# Distance Formula

The distance between two points A (x_{1},y_{1}) and B (x_{2},y_{2}) 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 (x_{2},y_{1}). As explained earlier, we have AC = |x_{2 }- x_{1}| and CB = |y_{2 } - y_{1}|. By the Pythagorean Theorem,

AB^{2 }= AC^{2} + CB^{2 }

â‡’ AB^{2 }= |x_{2 } âˆ’ x_{1}|^{2} + |y_{2 } âˆ’ y_{1}|^{2 }

But since |k|^{2} = k^{2} for any real number k, we have

AB^{2} = (x_{2 } âˆ’ x_{1})^{ 2} + (y_{2 } âˆ’ y_{1})^{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**

Find the distance between the following pairs of points.

(i) A(âˆ’2,5) and B(3,âˆ’7)

(ii) A(4,5) and B(âˆ’3,2)

(iii) A(âˆ’4,âˆ’4) and B(3,5)

(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**

(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), (âˆ’1,âˆ’1)

(ii) (12,8), (âˆ’2,6), (6,0)

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

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

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

AB^{2} = (3 âˆ’ 4)^{2 }+ (5 âˆ’ 4)^{ 2 }= 1^{2 }+ 1^{2 }= 2

BC^{2} = (âˆ’1 âˆ’ 3)^{ 2 }+ (âˆ’1 âˆ’5)^{ 2 }= (âˆ’4)^{ 2 }+ (âˆ’6)^{ 2 }= 52

and CA^{2} = (4 âˆ’ (âˆ’1))^{ 2 }+ (4 âˆ’ ( âˆ’1))^{ 2} = 5^{2 }+ 5^{2 }= 50.

Since BC^{2} = AB^{2} + CA^{2}, 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), (âˆ’2,6), (6,0) represent the points A, B and C, respectively. Then

AB^{2} = (âˆ’2 âˆ’ 12)^{ 2 }+ (6 âˆ’ 8)^{ 2 }= (âˆ’14)^{2 }+ (âˆ’2)^{2 }= 200

BC^{2} = (6 âˆ’ (âˆ’2))^{ 2 }+ (0 âˆ’ 6)^{ 2 }= 8^{2 }+ (âˆ’6)^{2 }= 100

and CA^{2} = (12 âˆ’ 6)^{ 2 }+ (8 âˆ’ 0)^{ 2 }= 6^{2 }+ (8)^{2 }= 100.

Since AB^{2 = }BC^{2 }+ CA^{2 , }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,âˆ’1), (9,3) represent the points A, B and C, respectively. Then

AB^{2} = (7 âˆ’ 4)^{2 }+ (âˆ’1 âˆ’ 3)^{2 }= 3^{2 }+ (âˆ’4)^{2 }= 25,

BC^{2} = (9 âˆ’ 7)^{2 }+ (3 âˆ’ (âˆ’1))^{2 }= 2^{2 }+ 4^{2 }= 20,

and CA^{2} = (9 âˆ’ 4)^{2 }+ (3 âˆ’ 3)^{2 }= 5^{2 }+ 0^{2 }= 25

Since AB^{2} = CA^{2}, we get AB = CA.

Therefore, the triangle ABC is an isosceles triangle.

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

Then AB^{2} = (âˆ’a âˆ’ a)^{2 }+ (âˆ’a âˆ’ a)^{2}

= (âˆ’2a)^{2 }+ (âˆ’2a)^{2}

= 4a^{2 }+ 4a^{2} = 8a^{2},

BC^{2} = (âˆ’a âˆ’ (âˆ’a))^{2 }+ (a âˆ’ (âˆ’a))^{ 2}

= (âˆ’a+ a)^{2 }+ (a+a)^{2 }

= 2{(âˆ’a)^{2 }+ a^{2}} [(a âˆ’ b)^{2 }+ (a + b)^{2 }= 2(a^{2 }+ b^{2})]

=2(3a^{2 }+ a^{2}) = 8a^{2 }

and CA^{2} = (a âˆ’ (âˆ’a))^{2 }+ (a âˆ’ a)^{2 }= (a + a)^{2 }+ (a âˆ’ a)^{2 }

= 2{a^{2 }+ (a)^{2}} [(a + b)^{2 }+ (a âˆ’ b)^{2 }= 2(a^{2 }+ b^{2})]

= 2(a^{2 }+ 3a^{2}) = 8a^{2}.

