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Distance Formula

The distance between two points A (x1,y1) and B (x2,y2) is given by the formula


Proof


We plot points A and B and construct right triangle ACB.  Note that C is the point of intersection of the line, drawn through A and parallel to x-axis and the line drawn through B and parallel to y-axis. Also, note that C has coordinates (x2,y1). As explained earlier, we have AC = |x2 - x1| and CB = |y2 - y1|. By the Pythagorean Theorem,

    AB2 = AC2 + CB2

⇒ AB2 = |x2 − x1|2 + |y2 − y1|2

But since |k|2 = k2 for any real number k, we have

AB2 = (x2 − x1) 2 + (y2 − y1)2

Taking square root of both sides, we get


Corollary
The distance of any point A(x,y) from the origin is given by .

Very Short Answers

 

Example

 Find the distance between the following pairs of points.
 (i) A(−2,5) and B(3,−7)

 (ii) A(4,5) and B(−3,2)

 (iii) A(−4,−4) and B(3,5)

Solution

(i) Using the distance formula AB = we get

AB = == = = 13 units.

(ii) Using the distance formula AB = we get

AB = = = = units

(iii) We have

AB = = = = units

 

Short Answers
 

Example

(a) By using the distance formula, prove that each of the following sets of points are vertices of a right triangle:

(i) (4,4), (3,5), (−1,−1) 

(ii) (12,8), (−2,6), (6,0)

(b) Show that the triangle with vertices (4,3), (7,−1) and (9,3) is an isosceles triangle.

(c) Show that the triangle with vertices (a,a), (−a,−a) and (−a,a) is an equilateral triangle.

Solution

(a) (i) Let the points (4,4), (3,5), (−1,−1) represent the points A, B and C, respectively. Then,

AB2 = (3 − 4)2 + (5 − 4) 2 = 12 + 12 = 2

BC2 = (−1 − 3) 2 + (−1 −5) 2 = (−4) 2 + (−6) 2 = 52

and CA2 = (4 − (−1)) 2 + (4 − ( −1)) 2 = 52 + 52 = 50.

Since BC2 = AB2 + CA2, it follows from the converse of the Pythagorean theorem that the triangle ABC is a right triangle, with right angle at A.

(ii) Let the points (12,8), (−2,6), (6,0) represent the points A, B and C, respectively. Then

AB2 = (−2 − 12) 2 + (6 − 8) 2 = (−14)2 + (−2)2 = 200

BC2 = (6 − (−2)) 2 + (0 − 6) 2 = 82 + (−6)2 = 100

and CA2 = (12 − 6) 2 + (8 − 0) 2 = 62 + (8)2 = 100.

Since AB2 = BC2 + CA2 , it follows the converse of the Pythagorean Theorem in which the triangle ABC is a right triangle with right angle at C.


(b) Let the points (4,3), (7,−1), (9,3) represent the points A, B and C, respectively. Then

AB2 = (7 − 4)2 + (−1 − 3)2 = 32 + (−4)2 = 25,

BC2 = (9 − 7)2 + (3 − (−1))2 = 22 + 42 = 20,

and CA2 = (9 − 4)2 + (3 − 3)2 = 52 + 02 = 25

Since AB2 = CA2, we get AB = CA.

Therefore, the triangle ABC is an isosceles triangle.


(c) Let the points (a,a), (−a,−a), (−a,a) represent the points A, B and C, respectively.

Then AB2 = (−a − a)2 + (−a − a)2

              = (−2a)2 + (−2a)2

              = 4a2 + 4a2 = 8a2,

       BC2 = (−a − (−a))2 + (a − (−a)) 2

             = (−a+ a)2 + (a+a)2

             = 2{(−a)2 + a2}            [(a − b)2 + (a + b)2 = 2(a2 + b2)]

             =2(3a2 + a2) = 8a2

and CA2 = (a − (−a))2 + (a − a)2 = (a + a)2 + (a − a)2

            = 2{a2 + (a)2}                [(a + b)2 + (a − b)2 = 2(a2 + b2)]

            = 2(a2 + 3a2) = 8a2.

Since, AB2 = BC2 = CA2, we get AB = BC = CA.

Thus, the triangle ABC is an equilateral triangle.

 
Example

(a) Find the value of x such that AB = BC, where A, B and C are (6,−1), (1,3) and (x,8), respectively.

(b) Which point on the x-axis is equidistant from (7,6) and (−3,4).

(c) An equilateral triangle has one vertex at the point (3,4) and another at (−2,3). Find the coordinates of the third vertex.

(d) Find the abcissa of points whose ordinate is 4 and which are at a distance of 5 from (5, 0).

(e) If the point (x, y) is equidistant from the points (a + b,b − a) and (a − b,a + b), prove that ay = bx.

(f) Find the relation that must exist between x and y, so that C (x,y) is equidistant from A (6,−1) and B (2,3).

(g) Find the circumcentre of the triangle whose angular points are A (4,3), B(−2,3) and C(6,−1).

Solution

(a) A(6, −1), B(1, 3) and C (x, 8) such that AB = BC.

We have AB2 = (1 − 6) 2 + (3 − (−1)) 2 = (−5)2 + 42 = 41.

and BC2 = (x − 1)2 + (8 − 3) 2 = (x − 1)2 + 25

Since AB = BC, we have AB2 = BC2

(x − 1)2 + 25 = 41

        (x − 1)2 = 16

           x − 1 = 4

                 x = 14

Therefore, x = 5, −3.
 

