Distance Formula

Distance between two points P(x,y) and Q(x,y) is
|PQ| = [(x₂ -x₁)² +(y₂ -y₁)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

1. A is a point on y-axis whose ordinate is 5 and B is the point (-3, 1). Compute the length of AB.
2. The distance between A (1, 3) and B (x, 7) is 5. Find the values of x.
3. What point (or points) on the y-axis are at a distance of 10 units from the point (8, 8)?
4. 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.
5. Find the abscissa of points whose ordinate is 4 and which are at a distance of 5 units from (5, 0).
6. What point on x-axis is equidistant from the points (7, 6) and (-3, 4)?
7. 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?
8. Show that the points (4, 2), (7, 5) and (9, 7) are collinear.
9. 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.]
10. 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.
11. 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.
12. 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.

13. 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.

14. 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