Since, AB^{2} = BC^{2} = CA^{2}, we get AB = BC = CA.

Thus, the triangle ABC is an equilateral triangle.

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

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

(c) An equilateral triangle has one vertex at the point (3,4) and another at (âˆ’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 âˆ’ a) and (a âˆ’ 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,âˆ’1) and B (2,3).

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

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

We have AB^{2} = (1 âˆ’ 6)^{ 2} + (3 âˆ’ (âˆ’1))^{ 2} = (âˆ’5)^{2} + 4^{2 }= 41.

and BC^{2} = (x âˆ’ 1)^{2 }+ (8 âˆ’ 3)^{ 2 }= (x âˆ’ 1)^{2 }+ 25

Since AB = BC, we have AB^{2} = BC^{2 }

(x âˆ’ 1)^{2 }+ 25 = 41

(x âˆ’ 1)^{2} = 16

x âˆ’ 1 = 4

x = 14

Therefore, x = 5, âˆ’3.

(b) Let us take (7,6) and (âˆ’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 AC^{2} = BC^{2}

â‡’ (x âˆ’ 7)^{ 2 }+ (0 âˆ’ 6)^{ 2 }= (x âˆ’ (âˆ’3))^{2 }+ (0 âˆ’ 4)^{ 2 }

⇒ x^{2 } âˆ’ 14x + 49 + 36 = x^{2 }+ 6x + 9 + 16

â‡’ âˆ’14x + 85 = 6x + 25

â‡’ 60 = 20x

â‡’ x = 3.

Thus, the required point is (3,0).

(c) Let us take the two given vertices to be A(3,4) and B(âˆ’2,3). Let the third vertex of the equilateral triangle be C(x, y). Then AB = BC = CA. This implies AB^{2} = BC^{2} = CA^{2}.

Since BC^{2} = CA^{2}, we get

(x + 2)^{2 }+ (y âˆ’ 3)^{2 }= (x âˆ’ 3)^{2 }+ (y âˆ’ 4)^{2 }

x^{2 }+ 4x + 4 + y^{2 } âˆ’ 6y + 9 = x^{2 } âˆ’ 6x + 9 + y^{2 } âˆ’ 8y + 16

10x + 2y âˆ’ 12 = 0

5x + y âˆ’ 6 = 0 (1)

Next, since AB^{2} = BC^{2}, we get

(âˆ’2 âˆ’ 3)^{ 2 }+ (3 âˆ’ 4)^{2 }= (x + 2)^{2 }+ (y âˆ’ 3)^{2 }

â‡’ 25 + (âˆ’1)^{2 }= (x + 2)^{2 }+ (y âˆ’ 3)^{2 }

â‡’ 26 = (x + 2)^{2 }+ (y âˆ’ 3)^{2 }(2)

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

(x + 2)^{2 }+ (6 âˆ’ 5x âˆ’ 3)^{2 }= 26

(x + 2)^{2 }+ (3 âˆ’ 5x)^{2 }= 26

x^{2} + 4x + 4 + 9 âˆ’ 30x + 25x^{2 }= 26

26x^{2 } âˆ’ 26x âˆ’ 13 = 0 2x^{2 } âˆ’ 2x âˆ’ 1 = 0

= = =

when x = , y = 6 âˆ’ 5 = = (7 âˆ’ 5)

when x = , y = 6 âˆ’ 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

â‡’ AB^{2} = 5^{2}

â‡’ (x âˆ’ 5)^{2} + (4 âˆ’ 0)^{2} = 25

â‡’ (x âˆ’ 5)^{2} = 25 âˆ’ 16 = 9

â‡’ x âˆ’ 5 = 3

â‡’ x = 5 3

â‡’ x = 2 or 8.

