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

Explanation

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.

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