Example 1 :
Find the equation of the straight line passing through the point (3, 4) and has intercepts which are in the ratio 3:2.
Solution :
Intercept form :
(x/a) + (y/b) = 1
Intercepts are in the ratio 3:2
So, x-intercept (a) = 3t
y-intercept (b) = 2t
(3/3t)+(4/2t) = 1
(1/t)+(2/t) = 1
3/t = 1
3 = t
t = 3
So, x-intercept (a) = 3t = 3(3) ==> 9
y-intercept (b) = 2t = 2(3) ==> 6
(x/9)+(y/6) = 1
(2x+3y)/18 = 1
2x+3y = 18
2x+3y-18 = 0
Example 2 :
Find the equation of the straight line passing through the point (5, -3) and whose intercepts on the axes are equal in magnitude but opposite in sign.
Solution :
Intercept form :
(x/a) + (y/b) = 1
intercepts are equal in magnitude but opposite in sign.
x-intercept (a) = t
y-intercept (b) = -t
The required line is passing through the point (5,-3)
(5/t)+(-3/(-t)) = 1 ---(1)
(5/t)+(3/t) = 1
(5+3)/t = 1
8/t = 1
8 = t
t = 8
By applying the value of t in (1), we get
(x/8)+(y/(-8)) = 1
(x-y)/8 = 1
x-y = 8
x-y-8 = 0
Example 3 :
Find the equation of the straight lines passing through the point (2, 2) and sum of the intercepts is 9.
Solution :
Intercept form :
(x/a) + (y/b) = 1
Sum of intercept = 9
a+b = 9
b = 9-a
The required line is passing through the point (2, 2)
(2/a)+2/(9-a) = 1
[2(9-a)+2a]/[a(9-a)] = 1
[18-2a+2a]/9a-a2 = 1
18/(9a-a2) = 1
18 = 9a-a2
a2-9a+18 = 0
(a-3) (a-6) = 0
a = 3 and a = 6
If a = 3 b = 6 |
If a = 6 b = 3 |
Equation of the line when a = 3 and b = 6 :
(x/3)+(y/6) = 1
(2x+y)/6 = 1
2x+y = 6
2x+y-6 = 0
Equation of the line when a = 6 and b = 3 :
(x/6)+(y/3) = 1
(x+2y)/6 = 1
x+2y = 6
x+2y-6 = 0
So, the required equations are
2x+y-6 = 0 or x+2y-6 = 0
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