# Distance Formula

Distance between two points P(x,y) and Q(x,y) is

|PQ| = [(x_{2} -x_{1})²
+(y_{2} -y_{1})²].

Hence the distance of the point P (x, y) from the origin (0, 0)

= [(x -0)² +(y -0)²] =
[x² +y²]

### Remarks

To prove that a quadrilateral is a

(i) *rhombus*, show that all the sides are equal.

(ii) *square*, show that all the sides are equal and the diagonals are also equal.

(iii) *parallelogram*, show that the opposite sides are equal.

(iv) *rectangle*, show that the opposite sides are equal and the diagonals are also equal.

## Illustrative Examples

### Example

Show that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right angled triangle.

### Solution

Let the points be A (7, 10), B (-2, 5) and C (3, -4), then

|AB| = [(- 2 -7)² +(5 -10)²] =
[81 +25] =
106,

|BC| = [(3 +2)² +(11 -3)²] =
[25 +81] =
106 and

|CA| = [(7- 3)² +(10 -(-4))²] =
[16 +196] =
212

=> AB² = 106, BC² = 106 and CA² = 212,

Hence AB² +BC² = 106 +106 = 212 = CA²

=> ABC is right angled and it is
right angled at B.

Also |AB| = 106 = |BC|
=> ABC is isosceles.

### Example

Show that the points (-1, -1), (2, 3) and (8, 11) are collinear.

### Solution

Let the points be A (-1, -1), B (2, 3) and C (8, 11), then

|AB| = [(2 -(-1))² +(3 -(-1))²] =
[32 +42] =
[9 +16] =
25 =5,

|BC| = [(8 -2)² +(11-3)²] =
[36 +64] =
100 = 10 and

|CA| = [(8 -(-1))² +(11 -(-1))²] =
[92 +122] =
[81 +144] =
225 =15.

Hence |AB| +|BC| = 5 +10 = 15 = |CA|

=> the given points are collinear.

### Example

The vertices of a triangle are A (1, 1), B (4, 5) and C (6, 13). Find cos A.

### Solution

If a, b, c are the sides of ABC, then

a = |BC| = [(6 -4)² +(13 -5)²] =
[4 +64] =
68 = 2.17,

b = |CA| = [(6 -1)² +(13 -1)²] =
[25+144] = 169
= 13, and

c = |AB| = [(4 -1)² +(5 -1)²] =
[9 +16] =
25 = 5

By cosine formula, we have

cos A = (b² +c² -a²)/ 2bc = (169 +25 -68)/(2.13.5) = 126/130 = 63/65

## Exercise

- A is a point on y-axis whose ordinate is 5 and B is the point (-3, 1). Compute the length of AB.
- The distance between A (1, 3) and B (x, 7) is 5. Find the values of x.
- What point (or points) on the y-axis are at a distance of 10 units from the point (8, 8)?
- Find point (or points) which are at a distance of 10 from the point (4, 3) given that the ordinate of the point (or points) is twice the abscissa.
- Find the abscissa of points whose ordinate is 4 and which are at a distance of 5 units from (5, 0).
- What point on x-axis is equidistant from the points (7, 6) and (-3, 4)?
- Find the value of x such that |PQ| = |QR| where P, Q, R are (6, -1), (1, 3) and (x, 8) respectively. Are the points P, Q, R collinear?
- Show that the points (4, 2), (7, 5) and (9, 7) are collinear.
- Show that the points (2, 3), (-4, -6) and (1, 3/2) do not form a triangle.

[**Hint.**Show that the given points are collinear.] - Show that the points A (2, 2), B (-2, 4) and C (2, 6) are the vertices of a triangle. Prove that ABC is an isosceles triangle.
- Show that the points:

(i) (-2, 2), (8, -2) and (-4, -3) are the vertices of a right-angled triangle.

(ii) (0, 0), (5, 5) and (-5, 5) are the vertices of a right-angled isosceles triangle.

(iii) (1, 1), (-1, -1),(-3 , 3)are the vertices of an equilateral triangle.

(iv) (2 a, 4 a), (2 a, 6 a), (2 a +3 a, 5 a) are the vertices of an equilateral triangle. - Show that the points:

(i) (2, 1), (5, 4), (4, 7), (1, 4) are the angular points of a parallelogram.

(ii) (7, 3), (3, 0), (0, -4), (4, -1) are the vertices of a rhombus.

(iii) (2, -2), (8, 4), (5, 7), (-1, 1) are the vertices of a rectangle.

(iv) (3, 2), (0, 5), (-3, 2), (0, -1) are the vertices of a square. -
The points A (0, 3), B (-2, a) and C (-1, 4) are the vertices of a right-angled triangle at A, find the value of a.

- Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of equal sides is 3 units.

## Answers

**1.**5

**2.**4 or -2

**3.**(0, 2), (0, 14)

**4.**(1, 2), (3, 6)

**5.**2 or 8

**6.**(3, 0)

**7.**5 or -3; No

**13.**1

**14.**(2 +(11)/2, 5/2) or (2 -(11)/2, 5/2)