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.

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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