Problem 1 :
Population of a city in the years 2005 and 2010 are 1,35,000 and 1,45,000 respectively. Find the approximate population in the year 2015. (assuming that the growth of population is constant)
Solution :
Let us write year and population as points
(2005, 135000) and (2010, 145000)
(y−y1)/(y2−y1) = (x−x1)/(x2-x1)
(y-135000)/(145000-135000) = (x-2005)/(2010-2005)
(y-135000)/10000 = (x-2010)/5
(y-135000) = 2000(x - 2010)
y - 135000 = 2000x - 4020000
y = 2000 x - 4020000 + 135000
y = 2000x - 3875000
Here x and y stands for year and population respectively.
x = 2015
y = 2000(2015) - 3875000
y = 4030000 - 3875000
y = 155000
Problem 2 :
Find the equation of the line, if the perpendicular drawn from the origin makes an angle 30◦ with x-axis and its length is 12.
Solution :
p = 12
Angle formed by the perpendicular = 30◦
Equation of the line :
x cos α + y sin α = p
x cos 30◦ + y sin 30◦ = 12
x (√3/2) + y (1/2) = 12
√3x + y = 24
Problem 3 :
Find the equation of the straight lines passing through (8, 3) and having intercepts whose sum is 1
Solution :
Let "a" and "b" be x and y intercepts respectively.
a + b = 1
b = 1 - a
Equation of the line :
x/a + y/b = 1
8/a + 3/(1-a) = 1
8(1 - a) + 3a = a(1 - a)
8 - 8a + 3a = a - a2
8 - 5a = a - a2
a2 - a - 5a + 8 = 0
a2 - 6a + 8 = 0
(a - 4) (a - 2) = 0
a = 4 and a = 2
If a = 4, then b = 1 - 4 = -3
If a = 2, then b = 1 - 2 = -1
a = 4, b = -3 x/4 - y/3 = 1 3x - 4 y = 12 |
a = 2, b = -1 x/2 - y/1 = 1 x - 2 y = 2 |
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