Q. 18
Find the equation of the straight line passing through the point (2, 1) and bisecting the portion of the straight line 3x – 5y = 15 lying between the axes.
Answer :
Concept Used:
The equation of a line in intercept form is 
Given:
The line passes through (2, 1)
……(i)
Assuming:
The line 3x – 5y = 15 intercept the x-axis and the y-axis at A and B, respectively.
Explanation:
At x = 0 we have,
0– 5y = 15
⇒ 5y = -15
⇒ y = -3
At y = 0 we have,
3x – 0 =15
⇒ x = 5
A= (0, -3) and B = (5, 0)
The midpoint of AB is (
)
Clearly, the point (
) lies on the line 
……(ii)
Using 
eq(i) + eq(ii) we get,
⇒a 
For a = 
we have,
⇒ b = 11
Therefore, the equation of the required line is:
⇒ 5x + y = 11
Hence, the equation of line is 5x + y = 11
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