Find the equation of the straight line drawn through the point of inte

1.	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: The lines x + y = 4 and 2x – 3y = 1 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 = – [(1 + 2λ) / (1 – 3λ)]x + [(4 + λ) / (1 – 3λ)] The equation of the line with intercepts 5 and 6 on the axis is x/5 + y/6 = 1 …. (2) So, the slope of this line is -6/5 The lines (1) and (2) are perpendicular. ∴ -6/5 × [(-1 + 2λ) / (1 – 3λ)] = -1 λ = 11/3 Now, substitute the values of λ in (1), we get the equation of the required line. (1 + 2(11/3))x + (1 – 3(11/3))y − 4 – 11/3 = 0 (1 + 22/3)x + (1 – 11)y – 4 – 11/3 = 0 25x – 30y – 23 = 0 ∴ The required equation is 25x – 30y – 23 = 0

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:

The lines x + y = 4 and 2x – 3y = 1

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 = – [(1 + 2λ) / (1 – 3λ)]x + [(4 + λ) / (1 – 3λ)]

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

x/5 + y/6 = 1 …. (2)

So, the slope of this line is -6/5

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

∴ -6/5 × [(-1 + 2λ) / (1 – 3λ)] = -1

λ = 11/3

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

(1 + 2(11/3))x + (1 – 3(11/3))y − 4 – 11/3 = 0

(1 + 22/3)x + (1 – 11)y – 4 – 11/3 = 0

25x – 30y – 23 = 0

∴ The required equation is 25x – 30y – 23 = 0