# Find the equation of the straight line drawn through the point of intersection of the lines x + y = 4 and 2x – 3y = 1 and perpendicular to the line cutting off intercepts 5, 6 on the axes.

Given:

lines x + y = 4 and 2x – 3y = 1

To find:

The equation of the straight line drawn through the point of intersection of the lines x + y = 4 and 2x – 3y = 1 and perpendicular to the line cutting off intercepts 5, 6 on the axes.

Explanation:

The equation of the straight line passing through the point of intersection of x + y = 4 and 2x 3y = 1 is

x + y 4 + λ(2x 3y 1) = 0

(1 + 2λ)x + (1 3λ)y 4 λ = 0 … (1)

y

The equation of the line with intercepts 5 and 6 on the axis is

… (2)

The slope of this line is

The lines (1) and (2) are perpendicular.

λ

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

25x – 30y – 23 = 0

Hence, required equation is 25x – 30y – 23 = 0

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