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Coordinate Geometry Ex- 7.1

 Hi Friends and Champs,

Before proceeding to the next exercise , let's recall  the position of a point in a xy plane , which you have already read in class ix .

Abscissa - or x-coordinate , the distance of a point form y-axis. and 

ordinate - or y-coordinate, the distance of a point from x- axis .

Distance between two points in a plane xy. 

Lets two points are P(x1,y1 ) and Q (x2,y2), then distance between them is :

PQ =  √ [(x2 - x1)2 + (y2 - y1)2]



Exercise - 7.1


 1. Find the distance between the following pairs of points : 
 (i) (2, 3), (4, 1) (ii) (– 5, 7), (– 1, 3) (iii) (a, b), (– a, – b) 

Solution: 
(i) Distance = √ [(x2 - x1)2 + (y2 - y1)2]
                        => √ [(4 - 2)2 + (1 - 3)2]

                        =>√ [(22 + (- 2)2]

                        =>√ [(8]

                        => 2√2

(ii) (– 5, 7), (– 1, 3)
 Distance = √ [(x2 - x1)2 + (y2 - y1)2]
                        => √ [(-1 - (-5))2 + (3 - 7)2]

                        =>√ [(42 + (- 4)2]

                        =>√ [(32)]

                        => 4√2
(iii) (a, b), (– a, – b) 
Distance = √ [(x2 - x1)2 + (y2 - y1)2]
                        => √ [(-a - a)2 + (-b - b)2]

                        =>√ [(4a2 + 4b2)]

                        => 2√ (a2 + b2)
                        
 2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2. 

Solution:
Distance = √ [(x2 - x1)2 + (y2 - y1)2]
                        => √ [(36 - 0)2 + (15 - 0)2]

                        =>√ [(1296 + 225)]

                        => √ 1521
                        => 39   ( question is same for section 7.2 also ).

 
3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear. 

Solution: Let's the point are P(1, 5), Q(2, 3) and R(– 2, – 11)

PQ = √ [(2 - 1)2 + (3 - 5)2]
PQ = √ [(1)2 + (- 2)2]
PQ = √ 5

QR = √ [(-2 - 2)2 + (-11 - 3)2]
QR = √ [(-4)2 + (- 14)2]
QR = 2√ 53

PR = √ [(-2 - 1)2 + (-11 - 5)2]
PR = √ [(-3)2 + (- 16)2]
PR = √ 265

Clearly seen the sum of any two distance pair  is not equal to third , so the points are not collinear.

4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle. 

Solution:
Let the points are P(5, – 2), Q(6, 4) and R(7, – 2).
PQ = √ [(6 - 5)2 + (4 - (-2))2]
PQ = √ [(1)2 + (6)2]
PQ = √ 37 =6.08

QR = √ [(7 - 6)2 + (-2 - 4)2]
QR = √ [(1)2 + (-6)2]
QR = √ 37 =6.08

PR = √ [(7 - 5)2 + (-2 -(-2)2]
PR = √ [(2)2 + (0)2]
PR = √ 4 = 2

Since the sum of any two sides are more than third one , so the points make a triangle. Also 2 sides are of same length , the triangle is an isosceles triangle.

 5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct. 



Soluion: Points are A(3,4), B(6,7) , C(9,4), D(6,1)

AB = √ [(6 - 3)2 + (7 - 4)2]
AB = √ (9 + 9)
AB = 3√2

BC = √ [(9 - 6)2 + (4 - 7)2]
BC = √ (9 + 9)
BC = 3√2

CD = √ [(6 - 9)2 + (1 - 4)2]
CD = √ (9 + 9)
CD = 3√2

DA = √ [(3 - 6)2 + (4 - 1)2]
DA = √ (9 + 9)
DA = 3√2

AC = √ [(9 - 3)2 + (4 - 4)2]
AC = √ (36 + 0)
AC = 6

BD = √ [(6 - 6)2 + (1 - 7)2]
BD = √ (0 + 36)
BD = 6

=> all the sides are equal and also both the diagonals are equal , 
and AB2 + BC2 = AC2
which prove ABC is a right angle , so satisfy all the property of square . So champa is right. 

