Question -
Answer -
Given:
The equations, 2x + 3y+ 1 = 0 and 3x тАУ 5y тАУ 5 = 0
The equation of thestraight line passing through the points of intersection of 2x + 3y + 1 = 0 and3x┬атИТ┬а5y┬атИТ┬а5 = 0 is
2x + 3y + 1+┬а╬╗(3x┬атИТ┬а5y┬атИТ┬а5) = 0
(2 + 3╬╗)x +(3┬атИТ┬а5╬╗)y + 1┬атИТ┬а5╬╗┬а= 0
y = тАУ [(2 + 3╬╗) / (3 тАУ5╬╗)] тАУ [(1 тАУ 5╬╗) / (3 тАУ 5╬╗)]
The required line isequally inclined to the axes. So, the slope of the required line is either 1or┬атИТ┬а1.
So,
тАУ [(2 + 3╬╗) / (3 тАУ5╬╗)] = 1 and тАУ [(2 + 3╬╗) / (3 тАУ 5╬╗)] = -1
-2 тАУ 3╬╗┬а= 3 тАУ5╬╗┬аand 2 + 3╬╗┬а= 3 тАУ 5╬╗
╬╗ = 5/2 and 1/8
Now, substitute thevalues of┬а╬╗┬аin (2 + 3╬╗)x + (3┬атИТ┬а5╬╗)y +1┬атИТ┬а5╬╗┬а= 0, we get the equations of the required lines as:
(2 + 15/2)x + (3 тАУ25/2)y + 1 тАУ 25/2 = 0 and (2 + 3/8)x + (3 тАУ 5/8)y + 1 тАУ 5/8 = 0
19x тАУ 19y тАУ 23 = 0 and19x + 19y + 3 = 0
тИ┤ The required equationis 19x тАУ 19y тАУ 23 = 0 and 19x + 19y + 3 = 0