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)
Problem 4 :
The coordinate of the midpoint of the line joining the point (2p, 4) and (-2, 2q) and (3, p). Find the value of q.
Solution :
The given problem can be done in two ways.
i) Using midpoint formula
ii) Using section formula
i) Finding the value of q, using midpoint formula :
(x_{1} + x_{2})/2, (y_{1} + y_{2})/2 = (3, p)
(2p + (-2))/2, (4 + 2q)/2 = (3, p)
(p - 1), (2 + q) = (3, p)
Equating the x and y-coordinates, we get
p - 1 = 3 p = 3 + 1 p = 4 |
2 + q = p 2 + q = 4 q = 4 - 2 q = 2 |
So, the value of q is 2.
ii) Finding the value of q using section formula :
The point (3, p) is dividing the line segment in the ratio 1 : 1 internally.
= (lx_{2} + mx_{1}) / (l + m), (my_{2} + ny_{1}) / (m + n)
x_{1} = 2p, y_{1} = 4, x_{2} = -2, y_{2} = 2q
1(-2) + 1(2p) / (1 + 1), 1(2q) + 1(4) / (1 + 1) = (3, p)
(-2 + 2p) / 2, 2q + 4 / 2 = (3, p)
(-1 + p), (q + 2) = (3, p)
Equating x and y-coordinates, we get
-1 + p = 3
p = 3 + 1 ==> 4
q + 2 = p
Applying the value of p, we get
q + 2 = 4
q = 4 - 2
q = 2
Problem 5 :
If the points P(a, -11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of the parallelogram. Find the value of a and b.
Solution :
For a parallelogram, the midpoint of the diagonals will be equal.
The point of intersection of diagonals will be the in the ratio 1 : 1. PQ and QS are the diagonals of the parallelogram.
= (lx_{2} + mx_{1}) / (l + m), (my_{2} + ny_{1}) / (m + n)
P(a, -11) and R(2, 15)
= 1(2) + 1(a) / (1 + 1), 1(15) + 1(-11) / (1 + 1)
= (2 + a)/2, (15 - 11) / 2
= (2 + a)/2, 4/2
= (2 + a)/2, 2 --------(1)
Q(5, b) and S(1, 1)
= 1(1) + 1(5) / (1 + 1), 1(1) + 1(b) / (1 + 1)
= (1 + 5)/2, (1 + b) / 2
= 6/2, (1 + b)/2
= 3, (1 +b) / 2 --------(2)
(1) = (2)
(2 + a) / 2 = 3 2 + a = 6 a = 6 - 2 a = 4 |
2 = (1 + b)/2 4 = 1 + b b = 4 - 1 b = 3 |
So, the value of a and b is 4 and 3 respectively.
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