Example 1 :
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 :
Intercepts on the axes are equal in magnitude but opposite in sign.
x-intercept(a) = k and y-intercept(b) = -k
(x/a) + (y/b) = 1
(x/k) + (y/(-k)) = 1
The straight line is passing through the point (5, -3)
(5/k) + (-3)/(-k) = 1
(5/k) + (3/k) = 1
(5 + 3)/k = 1 ==> 8/k = 1 ==> k = 8
a = 8 , b = -8
Equation of the line :
(x/a) + (y/b) = 1
(x/8) + (y/(-8)) = 1
(x - y)/8 = 1
x - y = 8
x - y - 8 = 0
Example 2 :
Find the equation of the line passing through the point (9, -1) and having its x-intercept thrice as its y-intercept.
Solution :
x - intercept (a) = 3b
(x/a) + (y/b) = 1
(x/3b) + (y/b) = 1
The straight line is passing through the point (9, -1)
(9/3b) + (-1/b) = 1
(3/b) + (-1/b) = 1
(3 - 1)/b = 1
2/b = 1
b = 2
So a = 3(2) = 6
Equation of the line :
(x/a) + (y/b) = 1
(x/6) + (y/2) = 1
(x + 3y)/6 = 1
x + 3y = 6
x + 3y - 6 = 0
Example 3 :
A straight line cuts the coordinate axes at A and B. If the midpoint of AB is (3, 2), then find the equation of AB.
Solution :
Let us a draw rough diagram for the given information.
Let "C" be the midpoint of the line joining the points A and B.
The required line intersects the x-axis at the point B and y-axis at A. So let us consider the point B as (a, 0) and A as (0, b)
Midpoint of AB = (x_{1} + x_{2})/2, (y_{1} + y_{2})/2
(a + 0)/2, (0 + b)/2 = (3, 2)
a/2, b/2 = (3, 2)
equating the x and y co-ordinates
a/2 = 3 b/2 = 2
a = 6 and b = 4
A (0,4) and B (6, 0)
Equation of the line AB :
(y - y_{1})/(y_{2 }- y_{1}) = (x - x_{1})/(x_{2 }- x_{1})
(y-4)/(0-4) = (x-0)/(6-0)
(y-4)/(-4) = x/6
6(y - 4) = -4x
6y - 24 = -4x
4x + 6y - 24 = 0
Divide the whole equation by 2, we get
2x + 3y - 12 = 0
Hence the required equation is 2x + 3y - 12 = 0
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