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 = - \(\Big(\frac{1+2λ}{1-3λ}\Big)x+\Big(\frac{4+λ}{1-3λ}\Big)\)
The equation of the line with intercepts 5 and 6 on the axis is
\(\frac{x}{5}+\frac{y}{6}=1\) … (2)
The slope of this line is \(-\frac{6}{5}\)
The lines (1) and (2) are perpendicular.
∴ \(-\frac{6}{5}\)\(\times\Big(\frac{-1+2λ}{1-3λ}\Big)\) = -1
⇒ λ = \(\frac{11}{3}\)
Substituting the values of λ in (1), we get the equation of the required line.
⇒ \(\Big(1+\frac{22}{3}\Big)x\) + (1-11)y - 4 - \(\frac{11}{3}\) = 0
⇒ 25x – 30y – 23 = 0
Hence, required equation is 25x – 30y – 23 = 0
