Question -
Answer -
Given:
The lines x + y = 4and 2x тАУ 3y = 1
The equation of thestraight line passing through the point of intersection of x + y = 4 and2x┬атИТ┬а3y = 1 is
x + y┬атИТ┬а4+┬а╬╗(2x┬атИТ┬а3y┬атИТ┬а1) = 0
(1 + 2╬╗)x +(1┬атИТ┬а3╬╗)y┬атИТ┬а4┬атИТ┬а╬╗┬а= 0 тАж (1)
y = тАУ [(1 + 2╬╗) / (1 тАУ3╬╗)]x + [(4 + ╬╗) / (1 тАУ 3╬╗)]
The equation of theline with intercepts 5 and 6 on the axis is
x/5 + y/6 = 1 тАж. (2)
So, the slope of thisline is -6/5
The lines (1) and (2)are perpendicular.
тИ┤ -6/5 ├Ч [(-1+2╬╗) / (1тАУ 3╬╗)] = -1
╬╗ = 11/3
Now, substitute thevalues of┬а╬╗┬аin (1), we get the equation of the required line.
(1 + 2(11/3))x +(1┬атАУ┬а3(11/3))y┬атИТ┬а4┬атАУ┬а11/3┬а= 0
(1 + 22/3)x + (1 тАУ11)y тАУ 4 тАУ 11/3 = 0
25x тАУ 30y тАУ 23 = 0
тИ┤ The required equationis 25x тАУ 30y тАУ 23 = 0