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)
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Jul 27, 24 04:58 AM
Jul 27, 24 04:44 AM
Jul 27, 24 04:15 AM