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 and y-intercept (b) = 2t.
3/3t + 4/2t = 1
1/t + 2/t = 1
3/t = 1
3 = t
So,
x-intercept (a) = 3(3) = 9
y-intercept (b) = 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
Substitute t = 8 in (1).
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 equation of a line :
x/a + y/b = 1
Here, a and b are x and y-intercepts respectively.
Sum of the intercepts = 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 = 9 - 3 b = 6 |
If a = 6, b = 9 - 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 equation of the required line is
2x + y - 6 = 0 or x + 2y - 6 = 0
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