# Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax+2ay+c=0 and 5bx+2by+d=0 lies in the fourth quadrant and is equidistant from the two axes, then

## Solution :

Since, ($$\alpha$$, -$$\alpha$$) lie on 4ax+2ay+c=0 and 5bx+2by+d=0.

$$\therefore$$ 4a$$\alpha$$ + 2a$$\alpha$$ + c = 0 $$\implies$$ $$\alpha$$ = $$-c\over 2a$$  …..(i)

Also, 5b$$\alpha$$ – 2b$$\alpha$$ + d = 0 $$\implies$$ $$\alpha$$ = $$-d\over 3b$$    …..(i)

from equation (i) and (ii),

$$-c\over 2a$$ = $$-d\over 3b$$

If p is the length of the perpendicular from the origin to the line $$x\over a$$ + $$y\over b$$ = 1, then prove that $$1\over p^2$$ = $$1\over a^2$$ + $$1\over b^2$$