## Section Formula Worksheet Solution1

In this page section formula worksheet solution1 we are going to see solution for each questions with detailed explanation.

(1) Find the midpoint of the line segment

(i)  (1 ,-1) and (-5 , 3)

Solution

Midpoint of the triangle =   (x₁+ x₂)/2 , (y₁ +y₂)/2

=   (1 + (-5))/2 , (-1 + 3)/2

=   (-4/2) , (2/2)

=   (-2 , 1)

(ii)  (0 , 0) and (0 , 4)

Solution

Midpoint of the triangle =   (x₁+ x₂)/2 , (y₁ +y₂)/2

=   (0 + 0)/2 , (0 + 4)/2

=   (0/2) , (4/2)

=   (0 , 2)

(2) Find the centroid of the triangle whose vertices are

(i) (1 , 3) (2 , 7) and (12 , -16)

Solution

Centroid of the triangle = (x₁ + x₂ + x₃)/3 , (y₁ + y₂ + y₃)/3

= (1 + 2 + 12)/3 , (3 + 7 + (-16))/3

=  (15/3) , (10-16)/3

=  5 , (-6/3)

= (5 , -2)

(ii) (3 , 5) (-7 , 4) and (10 , -2)

Solution

Centroid of the triangle = (x₁ + x₂ + x₃)/3 , (y₁ + y₂ + y₃)/3

= (3 + (-7) + 10)/3 , (-5 + 4 + (-2))/3

=  (13 - 7)/3 , (- 7 + 4)/3

=  (6/3) , (-3/3)

=  (2 , -1)

(3) The centre of the circle is at (-6 , 4). If one end of the diameter of the circle is at origin, then find the other end.

Here one end of the diameter is at origin that is A (0,0) and let B (a, b) is the required endpoint of the diameter

Center of the circle will be exactly middle of the diameter. So to find another endpoint we have to use the midpoint formula

Midpoint = (x₁ + x₂)/2 , (y₁+y₂)/2

(-6 , 4) = (0 + a)/2 , (0 + b)/2

(-6 , 4) = a/2 , b/2

a/2 = - 6                  b/2 = 4

a = -6 x 2                 b = 4 x 2

a = -12                    b = 8

Therefore another end point of the diameter (-12 ,8)

section formula worksheet solution1 section formula worksheet solution1