(b) Let us take (7,6) and (−3,4) to be points A and B respectively. Let C (x,0) on x-axis be the point which is equidistant from A and B, that is AC = BC. This implies AC2 = BC2  

⇒ (x − 7) 2 + (0 − 6) 2 = (x − (−3))2 + (0 − 4) 2

⇒  x2 − 14x + 49 + 36 = x2 + 6x + 9 + 16

⇒ −14x + 85 = 6x + 25

60 = 20x

⇒ x = 3.

Thus, the required point is (3,0).
 

(c) Let us take the two given vertices to be A(3,4) and B(−2,3). Let the third vertex of the equilateral triangle be C(x, y). Then AB = BC = CA. This implies AB2 = BC2 = CA2.

Since BC2 = CA2, we get

(x + 2)2 + (y − 3)2 = (x − 3)2 + (y − 4)2

x2 + 4x + 4 + y2 − 6y + 9 = x2 − 6x + 9 + y2 − 8y + 16

10x + 2y − 12 = 0

     5x + y − 6 = 0                                   (1)

Next, since AB2 = BC2, we get

(−2 − 3) 2 + (3 − 4)2 = (x + 2)2 + (y − 3)2

⇒ 25 + (−1)2 = (x + 2)2 + (y − 3)2  

⇒ 26 = (x + 2)2 + (y − 3)2                                  (2)

From (1) we have y = 6 − 5x. Putting this in (2), we get

(x + 2)2 + (6 − 5x − 3)2 = 26

(x + 2)2 + (3 − 5x)2 = 26

x2 + 4x + 4 + 9 − 30x + 25x2 = 26

26x2 − 26x − 13 = 0 2x2 − 2x − 1 = 0

= = =

when x = , y = 6 − 5 = = (7 − 5)

when x = , y = 6 − 5 = = (7 + 5)

Therefore, the coordinates of C are


(d) Let the abcissa of the required point be x. Then A(x,4) is at a distance of 5 from

 B(5, 0). That is AB = 5

           ⇒ AB2 = 52

(x − 5)2 + (4 − 0)2 = 25

⇒ (x − 5)2 = 25 − 16 = 9

      ⇒ x − 5 =  3

         ⇒ x = 5 3

             ⇒ x = 2 or 8.
 

(e) Let us suppose that (x, y), (a + b, b − a), and (a − b, a + b) represent the points A, B and C, respectively. Since AB = AC, we get AB2 = AC2

⇒ (a + b − x)2 + (b − a − y)2 = (a − b − x)2 + (a + b − y)2               (1)

⇒ (a + b)2 – 2(a + b)x + x2 + (b − a)2 – 2(b − a)y + y2                                         

                                 = (a − b)2 – 2(a − b)x + x2 + (a + b)2 – 2(a + b)y + y2

Cancelling (a + b)2, (a − b)2, x2 and y2 from both the sides, we get

−2(a + b)x – 2(b − a)y = -2(a − b)x – 2(a + b)y

⇒ −2x{a + b − (a − b)} = 2y{b − a −(a + b)}

⇒ −2x(2b) = 2y(−2a) −4bx = −4ay ⇒ bx = ay.

Alternatively, we can write (1) as

(a + b − x)2 – (a − b − x)2 = (a + b − y)2 – (b − a − y)2

⇒ (a + b − x + a − b − x) [a + b − x − (a − b − x)] = (a + b − y + b − a − y)[a + b − y −(b − a − y)]

⇒ (2a − 2x)(a + b − x − a + b + x) = (2b − 2y)(a + b − y − b + a + y)

⇒ (2a − 2x)(2b) = (2b − 2y) (2a)

⇒ 4ab − 4bx = 4ab − 4ay

 ⇒  −4bx = −4ay ⇒  bx = ay
 

(f) Since CA = CB, we have CA2 = CB2

⇒ (x − 6)2 + (y + 1)2 = (x − 2)2 + (y − 3)2

⇒ (x − 6)2 − (x − 2)2 = (y − 3)2 − (y + 1)2

⇒ (x − 6 + x − 2)(x − 6 − x + 2) = (y − 3 + y + 1)(y − 3 − y − 1)

⇒ (2x − 8)(−4) = (2y − 2)(−4)

2x − 8 = 2y − 2

x − 4 = y − 1 or x − y = 3 is the relation between x and y.

(g) A (4, 3), B(−2, 3) and C(6, −1) are the vertices of the triangle ABC.

Let D(x, y) be the circumcentre of the triangle ABC. By definition

DA = DB = DC ⇒ DA2 = DB2 = DC2

⇒ (x − 4)2 + (y − 3)2 = (x + 2)2 + (y − 3)2 = (x − 6)2 + (y + 1)2       (1)

The first two relations give us

(x − 4)2 = (x + 2)2 ⇒ x − 4 = (x + 2)

But x - 4 = x + 2 leads to the absurd relation –4 = 2, and x − 4 = −(x + 2) gives us 2x = 2 or x =1.

From the last two relations in (1), we get

   (x + 2)2 + (y − 3)2 = (x − 6)2 + (y + 1)2

⇒ (1 + 2)2 – (1 − 6)2 = (y + 1)2 – (y − 3)2 [ x = 1]

                 ⇒ 9 − 25 = (y + 1 + y − 3)(y + 1 − y + 3)

                     ⇒ −16 = (2y − 2)(4) = 8y − 8

                      −8 = 8y or y= −1.

Thus, the required circumcentre is (1,−1).

 
 
 





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