRY
131