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

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