(e) Let us suppose that (x, y), (a + b, b âˆ’ a), and (a âˆ’ b, a + b) represent the points A, B and C, respectively. Since AB = AC, we get AB^{2} = AC^{2 }

â‡’ (a + b âˆ’ x)^{2} + (b âˆ’ a âˆ’ y)^{2} = (a âˆ’ b âˆ’ x)^{2} + (a + b âˆ’ y)^{2} (1)

â‡’ (a + b)^{2} â€“ 2(a + b)x + x^{2} + (b âˆ’ a)^{2} â€“ 2(b âˆ’ a)y + y^{2}

= (a âˆ’ b)^{2} â€“ 2(a âˆ’ b)x + x^{2} + (a + b)^{2} â€“ 2(a + b)y + y^{2 }

Cancelling (a + b)^{2}, (a âˆ’ b)^{2}, x^{2} and y^{2} from both the sides, we get

âˆ’2(a + b)x â€“ 2(b âˆ’ a)y = -2(a âˆ’ b)x â€“ 2(a + b)y

â‡’ âˆ’2x{a + b âˆ’ (a âˆ’ b)} = 2y{b âˆ’ a âˆ’(a + b)}

â‡’ âˆ’2x(2b) = 2y(âˆ’2a) â‡’ âˆ’4bx = âˆ’4ay â‡’ bx = ay.

Alternatively, we can write (1) as

(a + b âˆ’ x)^{2} â€“ (a âˆ’ b âˆ’ x)^{2} = (a + b âˆ’ y)^{2} â€“ (b âˆ’ a âˆ’ y)^{2 }

â‡’ (a + b âˆ’ x + a âˆ’ b âˆ’ x) [a + b âˆ’ x âˆ’ (a âˆ’ b âˆ’ x)] = (a + b âˆ’ y + b âˆ’ a âˆ’ y)[a + b âˆ’ y âˆ’(b âˆ’ a âˆ’ y)]

â‡’ (2a âˆ’ 2x)(a + b âˆ’ x âˆ’ a + b + x) = (2b âˆ’ 2y)(a + b âˆ’ y âˆ’ b + a + y)

â‡’ (2a âˆ’ 2x)(2b) = (2b âˆ’ 2y) (2a)

â‡’ 4ab âˆ’ 4bx = 4ab âˆ’ 4ay

â‡’ âˆ’4bx = âˆ’4ay â‡’ bx = ay

(f) Since CA = CB, we have CA^{2} = CB^{2 }

â‡’ (x âˆ’ 6)^{2} + (y + 1)^{2} = (x âˆ’ 2)^{2} + (y âˆ’ 3)^{2 }

â‡’ (x âˆ’ 6)^{2} âˆ’ (x âˆ’ 2)^{2} = (y âˆ’ 3)^{2} âˆ’ (y + 1)^{2 }

â‡’ (x âˆ’ 6 + x âˆ’ 2)(x âˆ’ 6 âˆ’ x + 2) = (y âˆ’ 3 + y + 1)(y âˆ’ 3 âˆ’ y âˆ’ 1)

â‡’ (2x âˆ’ 8)(âˆ’4) = (2y âˆ’ 2)(âˆ’4)

â‡’ 2x âˆ’ 8 = 2y âˆ’ 2

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

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

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

DA = DB = DC â‡’ DA^{2} = DB^{2} = DC^{2 }

â‡’ (x âˆ’ 4)^{2} + (y âˆ’ 3)^{2} = (x + 2)^{2 }+ (y âˆ’ 3)^{2} = (x âˆ’ 6)^{2} + (y + 1)^{2} (1)

The first two relations give us

(x âˆ’ 4)^{2} = (x + 2)^{2} â‡’ x âˆ’ 4 = (x + 2)

But x - 4 = x + 2 leads to the absurd relation â€“4 = 2, and x âˆ’ 4 = âˆ’(x + 2) gives us 2x = 2 or x =1.

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

(x + 2)^{2} + (y âˆ’ 3)^{2} = (x âˆ’ 6)^{2} + (y + 1)^{2 }

â‡’ (1 + 2)^{2} â€“ (1 âˆ’ 6)^{2} = (y + 1)^{2} â€“ (y âˆ’ 3)^{2} [ x = 1]

â‡’ 9 âˆ’ 25 = (y + 1 + y âˆ’ 3)(y + 1 âˆ’ y + 3)

â‡’ âˆ’16 = (2y âˆ’ 2)(4) = 8y âˆ’ 8

âˆ’8 = 8y or y= âˆ’1.

Thus, the required circumcentre is (1,âˆ’1).