**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

= (lx_{2}+mx_{1})/(l+m), (ly_{2}+my_{1})/(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

= (lx_{2}+mx_{1})/(l+m), (ly_{2}+my_{1})/(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

= (lx_{2} - mx_{1})/(l-m) , (ly_{2}-my_{1})/(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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