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


5