 6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: 
 (i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0) 
 (ii) (–3, 5), (3, 1), (0, 3), (–1, – 4) 
 (iii) (4, 5), (7, 6), (4, 3), (1, 2) 

Solution:
(i) A(– 1, – 2), B(1, 0), C(– 1, 2), D(– 3, 0) 

AB = √ [(1 - (-1))2 + (0 - (-2))2]
AB = √ (4 + 4)
AB = 2√2

BC = √ [(-1 - 1)2 + (2 - 0)2]
BC = √ (4 + 4)
BC = 2√2

CD = √ [(-3 - (-1))2 + (0 - 2)2]
CD = √ (4 + 4)
CD = 2√2

DA = √ [(-1 - (-3))2 + (-2 - 1)2]
DA = √ (4 + 4)
DA = 2√2

AC = √ [(-1 - (-1))2 + (2 - (-2))2]
AC = √ (0 + 16)
AC = 4

BD = √ [(-3 - 1)2 + (0 - 0)2]
BD = √ (16 + 0)
BD = 4

and AB2 + BC2 = 8 + 8 =16 = AC2

so the given points make a square Parellogram.

(ii) A(–3, 5), B(3, 1), C(0, 3), D(–1, – 4) 

AB = √ [(3 - (-3))2 + (1 - 5)2]
AB = √ (36 + 16)
AB = √52 = 2√13

BC = √ [(0 - 3)2 + (3 - 1)2]
BC = √ (9 + 4)
BC = √13

CD = √ [(-1 - 0)2 + (-4 - 3)2]
CD = √ (1 + 49)
CD = √50 = 5√2

DA = √ [(-3 - (-1))2 + (5 - (-4))2]
DA = √ (4 + 81)
DA = √85

AC = √ [(0 - (-3))2 + (3 - 5)2]
AC = √ (9 + 4)
AC = √13

BD = √ [(-1 - 3)2 + (-4 - 1)2]
BD = √ (16 + 25)
BD = √41

No sides and also diagonals are equal ,

so the given points do not make a Parellogram..

(iii) A(4, 5), B(7, 6), C(4, 3), D(1, 2) 
AB = √ [(7 - 4)2 + (6 - 5)2]
AB = √ (9 + 1)
AB = √10

BC = √ [(4 - 7)2 + (3 - 6)2]
BC = √ (9 + 9)
BC = 3√2

CD = √ [(1 - 4)2 + (2 - 3)2]
CD = √ (9 + 1)
CD = √10

DA = √ [(4 - 1)2 + (5 - 2)2]
DA = √ (9 + 9)
DA = 3√2

AC = √ [(4 - 4)2 + (3 - 5)2]
AC = √ (0 + 4)
AC = 2

BD = √ [(1 - 7)2 + (2 - 6)2]
BD = √ (36 + 16)
BD = √52 = 2√13

opposite sides are equal ,so the given points make a Parellogram..


7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9). 

Solution :
Let the point P (x, 0) is at equidistance from both the given points.
then, √ [(x - 2)2 + (0 - (-5))2] = √ [(-2 - x)2 + (9 - 0)2]
=> (x - 2)2 + 25 = (-2 - x)2 + 81            --square at both sides
=> x2 - 4x + 4 +25 = 4 + 4x + x2 + 81
=>  -4x + 29 = 4x + 85
=> 8x = -56
=> x = -7

so the point is ( -7 , 0 )

8. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Solution :

PQ = 10
=> √ [(10 - 2)2 + (y - (-3))2]  = 10
=> [ 82 + (y +3))2] = 100                     ----square at both sides
=> 64 + y2 + 6y + 9 = 100
=> y2 + 6y - 27 = 0
=> y2 + 9y - 3y - 27 = 0
=> y( y + 9 ) - 3(y+9) =0
=>(y + 9)(y - 3) =0

So the points are ( 10, -9) and ( 10 , 3) are at equidistance from the point P .

9. If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.

Solution:
Since Q is at euidistance from the points P and R , therefore,
√ [(5 - 0)2 + (-3 - 12] = √ [(x - 0)2 + (6 - 1)2]
=> ( 25 + 16 ) = ( x2 + 25 )                 ---- Square at both sides
=> 41 = x2 + 25
=> x2 = 16
=> x = ± 4

QR = √ [(± 4 - 0)2 + (6 - 1)2]
QR = √(16 + 25)
QR = √41

PR = √ [(± 4 - 5)2 + (6 - (-3))2]

PR = √ [ (4-5)2 + 81 ] =  √82 at x = +4
and
PR = √ [ (-4-5)2 + 81 ] =  9√2 at x = -4
  
10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4)

Solution:
Since ( x , y) is at equidistance from both te points , therefore

√ [(x - 3)2 + (y - 6)2] = √ [(x + 3)2 + (y - 4)2]
=>(x - 3)2 + (y - 6)2 = (x + 3)2 + (y - 4)2            --square at both sides
=> x2 - 6x + 9 + y2 - 12y + 36= x2 + 6x + 9  + y2 - 8y + 16
=>  - 6x + 9 -12y + 36 = 6x - 8y + 9 + 16
=> 12x + 4y = 20
=> 3x + y = 5  ( required relation )

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