Question 1 :
Using section formula, show that the points A(7, −5), B(9, −3) and C(13, 1) are collinear.
Solution :
Let point C(13, 1) divide the other ratio k : 1. So we can take as m = k and n = 1
So x_{1} = 7, y_{1} = - 5, x_{2} = 9, y_{2} = - 3
We know that section formula is given by
(mx_{2} + nx_{1} /m + n , my_{2} + ny_{1} /m + n)
(13 , 1) = (9k + 7(1) / k +1 , -3k + (-5)1 / k + 1)
Equating the co ordinates to the respective numbers we get
9k + 7 / k + 1 = 13 , -3k - 5 / k + 1 = 1
- 4k = 6 -4k = 6
k = - 3/2 k = - 3/2
Since k is negative , it divides externally in the ratio 3 :2. Thus the points A,B and C are collinear.
Question 2 :
A line segment AB is increased along its length by 25% by producing it to C on the side of B. If A and B have the coordinates (−2, −3) and (2, 1) respectively, then find the coordinates of C.
Solution :
Let the length of line segment AB is x units.
It is given that AB is increased by 25% extended upto C.
Hence, BC = (25/100) x = x/4
AB is externally divided into m : n i.e 5 : 1
By Section formula for external division
= (mx_{2} - nx_{1} /m - n , my_{2} - ny_{1} /m - n)
= (5(2) - 1(-2))/(5-1), (5(1) - 1(-3))/(5-1)
= (10 + 2)/4, (5 + 3)/4
= 12/4, 8/4
= (3, 2)
Hence the required point C is (3, 2).
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