PRACTICAL PROBLEMS ON EQUATION OF LINE FOR GRADE 11

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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