Q. 55.0( 4 Votes )
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.
Answer :
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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