PRACTICAL PROBLEMS USING SECTION FORMULA

Problem 1 :

Using the section formula, show that the points

A(1, 0),B (5, 3), C (2, 7) and D(-2, 4)

are vertices of a parallelogram taken in order.

Solution :

The midpoint of the diagonals AC and the diagonal BD coincide

Section formula internally

= (lx2+mx1)/(l+m), (ly2+my1)/(l+m)

l = 1 and m = 1

The midpoint of the diagonals AC. The midpoint of diagonal is in the ration 1:1

=  (1(2)+1(1))/(1+1) , ((1(7) + 1(0))/(1+1)

=  (2+1)/2 , (7+0)/2

=  (3/2, 7/2)   ------- (1)

The midpoint of the diagonals BD. The midpoint of diagonal is in the ration 1:1

=  ((1(-2)+1(5))/(1+1) , ((1(4) + 1(3))/(1+1)

=  (-2+5)/2 , (4+3)/2

=  (3/2, 7/2)    ------- (2)

Two diagonals are intersecting at the same point. So the given vertex forms a parallelogram.

Problem 2 :

The 4 vertices of a parallelogram are

A(-2, 3), B(3, -1), C(p, q) and D(-1, 9)

Find the value of  p and q.

Solution :

The midpoint of the diagonals AC and the diagonal BD coincide

Section formula internally

= (lx2+mx1)/(l+m), (ly2+my1)/(l+m)

l = 1 and m = 1

The midpoint of the diagonals AC. The midpoint of diagonal is in the ratio 1:1

=  (1(p)+1(-2))/(1+1) , ((1(q) + 1(3))/(1+1)

=  (p-2)/2 , (q+3)/2  ------- (1)

The midpoint of the diagonals BD. The midpoint of diagonal is in the ratio 1:1

=  ((1(-1)+1(3))/(1+1) , ((1(9) + 1(-1))/(1+1)

=  (-1+3)/2 , (9-1)/2

=  (1, 4)    ------- (2)

(1)  =  (2)

Equating x and y -coordinates, we get 

(p-2)/2  =  1

p-2  =  2

p  =  4

(q+3)/2  =  1

q+3  =  2

q  =  -1

Problem 3 :

Find the coordinates of the point which divides the line segment joining

(3, 4) and (-6, 2)

in the ratio 3:2 externally.

Solution :

Section formula externally

=  (lx2 - mx1)/(l-m) , (ly2-my1)/(l-m)

A (3, 4) B (-6, 2)     3 : 2

l = 3 and m = 2    

=  (-18 - 6)/1 , (6 - 8)/1

=   (-24 , -2)

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