Question 1 :
If P(r, c) is mid point of a line segment between the axes, then show that (x/r) + (y/c) = 2.
Solution :
Midpoint of the line segment = (r, c)
(x_{1} + x_{2})/2, (y_{1} + y_{2})/2 = (r, c)
(a + 0)/2, (0 + b)/2 = (r, c)
a/2 = r b/2 = c
a = 2r and b = 2c
Equation of the line
x/a + y/b = 1
x/2r + y/2c = 1
x/r + y/c = 2
Question 2 :
Find the equation of the line passing through the point (1, 5) and also divides the co-ordinate axes in the ratio 3:10.
Solution :
Let A (a , 0) and B(0, b) are the points on the coordinate axis
The required line is passing through the point (1, 5).
Here a = 3k and b = 10k
x/a + y/b = 1
1/3k + 5/10k = 1
1/3k + 1/2k = 1
(2 + 3)/6k = 1
k = 5/6
a = 3(5/6) = 5/2 and b = 10k = 10(5/6) = 25/3
x/(5/2) + y/(25/3) = 1
(2x/5) + (3y/25) = 1
10x + 3y = 25
Question 3 :
If p is length of perpendicular from origin to the line whose intercepts on the axes are a and b, then show that 1/p^{2} = 1/a^{2} + 1/b^{2} .
Solution :
Equation of the given line is x/a + y/b = 1
Now let us find the distance between the point origin to the line x/a + y/b - 1 = 0
d = |Ax + By + C|/√A^{2} + B^{2}
A = Coefficient of x in the given line = 1/a
B = Coefficient of y in the given line = 1/b
C = -1
P = |(1/a)x + (1/b)y - 1|/√(1/a)^{2} + (1/b)^{2}
Since the line is passing through the (0, 0)
P = |(1/a)0 + (1/b)0 - 1|/√(1/a)^{2} + (1/b)^{2}
p = 1/√(1/a)^{2} + (1/b)^{2}
Taking squares and reciprocals on both sides,
1/p^{2} = (1/a)^{2} + (1/b)^{2}
1/p^{2} = (1/a^{2}) + (1/b^{2})
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