DISTANCE BETWEEN TWO POINTS

 

Consider the figure below:

 

Let P and Q be points with coordinates (x₁,y₁) and (x₂,y₂) respectively as shown above.

PQR is right angled triangle. By Pythagoras theorem:

PQ² = PR² + RQ² = (x₂-x₁)² + (y₂ – y₁)²

 

Therefore:  PQ² =  (x₂-x₁)² + (y₂ – y₁)²

 

Example #1

Find the distance between points (5,^-2) and (2,2)

Solution:

From the above points:

x₁ = 5, y₁ = -2   and  x₂ = 2, y₂ = 2  

Therefore by distance formula:

PQ² =  (x₂-x₁)² + (y₂ – y₁)²

Distance =√[ (5 – 2)² + (-2 -2)²]

Distance =√[ 3² + (-4)²]  = √ (9 + 16) = √[ 25]

Therefore, the distance = 5 units

 

Example #2

Prove that the Triangle with vertices given by A(3,5), B(-1,-1) and C(4,4) is a right angled triangle.

Solution:

AB² = (-1 - 3)² + (-1 - 5)²

AB²  = 16 + 36 = 52

 

BC² = (-1 - 4)² + (-1 - 4)²

BC²  = 25 + 25 = 50

 

AC² = (4 - 3)² + (4 - 5)²

AC²  = 1 + 1 = 2

 

Relating the three sides we found that:

AB² = BC² +  AC²

Hence Triangle ABC is the right angled triangle.

 

Try yourself:

 

1 .Find the distance between each of the following points:

                                (a)  A(6,2)   and  B(-2,4)

                                (b)  C(-2,2)  and  D(8,-2)

                                (c)  E(3,1)    and   F(-2,6)

                                (d)  G(3,7)   and   H(9,-2)

 2. Find the perimeter of triangle given by vertices (2,1) , (6,1) and (6,4)

 3. Find the area of the Triangle with vertices given by A(3,5), B(-1,-1) and C(4,4)

  4. Given points: A(3,1), B(0,6) and C(-5,3). Prove that triangle PQR is isosceles.

 

********Thanks for reading and have a nice day******

Added By (Yahyou M) - BADSHAH