Q. 95.0( 2 Votes )

# <span lang="EN-US

Given:

Lines 3x – y = 5 and x + 3y = 1

To find:

The equation of the straight line which passes through the point of intersection of the lines 3x – y = 5 and x + 3y = 1 and makes equal and positive intercepts on the axes.

Explanation:

The equation of the straight line passing through the point of intersection of 3x y = 5 and x + 3y = 1 is

3x y 5 + λ(x + 3y 1) = 0

(3 + λ)x + ( 1 + 3λ)y 5 λ = 0 … (1)
y The slope of the line that makes equal and positive intercepts on the axis is 1.

From equation (1), we have: λ = 2

Substituting the value of λ in (1), we get the equation of the required line.

3 + 2x + -1 + 6y – 5 – 2 = 0

5x + 5y – 7 = 0

Hence, equation of required line is 5x + 5y – 7 = 0

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